help-10

[full elm]

I'll try it tomorrow. If i cant solve it ill create a new channel

Thank you @twin sandal

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[obtuse pebble]

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[full elm]

.reopen

[obtuse pebble]

✅

[full elm]

if anyone wants the answer: its 178. I checked using a computer program

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[long topaz]

A 0.5 kg ball is struck by a bat causing it to experience an impulse of 10 N•S if the change in time is 0.2 seconds, calculate the average force exerted on the ball.
I know that f•t=m•v and so far have gotten f(0.2) =m(0.5) but confused on what to do after this

[timid silo]

How did you come to the conclusion that v equals 0.5?

[long topaz]

That was the weight I accidentally put there woops

[timid silo]

We know that
J (impulse) = F.time

We are given the value of impulse and the time for which the cat is in contact with the ball

[long topaz]

Ooh

10 n/s = 0.2 seconds

??

Is that right

[timid silo]

🤨

Write that again

[long topaz]

Or am I stupid

10 n•s

Mb

[timid silo]

You completely missed to mention force

10 = F . (0.2)

[long topaz]

Alr

F. Is force times right?

[timid silo]

Yeah

[long topaz]

So 10 = 0.5 times 0.2

[timid silo]

Eh

What?

10 = F (0.2)

What is F?

[long topaz]

Force?

[timid silo]

What is it's value

[long topaz]

Ooh

10 = 50(0.2)?

[timid silo]

Yeah so F is 50N

[long topaz]

F = 50?

[timid silo]

Yes

[long topaz]

oooh

Mb I'm super slow

Thank you

[timid silo]

It's alright

[long topaz]

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[magic geyser]

Why would 0<t<2 be increasing rather than 1<t<5?

[daring rock]

The graph shows the particle's position

on 1<t<5, the position is increasing

[magic geyser]

oh shoot

[daring rock]

But the question asks where the velocity is increasing

[magic geyser]

Wait

so how would I go about

solving this

[daring rock]

Are you familiar with concavity? Like if a graph is concave up/down

[magic geyser]

Yes

[daring rock]

velocity is the derivative of position

v(t) = x'(t)

Another way to say "velocity is increasing" is to say v'(t) > 0

[magic geyser]

i understand that part

[daring rock]

Awesome, so then

Since v(t) = x'(t), then v'(t) = x''(t)

So, "velocity is increasing" means v'(t) is positive, which means x''(t) is positive

[magic geyser]

which is a(t) = v'(t)

wait dont u use the 2nd der for concavity

[daring rock]

yes

[daring rock]

So you're looking for where x''(t) is positive

which you can see on the graph based on the concavity of x(t)

[magic geyser]

its only concave up on 0,2 bc it has a min

[daring rock]

yeah I mean we can kind of tell it's concave up from 0 to 2 because it's "U" shaped there

[magic geyser]

ohh and then you can apply that logic

[daring rock]

it's hard to tell exactly where it stops being concave up, but 2 is a good estimate

[magic geyser]

position is increasing because velocity > 0

then you derive both

to get

velocity is increasing bc acceleration > 0

[daring rock]

I think you're correct, but using f(x) is confusing

the position is x(t)

v(t) = x'(t) is increasing because a(t) = x''(t) > 0

yes

[magic geyser]

okay tysm

[daring rock]

np 👍

[magic geyser]

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[cosmic falcon]

I don’t get this one at all

[cosmic falcon]

What’s concave

I understood infection points are the second derivative points that’s are = to 0

So basically they are the critical points of the second derivative

But what’s concave how is it different than increasing and decreasing

This is the function I think

[worn yoke]

concave up is when the function looks like $\cup$

concave down is when the function looks like $\cap$

[warm shale]

cloud

[worn yoke]

concave up: second derivative is positive
concave down: second derivative is negative

[cosmic falcon]

So -INF to 0 looks like it

Concave up

But it says it’s wrong

[worn yoke]

did you find the second derivative?

[cosmic falcon]

Yup

[worn yoke]

what is it?

[cosmic falcon]

-12x^2-24x

[worn yoke]

can you factor it?

[cosmic falcon]

-12x(x+2)

So the 0’s are 0 and -2

The infection points

[worn yoke]

so let's consider:

1. where is -12x negative? where is it positive?
2. where is (x+2) negative? where is it positive?
3. based on that, where is (-12x)(x+2) negative? where is it positive?

[cosmic falcon]

Wut

-12x is already negative

Oh wait

[worn yoke]

is it negative for every value of x?

[cosmic falcon]

Ohh

For anything above 0

?

