help-23

[slate tapir]

hmm

sometimes

it is not even about keyword

[round egret]

for example you know HCF is the largest number that divides two or more numbers without leaving a remainder

[slate tapir]

ohhhhh

W

[round egret]

while the LCM, is the smallest number that is divisible by both numbers

[slate tapir]

so hcf is a factor

[round egret]

yeah it's in the name

[slate tapir]

lcm is always the bigger number

i will say one problem pls say it is lcm or hcf

[round egret]

yes LCM is always greater or equal to the HCF

[slate tapir]

there is a circular path around a sports field, sonia takes 18 mins to completes one round of the circular path while ravi takes 18 mins to complete it. suppose they both start at the same speed,same direction and same time . after how many mins they will meet again at starting point?

[round egret]

in this case it would be LCM

[slate tapir]

how?

bec 18 and 12 is big number?

or any other reasons?

[round egret]

how did you get 12?

[slate tapir]

.

sry

my mistake

ravi takes 12 min

not 18

my typing mistake

[round egret]

oh

in this case it would still be LCM

if they meet again at the starting point. Both must have completed a whole number of rounds and arrive back at the starting point simultaneously

To do that, you would need to find a time duration that is a multiple of both 12 and 18 minutes

and you want the lowest as well

(They also both start together at the same point and same time, moving in the same direction)

[safe radish]

@slate tapir Has your question been resolved?

[warm warren]

HCF-> finding the largest number that can divide both numbers.
LCM-> finding the smallest number that is divisible by both numbers.

(may be more than two numbers, we shall ignore that for now)

[safe radish]

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[round egret]

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[solar sand]

A circle of radius 1 is inscribed in an equilateral triangle of suitable size. Then three more circles are inscribed between the first circle and two sides of the triangle near each vertex. The process continues indefinitely, with progressively smaller circles. What is the sum of the radii of all the circles?

[jolly sail]

I may be wrong but isn't it just the length from center of triangle to a corner then divide by 2?

[pulsar pecan]

but then you have to consider the other circles

[pulsar pecan]

where are you stuck?

[solar sand]

I am confused on how to start the solution

never dealt with these types of question

[pulsar pecan]

You can first try drawing a picture

if you haven’t done that yet

always try drawing a picture when you encounter a geometry problem

[solar sand]

I have the picture

[pulsar pecan]

Ok, could you send it here?

[pulsar pecan]

right

notice that the centers are collinear

[solar sand]

Yeah

[timber karma]

intuitively you would find a general term for each radii

which would be some ratio

how would you get a ratio in geometry?

[pulsar pecan]

and then you would have a geometric series

so the problem boils down to finding the ratio

[pulsar pecan]

and also try to give things variable names

until you can get equations relating the variables

for example, give a variable name to the radius of the central circle, and all the other infinitely many circles

[solar sand]

hmm, I don't quite get the idea for finding the ratio

Let's say, ro be the radius of initial circle and r1 be the radius of second

Then we have ro=1 but we don't have any information of r1

so how do I find the ratio?

[pulsar pecan]

My point was that you can guess intuitively that r_0/r_1 = r_1/r_2 = r_2/r_3 and so on

and so you can find the sum of all the radii as a geometric series from this

so the hard part is just finding the ratio

[pulsar pecan]

Are you allowed to use trigonometry?

[solar sand]

Yes

There are no strict set of rules

[pulsar pecan]

Oh good

I’ve figured out the solution; that took too long

let me draw a diagram

[solar sand]

sure, really appreciated

[pulsar pecan]

@solar sand try and find equations relating r_0 and r_1 based on this

[solar sand]

alright wait

[solar sand]

I got AR=ro√3 and CR=r1√3

and from similarity I ended up getting ro/r1=ro/r1 😭

where did I go wrong

[pulsar pecan]

consider: OB = OA + BC

[solar sand]

huh

that would just be ro+r1

@pulsar pecan could u elaborate the essence behind doing OB=OA+BC which is basically ro+r1?

I tried to do OA/BC =AR/CR through similarity to find the ratio which didn't go well

[pulsar pecan]

yeah and OB is twice (OA - BC) too

I think this idea is used in finding radii of Ford circles or something

[solar sand]

Got the r1=1/3
so for rn it would basically be (1/3)^n which is geometric series?

But the thing that's bugging me is how ro+r1=2(ro-r1)

[pulsar pecan]

because of trig ratios

[pulsar pecan]

yeah, r_n would be that, because you can apply homotheties/similarities

[pulsar pecan]

this is why I made the angle red

[solar sand]

uhmm, so sorry but I don't understand anything about it besides being 60😓

[solar sand]

oh wait I think I get it

[pulsar pecan]

yeah there you go

can you say why, just to make sure we’re on the same page?

[solar sand]

sin30=ro/OR
which gives OR=2ro

Similarly, sin30=r1/BR which gives BR=2r1
and OB=OR-BR

so 2(ro-r1)

is this right?

[pulsar pecan]

Yes

[solar sand]

thank you so much

you are a real life saver

[pulsar pecan]

No worries

Next time, please send any working you’ve got so you can be helped faster

like in this case where you could send the picture

(which you had drawn)

[solar sand]

Sure!

[pulsar pecan]

And also say where exactly you are stuck

For the same reason

[solar sand]

Alright, I didn't know the approach to this. Thanks

[pulsar pecan]

Also generally you can just try giving things variable names and see how far you can get; if you’d noticed that trig ratio, then you’d be done with the ratio-finding and then have solved the problem

[solar sand]

Will definitely keep that in mind, thanks again

[pulsar pecan]

You’re welcome!

[pulsar pecan]

If you’re done for now, then you can type .close to close the channel

[solar sand]

.close

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[stiff nest]

how do you do D? do you need to do linear interpolation?

[pulsar pecan]

you can look at other past papers; do these kinds of questions usually ask you to do linear interpolation?

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[hybrid basin]

The ques, its of 11th standard maths ques…. Ik its easy but i wanted to ask like in main exam if this stuff of question comes, can i say like “as n is Natural number therefore let n=1 and solve the ques and then say “similarly with other” or have to do a proper shi? If have yo do a proper thing then can anyone help me out what to write or yk

[warm warren]

The solution seems to be of induction.

Do you know what induction is?

mathematical induction

[hybrid basin]

Nahh…

[timber karma]

what are you asking?

[warm warren]

The problem seems to be of induction; are you sure you haven't studied it before?

[hybrid basin]

Am js asking, can i do the ques in exam by taking the values of “n” or have to prove it by a yk format of idk

[hybrid basin]

I havent heard of it…. But this ques is of Sets chapter and idk more abt that…

[thorny dome]

are you sure you've never seen induction before

prove a property holds for n=0 or 1, then prove that it holds for n=k+1 given it holds for n=k

[warm warren]

Base Case and Inductive Step. it's better to write:\
Base Case: Let $n=1$.\
....end of the proof for $n=1$\\

Inductive step: Now, assume the assertion (write the question) to be true for all numbers $n=1,...k$. Consider the validity of the statement for $k=1$:\

(write proof)\\
Since the statement is true for $n=k+1$, thus by the principle of mathematical induction, it is true for all natural $n$. \\\
This will actually depend a lot on your prof, and his/her marking scheme.

[flat frigate]

Annie Maqionde

[thorny dome]

you can't just like say you do it like another question, you have to actually do the proof (is that what you're asking?)

[thorny dome]

can you show the rest of this solution? what are you confused about on it?

[hybrid basin]

Uhm wait a sec, let me show book solution which i yk

Uhm ig thats why i cant understand the proof… cuz i havent heard abt that yk…

[warm warren]

No then you can just say\
Consider first $n=1$.\
Then go on with the rest of your proof

[flat frigate]

Annie Maqionde

[warm warren]

it is my serious and grave error, it is not induction.

I apologize

[hybrid basin]

Ohhh….

[hybrid basin]

So i can do like this right? @warm warren

[thorny dome]

can you follow the proof? which line exactly is confusing you

[warm warren]

I feel OP has a doubt in how to present it.

[thorny dome]

do you know what the binomial theorem is

[safe radish]

@hybrid basin Has your question been resolved?

[safe radish]

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[quick wraith]

I have no idea what the complex numbers are

[quick wraith]

So I do not know how to work through this problem

I already tried doing usual algebraic manipulation

and just got to the trivial solution for the scalars

I am working on problem b)

[tardy mango]

Note that your scalars can be complex

I assume you implicitly assumed the scalars are real and set real part = imaginary part = 0

Note that your scalars can be complex
but that doesn't work because of this

[quick wraith]

yes

[quick wraith]

I am thinking of using reciprocals

Maybe that works

[tardy mango]

using reciprocals could work

[quick wraith]

why are reciprocals so important in complex analysis?

I have not studied complex numbers

All I remember vaguely is someone using reciprocals with complex numbers

to simplify

[tardy mango]

they allow division to be well-defined for points in a 2d plane ig

and that encodes things like scaling and rotation

[tardy mango]

tbf you don't need reciprocals though

[quick wraith]

When do I learn about this stuff formally?

[quick wraith]

mm

[tardy mango]

uhhhhh idk

they'd probably do it to some level in complex analysis

but it's not unreasonable to study complex numbers in high school or for comp math

[quick wraith]

I thought so too

[tardy mango]

australia, the uk, ireland, etc come to mind as countries that teach about complex numbers (albeit to varying levels) in high school

[quick wraith]

Well, let's not work with reciprocicals

unless they come off naturally to the solution

[tardy mango]

you can give a pair of (nonzero) scalars in ten seconds with no computation

that's my hint

think simple (and maybe a bit troll)

[quick wraith]

Okay

[quick wraith]

@tardy mango I had also tried rewriting the scalars in complex form

Would that have worked?

[tardy mango]

Hint: ||ab+(-ba)=0||

[tardy mango]

sure, but that is too much effort (and you get an undetermined system anyway so you'd need to take some creative liberty in picking values for certain components if you want a concrete numeric answer)

[quick wraith]

I just ended up with two unknowns

That hinted me at the proof being true, but I don't have the tools to prove it

What about choosing a = -i, b = i?

[tardy mango]

a pair of scalars in parametric form but again, too much work.

[tardy mango]

try it

[quick wraith]

(-i + -2i) + (i - 2i) = 0
(-4i) = 0

It is wrong

I should add a real part

[tardy mango]

is this supposed to relate to $ab+(-ba)=0$ or is this reusing $a$, $b$ for something else

[flat frigate]

Civil Service Pigeon

[quick wraith]

Something else

I multiplied a by 1+i, and b by 1-i

[tardy mango]

ok

[quick wraith]

a = -i, b = i

I got -4i = 0 lmao

Wait

-4i = -4(sqrt -1), right?

[tardy mango]

Eh ...

normally people say $i=\sqrt{-1}$, but I don't even really love that tbh

[quick wraith]

-4(sqrt -1) = sqrt 4?

[flat frigate]

Civil Service Pigeon

[tardy mango]

hold on

what are you doing

[quick wraith]

I mean

[tardy mango]

$\sqrt{ab}=\sqrt{a}\sqrt{b}$ works for $a,b \geq 0$

[flat frigate]

Civil Service Pigeon

[tardy mango]

you can't necessarily say the same if one of them is negative

[quick wraith]

Oh

okay

yeah, we gotta be careful

let's not do drugs

sorry

[tardy mango]

consider how you can make this "match" $c_1 (1+i)+c_2 (1-i)=0$

[flat frigate]

Civil Service Pigeon

[tardy mango]

again, be very lazy

[quick wraith]

c1 = -c2

okay

[tardy mango]

mmmmm

try to pick a and b first

[quick wraith]

Yeah

on it

[tardy mango]

no

no

uhhh

actually

eh

try it

[quick wraith]

10 + i = -(-(10 + i))

so c1 = 10 + i, c2 = -(10 + i)

seems reasonable

[tardy mango]

I feel like you're getting there but idk if the numbers work out

[quick wraith]

(10 + i)(1 +i) = (10 + 10i) + (i - 1)

-(10 + i)(1-i) = -[(10 - 10i) + (i + 1)]

Now we add them

(9 + 11i) + (9i - 11) @tardy mango

this looks really good

9 + 11i = 11 - 9i looks really bad

[tardy mango]

I need to go to class

Recall that ab+(-ba)=0

,texsp ||Pick $a=1+i$ and $b=1-i$.||

[flat frigate]

Civil Service Pigeon

[tardy mango]

Then,
,texsp ||$$ab+(-ba)=0 \implies (1+i)(1-i)+(-1+i)(1+i)=0.$$
Setting $c_1=1+i$ and $c_2=-1+i$ yields
$$c_1 (1-i)+c_2 (1+i)=0$$
and so we are done since $c_1$, $c_2 \neq 0$.||

[flat frigate]

Civil Service Pigeon

[quick wraith]

ah

okay

well bye

thanks

.close

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[copper shadow]

Any suggestions on how to proceed?

[quiet plume]

Hum. Maybe complete the square and trig sub eventually?

[copper shadow]

So my attempt at simplification won't be able to lead to anything?

[quiet plume]

Maybe you could set u = sqrt(what you have) and rewrite everything in u. Should be left with rational functions at least.

Then a bunch of partial fractions await

[copper shadow]

But when I sub in for dx I'll have a bunch of terms in x

[slate ingot]

I think you can take
X = tan²ø

Thatll turn the integration to sec³ø

[quiet plume]

You would be able to rewrite them in terms of u.

[slate ingot]

Either that or im tripping

[copper shadow]

So we would have $u=\sqrt{1+\frac{1}{x}$. And then $du= \frac{1}{-2}\frac{1+\frac{1}{x}}^{-1/2}$

[flat frigate]

BigBen
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[copper shadow]

NVM I'm just going to write

[copper shadow]

What about the 1's?

[slate ingot]

I got something, idk if its correct tho

[slate ingot]

Wdym?

[copper shadow]

Wait I think the trig sub works. Let me do it completely

[copper shadow]

I didn't think it through. Now I just need to integrate sec^3x

[slate ingot]

Yep

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@copper shadow Has your question been resolved?

[copper shadow]

.solved

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[stiff nest]

how da hell do i do dis

[plucky elk]

use (a+b)^2 = a^2 + 2ab + b^2

[stiff nest]

why would that help

[plucky elk]

do you see the quantity squared here

[plucky elk]

use this and expand

[stiff nest]

thaniks

.close

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[dusk walrus]

Was just wondering if this graph was correct, the question is asking to sketch one continuous curve with the following characteristics listed in image.

[plucky elk]

just fix f(1) = 1 requirement

[languid ermine]

your f(1) looks a tad high

[plucky elk]

maaaaaybe smooth this out a little more so that f'(4) = 0

[languid ermine]

everything else looks good though

[dusk walrus]

Ok ty but im just confused about this

[dusk walrus]

.

[plucky elk]

why

[dusk walrus]

oh wait nvm

had brain fog for a sec

thanks for the help

[languid ermine]

@dusk walrus if you're done with this channel, type ".close" to lock it

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[hallow agate]

I need help

[hallow agate]

With solving x

For angles

[fallow oracle]

Hi, the 3 angles inside a triangle add up to 180.

So 180-46-77 is 57, which is the last angle in the triangle.
A line has an 180 angle.
So X is 180-57 (123 degrees)

[hallow agate]

Ok thank u

.close

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[static light]

i am tring to make a ai that can learn amost like a human anyone there to make the math?

[static light]

idk anyone can help?

[plucky elk]

Not what help channels are for

[normal moss]

Yes

"making the math" is so simple, of course ask it in a math discord help channel

[static light]

ah ;C

but they to help

https://tenor.com/view/fahhhmeme-gif-5987802747989276454

[normal moss]

!done

[safe radish]

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[static light]

f

.close

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[ancient kernel]

hint for this problem pls. topic is pigeonhole principle

[ancient kernel]

hint i got:One country has 330 members. Take the largest number and subtract the others from that country.
It's multi step pigeonhole

[opaque fern]

Wait im confused

You are given a hint

And you want another hint, too?

[ancient kernel]

i can't solve it with the first hiint

💀

im dumb

[opaque fern]

This doesn't seem trivial anyway. Hmm...

[ancient kernel]

got it?

<@&286206848099549185>

nvm i got it

.close

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[lilac relic]

Would I use this formula for this question?

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[proud carbon]

is the img quality too bad

in this question i didnt understand where exactly i can begin so i just need a little kickstart i guess

[timber karma]

$\lim_{x\to \infty} \frac{(2x+1)^{40}(4x-1)^5}{(2x+3)^{45}}$

[flat frigate]

Copter

[proud carbon]

yeah

[opaque fern]

Have you tried anything?

[opaque fern]

Oh sorry just read this

Okay

[proud carbon]

no i didnt understand what i can take as my first step

[warm warren]

Hint: The sum of powers in the numerators is 40+5 = 45, and the denominator is 45 too.

[proud carbon]

ohh okay

[proud carbon]

how can i use it because the bases are different right

[opaque fern]

At infinity, the high order terms are the only ones that matter, right?

[warm warren]

Your first strategy would be to somehow replace the $x$ terms with $1/x$ by multiplying/dividing by a suitable number(hint: here, its some power of $1/x$)

[flat frigate]

Annie Maqionde

[proud carbon]

yeah i think

[opaque fern]

What is the highest order term in (2x+1)^40 you think?

[proud carbon]

oh x^40

[opaque fern]

No need for an exact answer

[proud carbon]

and some

coefficient

[opaque fern]

Yeah!

But it is x^40

What about (4x-1)^5

[proud carbon]

x^5 and its coefficient

well highest order is 5

[opaque fern]

So we gotta figure out those coefficients

[proud carbon]

2^40

and 4^5

[opaque fern]

Yeah

[proud carbon]

oh wait

[opaque fern]

So, lets ignore literally every other term and simplify the whole numerator to that product

What would it be

[proud carbon]

so since its tending to inf we can negate the other terms?

[opaque fern]

Yep!

[proud carbon]

ohh thats new to me

tysm

[opaque fern]

Coming back to this

[proud carbon]

can i ask 1 thing ?

if its okay

why exactly do we negate

like whys it allowed

[opaque fern]

So like

Maybe its better to use a simpler example

[proud carbon]

Also another little thing

if thats correct

umm @opaque fern ?

[proud carbon]

did i make an error somewhere

because according to our module's solution

the ans should be does not exist

[opaque fern]

Sorry I was distracted by something

[proud carbon]

oh okay np

[opaque fern]

This is exactly correct I yhink

It should be 32

But why did u write 3^2 at the end

[proud carbon]

Oh it's 32

[opaque fern]

Oh LOL

[proud carbon]

Sorry for the bad writing 😭

[opaque fern]

😭 😭 😭 😭

[proud carbon]

I guess error in the book then

[opaque fern]

If you are trying to evaluate [
\lim_{x\to\infty}\4{x^2+2x+3}{4x^2+5x+6}
]
You can factor $x^2$ from both the numerator and denominator to get
[
\lim_{x\to\infty} \4{\cancel{\0b{x^2}}\8{1 +\52x +\53{x^2}}}{\cancel{\0b{x^2}}\8{4 +\55x +\56{x^2}}}
]
The rest of the terms all converge to 0 as they approach infinity. Leaving you with [
\lim_{x\to\infty}\414
]

Writing this on phone was NOT it

[empty apex]

Polynomial of degree n/polynomial of degree n

Notice something?

[tall spade]

Just a small remark, you factor out x^2, not 1/x^2.

[opaque fern]

Ah thanks good catch

[proud carbon]

im back

sorry for the little delayed reply

[proud carbon]

no not rly

[proud carbon]

ohhh and how exactly did we use this here

[opaque fern]

Oh its the same situation really

Its just that your numerator and denominator's polynomial are reaaaaally big

If you multiply them out, you get a bunch of terms but if you factor out the highest-order term, all of the rest converge to 0

Similar to my example above

[proud carbon]

ohh

i get it

[opaque fern]

Yeppiee

[proud carbon]

well thats my doubt for now

tysm

[opaque fern]

Aighty np

[proud carbon]

i guess ill close this now

.close

[safe radish]

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[empty apex]

$$ Lim x-> inf ((a_n)x^n + (a_n-1)x^(n-1)+ ....... + a_0)/(((b_n)x^n + (b_n-1)x^(n-1)+ ....... + b_0) $$
Try this

[flat frigate]

wick

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[brave wolf]

<@&268886789983436800>

[near sky]

Bro will return in avengers Doomsday

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[bronze yoke]

A person walks in a straight line at a constant speed of 2 m/s.

a. Draw the position vs. time graph assuming the motion starts at the origin.

b. Draw the position vs. time graph assuming the motion starts 1 m from the origin and the object moves away from the origin.

c. Draw the velocity vs. time graph assuming the motion starts at the origin.

d. Draw the velocity vs. time graph assuming the motion starts 1 m from the origin and the object moves away from the origin.

the question was translated from another language so if something needs clarifying let me know
I tried doing a, and b, but I don't have the answers so can anyone help me? How do I know if my graph is correct

[bronze yoke]

this is my attempt for a. and b.

[warm warren]

Well since you haven't provided more context (is there anything else given?), the graphs are right

But, for future,

please either label which line is what

or use two graphs.

[bronze yoke]

that's the whole question

[bronze yoke]

label it as "a" ?

[warm warren]

I'd label it as $\mathrm{a)}$

[flat frigate]

Annie Maqionde

[bronze yoke]

the right one I mean

I see

[warm warren]

But its your personal choice

Have you tried c and d? Where are you stuck in them, if you are?

[bronze yoke]

I'm trying to do it now, but wouldn't they look the same as the first graphs but instead the graph represents V?

[warm warren]

c would be identical to a, I suppose

it's most probably given to confuse you with pos-time and vel-time

[bronze yoke]

it would be the same, so like this?

[warm warren]

the y-axis would be labeled v

and change the labellings on the y-axis

its no longer 1 m

its 1 m/s

[bronze yoke]

ohh wait I missed a line from the question

the beginning is
A person walks in a straight line at a constant speed of 2 m/s.

sorry

the second graph would be a straight line wouldn't it? since the speed is constant

[warm warren]

then im afraid you'll have to correct all four graphs.

[bronze yoke]

Ok

what's wrong about a?

[warm warren]

A is correct mb.

[bronze yoke]

I assume b is correct as well because it's the same graph with the same slope except that it starts from 1m instead of 0

if we change y to instead be X(v) in m/s
wouldn't that mean that the graph would be simply a straight line as the speed is constant?

c and d would be simply the function v(t) = 2 wouldn't it

[covert rain]

are you talking about uniform rectilinear motion?

[bronze yoke]

I have no idea what that is

[covert rain]

the motion with s(t)=vt and v constant

with zero initial conditions

Anyway, I think what you mean is that the body always remains in the same position.

[bronze yoke]

no

[covert rain]

and what is the question?

[bronze yoke]

in the question it's given that the person walks in a straight line at a constant speed of 2 m/s.

[covert rain]

to travel a straight line means that it moves with uniform rectilinear motion

if you put yourself in a real situation

Does saying you walk in a straight path mean that your position changes at the right time?

[bronze yoke]

I think this is what the graph would look like for c, and d

[bronze yoke]

here's the question @covert rain

[covert rain]

This graph means that your position does not change over time but always remains the same

[bronze yoke]

it's the very beginning of physics introduction so like highschool level

[bronze yoke]

sorry I meant to write 2m/s

[covert rain]

x(t) = x0

[bronze yoke]

so y represents velocity

[covert rain]

I repeat, it means that you always remain still in the same position

if this is the graph of the position

[bronze yoke]

c. Draw the velocity vs. time graph assuming the motion starts at the origin.

[bronze yoke]

how come if y represents velocity at m/s and not position

[covert rain]

so you use X(m) to represent position and X(v) to represent velocity?

[bronze yoke]

yeah

I'm not sure how to write that correctly

[covert rain]

ok

sorry for the apologize

[bronze yoke]

no worries, are my graphs correct for the questions?

[covert rain]

your graph means that the speed is always constant over time

so its like X(v) = v0

[bronze yoke]

yes, is that what was asked of me in c. and d.?

[covert rain]

perhaps origin means t=0

so v0=2m/s

c is correct

since the velocity is constant and does not depend on the position I assume that the graph of c is the same as that of d

[bronze yoke]

yeah that's what I also thought

[covert rain]

its just X(v)=v0

[bronze yoke]

maybe they said that in d to see if I'd confuse position with velocity

[covert rain]

could be

[bronze yoke]

"asssuming the motion starts at the origin" means I have to draw it starting at (0,0)? even though it's 2m/s?

or is this the correct way

[covert rain]

since the text clearly says that the speed is 2

I assume you mean that t0=0

this practically means that the body leaves at time t=0

[bronze yoke]

wait, wouldn't that mean that at d where it says "assuming the motion starts 1 m from the origin " the motion would start at t = 1?

[covert rain]

meters is not time

It means that you start from a slightly further forward position but still at time t=0

It's like saying that two people have to run but one of them already has a 1 meter advantage at the start, but both of them still start at the same time which in this case is t=0

[bronze yoke]

I see

[bronze yoke]

this solution is incorrect right?

d. would be correct but c wouldn't

[covert rain]

In your opinion, does solution c represent the motion of a person traveling at 2m/s?

[bronze yoke]

yeah to me it seems like it represents v = 0

[covert rain]

so does that mean the person is not moving at speed 2?

since v is costant

do you know derivatives?

[bronze yoke]

yeah that makes sense it would start at (0,2)

also is it fine that I wrote X(v)?

as y?

[covert rain]

How does your teacher write them?

[bronze yoke]

I'm not sure what X(v) means exactly and how it is different from writing v like the other person did

[bronze yoke]

for position I remember the teacher writing X(m)

but I'm not sure what she writes for velocity

[covert rain]

It seems to represent the position with the large letter and then inside the brackets you put a unit of measurement

[covert rain]

but here he only writes V

[near sky]

the gap here does not feel right

[bronze yoke]

true

[covert rain]

it is your point a right?

[bronze yoke]

here's my final solution for a and b

[covert rain]

notice that there is a connection between points a and b

[bronze yoke]

yeah they have the same slope

[covert rain]

no sorry

[bronze yoke]

which is velocity

[covert rain]

a and c

also b and d

[bronze yoke]

velocity = 2?

[covert rain]

this is an imporant thing

[bronze yoke]

so this would be my final graph for c and d

[covert rain]

so at point a he asked you to represent the motion starting from the origin, right?

[bronze yoke]

yeah

[covert rain]

what is the slope in that graph?

[bronze yoke]

in graph c? 1?

[covert rain]

in the graph a

[bronze yoke]

2

[covert rain]

ok

this means v0=2

so at point c why should it be 0?

i mean not the slope in this case

i mean v0

[bronze yoke]

origin means (0,0) which is what's asked in question c, since v = 2, (0,2) still satisfies what's asked of me?

[covert rain]

the origin in this case its t=0 i suppose

It could also mean that you have to represent the speed of point a) which is a motion that starts from the origin

[bronze yoke]

wait if x is t(s) then y should be v(m/s) instead of X(v) no?

[covert rain]

this is correct , as I said before I think it wants to represent the graph of the speed of the motion of point a) which starts from the origin

[covert rain]

this notation of V is from the teacher ?

[bronze yoke]

no that's a student

this labelling would be the correct way right?

[covert rain]

you coul write V(m/s) at this point

if the teacher writes X(m)

[bronze yoke]

so this

[covert rain]

Do you have any examples from the teacher?

you could see from there

or from the book if the teacher uses the same notations

[bronze yoke]

right the book

there's a graph where the book uses v[m/sec] and t[sec]

yeah I think that's the correct way

what I did should be correct

[covert rain]

why did you put a black dot?

[bronze yoke]

above 1? just some leftovers I didn't notice when I erased stuff

[covert rain]

it's not a problem

I'm just saying that it doesn't mean that the body leaves at time t=1s

[covert rain]

yes

[bronze yoke]

okay, thank you

[covert rain]

so a and b are the same graph except that for one the initial position changes, while c and d are the same

[bronze yoke]

thanks

.close

[safe radish]

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[wispy jacinth]

ok so

[wispy jacinth]

i js forgot the formula for the bloxpot

statistics thing

[summer berry]

Hi

[warm warren]

boxplot? Which formula? Do you know that $Q_2, Q_1, Q_3$ are?

[flat frigate]

Annie Maqionde

[summer berry]

Can anyone solve this question (i) and (ii) part

[wispy jacinth]

yes

[upbeat swan]

#❓how-to-get-help

[wispy jacinth]

my teacher is so deadass bruh i canr even understand her

[warm warren]

!occupied

[safe radish]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[warm warren]

do you know how to calculate Q_1, Q_2, Q_3?

which specific formula did you forget?

[azure trench]

Hi

[warm warren]

if you want to learn, khan academy exists

[warm warren]

!redir

[safe radish]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[wispy jacinth]

tbh all of em since my teacher is deadass

[azure trench]

Ohh

[wispy jacinth]

let me

show u the

ques

[warm warren]

@wispy jacinth

[wispy jacinth]

alright

i mean im also confused how to solve the ques

[warm warren]

k first study it from some source

then try to solve the qs.

[safe radish]

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[pulsar hill]

having some confusion graphing functions with transformations,
specifically on the order i'm supposed to do the transformatiosn in and i cant find any good sources

[fathom jewel]

Do you have a specific example?

[timid ridge]

its just pemdas

i assume you are taught the form
b ( a(x - h) ) + k or something similar

[fathom jewel]

[timid ridge]

[pulsar hill]

i flowkenuinley forgot how to use texit

[fathom jewel]

Just write it in text

[pulsar hill]

y = -3log_2 (-(x-4)) + 1

[pulsar hill]

ohh

so i do the h shift right

[fathom jewel]

I don't see the point being made about pemdas, but I am happy to let angel explain that

[pulsar hill]

then the y-axis reflection, then the x axis reflection and the vertical stretch, then the vertical shift?

[timid ridge]

apply your transformations in the order you would evaluate the function

so that means x - 4 then the -(...)

[pulsar hill]

oh thanks

.close

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[split moss]

why the eigenvalues of $HH^\dagger$ are always non-negative?

[flat frigate]

Custos Caeli Invictus

[split moss]

$H^\dagger = \overline{H}^T$

[flat frigate]

Custos Caeli Invictus

[split kayak]

Consider det(A*B) = det(A) * det(B)

And what are the properties of a hermitian matrix.

[split moss]

I don't remember why this is true 🙁

Anyway

$det(H H^\dagger) = det(H) det(H^\dagger)$

[flat frigate]

Custos Caeli Invictus

[split moss]

if det(H^T) = det(H) then

$det(H^\dagger) = \overline {det(H)}$

[split kayak]

Is H supposed to be a hermitian matrix or an arbitrary matrix?

[flat frigate]

Custos Caeli Invictus

[split moss]

Arb

[split moss]

$det(H) \overline{det(H)} = |det(H)|^2 \ge 0$

[flat frigate]

Custos Caeli Invictus

[split moss]

So the total determinant is non-negative

@split kayak and now?

<@&286206848099549185>

But in theory what I found only implies that the product of eigenvalues is not negative but it does not mean that all eigenvalues are ≥ 0

[safe radish]

@split moss Has your question been resolved?

[gusty pasture]

Does H have eigenvalues or is it not diagonalizable?

[minor goblet]

right, can you say anything about the eigenvalues of Hbar from the eigenvalues of H?

[safe radish]

@split moss Has your question been resolved?

[safe radish]

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[wise hill]

Considering the values you found for R2 and r, include in your paragraph a statement
regarding whether you believe there is a linear association, how well you believe the model
fits the data, and why you believe those things.

[tall hearth]

I mean r² indicates the strength of a linear association

the higher it is

[wise hill]

Ah ok. So if both are low, there is a weak linear association?

[tall hearth]

yup

although the sign of r just indicates the direction of the relationship

That's why we use r squared to determine the strength

[wise hill]

Hmmmm but I also have to see how well I believe the model fits the data. My scatter plot looks chaotic.

[safe radish]

@wise hill Has your question been resolved?

[safe radish]

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[warm warren]

im sorry, please translate

[tropic marlin]

ummmmm

[warm warren]

And you haven't told us where you're stuck, too.

[tropic marlin]

i need hel;p with ב.(1)

[warm warren]

Yes so may you translate?

[tidal imp]

angle ABD equal to angle ACD?

[tropic marlin]

Yes i need to know why it does

[tidal imp]

Have you heard of cyclic quadrilaterals?

[tropic marlin]

what about it

[warm warren]

Hint: AD is a chord and it subtends the angles ACD and ABD at two different points on the circumference.

Does a theorem.......now strike you as oddly familiar to what I wrote above?

[tropic marlin]

so they are equal because they both lean on AD?

so two angles thats lean on the same chord are equal

therefore ACD=ABD

[wary marten]

Hello, Fellow Companions. I'm Roagi's Partner and we would love to get some help.

[tropic marlin]

they helped👍

@warm warrenthanks

.close

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[opaque fern]

nice bunny

but also

what happened here

[fickle mantle]

I sent the divided x to the 37

[opaque fern]

you cant do that

you would have to multiply EVERYTHING by x

[fickle mantle]

Why

[burnt notch]

Because otherwise you would change the equation

[opaque fern]

1/2 + 1/2 = 1

according to your logic we can do

1/2 + 1 = 2

[thorn dragon]

I think he only multiplied the right hand side and not the left hand side too

[burnt notch]

Because: revise basics of equations...

[fickle mantle]

Ok

[safe radish]

@fickle mantle Has your question been resolved?

[stoic torrent]

whats the issue now?

[stoic torrent]

-# as said here, you cant just shift the x individually

[stoic torrent]

else ur doing this

[stoic torrent]

so here you would need to multiply by x on both sides!
which in turn will "remove" the x in denominator and give you a QE in x

[fickle mantle]

Is this better

[opaque fern]

yuh

[fickle mantle]

But how do I do it

Do I use the quadratic formula

[opaque fern]

you can

but you can also factor it

what are two numbers that multiply to 60 and add to -17

[fickle mantle]

Is t it 51 and. It 60

[opaque fern]

what

[fickle mantle]

I got 51

[opaque fern]

for?

[fickle mantle]

The work

Instead of 60

[opaque fern]

3*20 not 3*17

so no

it is 60

[fickle mantle]

How is it 20

[opaque fern]

for ax^2 + bx + c

you consider the things that multiply into a*c

not a*b

[fickle mantle]

Can u reword this

I don’t understand

[slate ingot]

So to speak
For ax² + bx + c = 0
Consider two numbers p and q
For factorizing the equation, you look for p and q such that
p+q = b
pq=ac
Then you write b = p + q

Honestly this is standard theory thats covered in class at some point

But its hard to understand the logic behind factorization of quadratics until youre familiarized with polynomials

[open wedge]

$(x - a)(x - b) = x^2 + (a + b)x + ab$, that's why you look for 2 numbers like so

[flat frigate]

1 divided by 0 equals Infinity

[slate ingot]

Precisely

[open wedge]

but sometimes you don't need to factor to solve for $x$, you could also complete the square

[flat frigate]

1 divided by 0 equals Infinity

[fickle mantle]

Nvm

Lex is right

I multiplied b

[opaque fern]

[opaque fern]

-20 and 3 multiply to -60 tho

not 60

[fickle mantle]

Where’s the - from

[opaque fern]

your picture lol

[fickle mantle]

Yes but the c is positive

[opaque fern]

exactly

so it cant be -20 and 3

[fickle mantle]

Yeah

[opaque fern]

so

other guesses?

[fickle mantle]

Quadratic formal

[opaque fern]

no

[opaque fern]

just think of this

[fickle mantle]

Ohh 5 and 12

[opaque fern]

almost

[fickle mantle]

5 and 12

[opaque fern]

5 + 12 = 17

not -17

[fickle mantle]

What💔

Dang it

[opaque fern]

you are sO close

hint: the numbers' magnitudes are right, check the signs 😭

[fickle mantle]

But that,l make it -60

[opaque fern]

what happens if i multiply a negative with a negative

[fickle mantle]

Positive

[opaque fern]

so...

[fickle mantle]

Oh

-5 -12

[opaque fern]

yes

gjob

[fickle mantle]

Ur smart ty everyone

[opaque fern]

u r not done tho

so like

3b^2 - 17b + 20 = 0 -> 3b^2 - 12b -5b + 20 = 0

how do you finish this

[fickle mantle]

My WiFi sucks

[opaque fern]

yaaaaaaay

you did it!!!

[fickle mantle]

[primal bone]

(incidentally, holy crap you structured this better than some of my students)

(A lot of them almost always ignore the "=0" part altogether )

[fickle mantle]

Wow a teacher

I’ve never met a real teacher on dc

[opaque fern]

a lot of people here are

mostly tutors though

[fickle mantle]

Oh

[opaque fern]

we used to have a very active highschool/middleschool teacher here but they removed her from the server

[fickle mantle]

Cool

[fickle mantle]

Why

[opaque fern]

drama

lots of it

[fickle mantle]

Ohh

Can I get help w question e

[covert rain]

what is the question?

[fickle mantle]

Question e

[slate ingot]

You have 2(x+y) = 39,2
-# i cant tell if its a decimal
And xy = 96

[opaque fern]

2*96 = 192 not 196 btw

[covert rain]

you can make a system between the area and the perimeter

[slate ingot]

-# This is the sort of mistakes i make on my tests 💀

[fickle mantle]

Oh

[opaque fern]

otherwise the logic is fine tbh

[slate ingot]

Yep

[fickle mantle]

A system?

[fickle mantle]

Oki

[covert rain]

yes

[opaque fern]

thats already what u r doing

[covert rain]

like this

[winter whale]

skittles always asks good questions

[opaque fern]

so like

u kinda

[opaque fern]

yeeted that denominator at the end

[desert juniper]

never forget to put which variable you're solving for

[covert rain]

write 0.64 as a fraction

[opaque fern]

ok imma dip there are too many people helping now

[fickle mantle]

Mk

Ty

[fickle mantle]

Ok

[safe radish]

@fickle mantle Has your question been resolved?

[fickle mantle]

Do uk the formula for speed pls

[fickle mantle]

For question f

[icy dome]

distance divided by time

[slate ingot]

In general, average speed is
Total distance/total time
Imma check f once tho, gimme a sec

[fickle mantle]

Ty both of u

[icy dome]

welcome do u need the solution for the f part?

[fickle mantle]

I think I subtracted 30 form 180 to find the rerunning distance

[icy dome]

no thats wrong

[slate ingot]

I dont think thats correct

[icy dome]

time to ceres is - 180/ x and time bck home is 180/x-30

so it the equation to do it will be 180/x-30 - 180/x = 1

[fickle mantle]

What

I’ll reread the question

[fickle mantle]

I have this but idk how to start

Can I assume the initial time was an 1h

[icy dome]

The problem says she travelled she travelled 30km/h slower while goign home

so 90-30= 60 so the speed was 60km/h

and her speed while going to Ceres was 90km/h sos its 60km/h

hope this helps you

[fickle mantle]

I don’t understand

[icy dome]

okay should i re explain it

[fickle mantle]

Yes pls

[icy dome]

okay wait im typing it

[fickle mantle]

Km is distance right

[icy dome]

yeah its kilometer

if u see the question says 180km to ceres and 180 km back

[fickle mantle]

Yes

[icy dome]

So we can make her normal speed as = x - 30km/h beacuse the questions said she was 30 km/h slower

now we use the formula Time=distacne divided to speed

time going there = 180 divided to x
time coming back = 180 divided to (x - 30)

and in the question it said she lost 1 hr on the trip back to home so

180 divided to (x - 30) = 180 divided to x + 1

u there?

[fickle mantle]

Yes

I’m just trying to understand

[icy dome]

oh also sorry im on laptop so the dividsion sign isnt there

u under standing ?

[fickle mantle]

It’s ok

[fickle mantle]

No sorry

[icy dome]

oh

[fickle mantle]

Erm so the returning part of the question is the speed right?

[icy dome]

yeah

[fathom jewel]

-# Can't you use /

[icy dome]

i thoguht she wouldnt understand

[fickle mantle]

Can u explain the question first ppls

[icy dome]

we have to find the speed the lady traveled back in

[fickle mantle]

Oh

How do I use this to form an equation like u said

[icy dome]

i told u the equation

[summer berry]

hi

what was the question i hope i can help you

if it is quadratic

[summer berry]

is this 39.2 or 39,2?

@fickle mantle

[fickle mantle]

It’s the same thing

Just my school using a the point as a comma

[summer berry]

oh

l = 4.26?

and b = 22.5 something is this the answer @fickle mantle ?

[fickle mantle]

I don’t understand

[summer berry]

you ask e) right

[fickle mantle]

MY phones charging

Late replies

[fickle mantle]

Yes

[summer berry]

so what you don't understand

like any equation or something

[fickle mantle]

Uh I can’t explain

[summer berry]

so have you formed the equation

[undone notch]

factorize

[primal bone]

I'm making a guess - South Africa?

[fickle mantle]

Howd uk

[primal bone]

(because then that would be the right way to write a decimal)

[primal bone]

One of the other questions uses the S. African rand

(the real check I wanted to do was confirm the decimalisation - that does apparently seem to be the standard in Afrikaans, so I wouldn't be surprised if that's the standard as a nation)

[fickle mantle]

I have no idea what ur saying

[primal bone]

Using a comma for a decimal point is typical in Afrikaans

So it's not a school thing to write it like that, as such, but a national thing, was my point

[primal bone]

Regarding question (f)? No, because that's not what the "time" in speed/distance/time equations is about

"time" refers to the time taken to travel a distance

[fickle mantle]

can u explain speed, distance and time to me

It s even a while since I pre die. Questions lie, these

[primal bone]

(average) speed, by definition, is distance travelled divided by time taken

[primal bone]

This equation

[fickle mantle]

oki

Then can u break down the questions w me

[primal bone]

Honestly, a pretty decent explanation is here

[primal bone]

'cept to clarify this last line -
180 / (x-30) = (180/x) + 1

[fickle mantle]

why do we use time

[primal bone]

(where x is the speed the lady travels at when heading to Ceres)

[primal bone]

As opposed to...?

[fickle mantle]

what they said

[primal bone]

?

What they said uses time, so I'm not sure I get what your question is

If you want, an equivalent approach to the question can go as follows:

[fickle mantle]

I kind of understand

[primal bone]

Let s be the speed she travels to Ceres (in km/h), and t the time it takes her (in hours)

[fickle mantle]

The distance form where she is is 180km and she was traveling at a slower speed when returning than when travelling to. So the unknown speed minus the slower speed

[primal bone]

Then since speed = distance/time i.e. that distance = speed x time, we have st = 180

[fickle mantle]

what’s st

[primal bone]

s times t

[primal bone]

from here

[fickle mantle]

Oh

[primal bone]

(it's what Aarna used x for)

[primal bone]

From this - if we look at the return journey and do the same thing, she's travelling at (s - 30) km/h, for an hour longer (i.e. (t + 1) hrs), across the same distance (180 km)

So (s - 30)(t + 1) = 180

So now we have two equations, and two unknowns -
st = 180
(s-30) (t + 1) = 180

Simultaneous equations - solve however you like

[fickle mantle]

why do we use the distance form,USA

Formula

I understand everything you’ve said so far

[primal bone]

I get the feeling you don't tbh

wdym why?

We're told information about speed, about distance and about time; what other formula is there to consider?

[fickle mantle]

Oh

No I do understand

I just dirnt know why u used the distance as a subject formula

Instead if the rest

Idk how to explain

It makes sense in my head

[primal bone]

Fundamentally it doesn't matter what the subject of the formula is

20 = 4 x 5 and 4 = 20 / 5 are practically speaking the same

The point is to get some equations with which we can attempt to solve for some variables

[fickle mantle]

Oh ok ok

Tysm

[fickle mantle]

.close

[safe radish]

Channel closed

Closed by @fickle mantle

Use .reopen if this was a mistake.

[safe radish]

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[gentle hedge]

how to evaluate this limit

[gentle hedge]

h^2 * ln h

i dont like the logic they've used