[worn yoke]

yes

[cosmic falcon]

But it does want look like the upside down U

[worn yoke]

we need to find where the second derivative is positive, and where it is negative

[cosmic falcon]

So the shape doesn’t matter basically

?

[worn yoke]

it does matter, but it will be more reliable to use an algebraic method than to rely on your graphing skills

[cosmic falcon]

I like that

So basically it’s concave up for x>0?

So (0,inf)

Or do you have to check all of it together

But that’s impossible because for x+2 it’s positive for x>-2

Which means -1 is an answer for this but not for the other

So it’s contradicting each other

So how do I know if it’s (0,inf) or (-2,inf)

[worn yoke]

one thing I like to do in these situations is to draw a sign chart

the top is a number line, we draw vertical lines at the zeroes of each factor

then add 0 on the line in the proper row

then in between each line, we will fill out either + where the factor is positive, or - where the factor is negative

[cosmic falcon]

Where is that?

Not sure if I understand than graph

And wouldn’t that be -2?

Instead of 2

[worn yoke]

it should be, yes

[cosmic falcon]

How do you combine both answers tho?

[worn yoke]

updated:

[cosmic falcon]

Each factor has a different answer

That contradicts the other

[worn yoke]

we know that (positive)*(positive) = positive, (negative)*(negative) = (positive), and (positive)*(negative) = (negative)

so we will fill in the bottom row based on our answers from the top two

[cosmic falcon]

Hmm

But x can’t be two different numbers

Both x of the 12 and x of the +2 has to be the same number

But one will make the other one false

Since it has to be positive or something

Like for -12x it has to be above 0 but for x+2 it can be below 0 too

But then if we choose x=-1 it’ll make -12x positive

[worn yoke]

-12x and (x+2) will have potentially the same sign or potentially different signs in different intervals

[cosmic falcon]

I think I’m lost, what are we trying to find again🤣

[worn yoke]

the purpose of this chart is to keep track of this

let's start with this:

1. between -infinity < x <-2, is -12x positive or negative?
2. between -2< x <-0, is -12x positive or negative?
3. between 0< x < infinity, is -12x positive or negative?

[cosmic falcon]

Both?

1
For any negatives its positive and for any positives its negative

And for 0 it’s 0

[worn yoke]

so out of (1), (2), and (3), on which intervals is it positive?

[cosmic falcon]

None

[worn yoke]

why none?

[cosmic falcon]

It’s (-inf,0)

[worn yoke]

yes

[cosmic falcon]

Cuz it’s positive for all negative

You tried to trick me? Haha

[worn yoke]

i'm not asking "which interval is the only interval where it's positive"

i'm asking "which intervals (not exclusive) make it positive"

[cosmic falcon]

There are more??

How

[worn yoke]

it's not a trick question

[cosmic falcon]

The only numbers left are 0 to infinity and it’s negative in all of them

[worn yoke]

(-inf, -2) is part of the interval (-inf,0)

[cosmic falcon]

But -inf,0 already includes -2

Why do we need to specify it

[worn yoke]

anything we say about a larger interval also applies to the intervals inside it

[worn yoke]

because those are the 3 we have on the chart

[cosmic falcon]

But it’s already included

Why don’t we do (-inf,-10) (-inf,-9) etc

[worn yoke]

because the chart is separated based on the zeros of each factor. -2 is a zero of another factor so we split it there

[cosmic falcon]

Ohh! Because-2 is not allowed?? Cuz it’ll make the other one 0

Ohhh

So (-inf,-2)U(-2,0)

[worn yoke]

so far we can fill out the first row of the chart, like this:

now let's look at the next row. where is (x + 2) positive, and where is it negative?

[cosmic falcon]

Wait is this correct?

[worn yoke]

no, -2 is still included. we just are not asking about it specifically for making this chart

we don't really care about the sign of -12x when we know that the (x+2) will make the entire function 0

[cosmic falcon]

Ohh cuz that’ll be times 0

Well for the second one it’ll be negative from -inf,-2

Then positive from -2,inf

So - + +

[worn yoke]

so now our chart looks like this:

[cosmic falcon]

Now you add the signs??

So - + - ?

[worn yoke]

going down the column, we will multiply them

so for the first one (-inf,-2) we do (+)(-)

[cosmic falcon]

My brain

What does that tell us

[worn yoke]

the sign of the overall function

[cosmic falcon]

Just because of the first column?

[worn yoke]

it tells us the sign of the function between (-inf,-2), yes

[cosmic falcon]

I don’t get the logic

Isn’t it basically that the function can’t be 0

[worn yoke]

if we know the sign of two numbers which are being multiplied together, then we can predict the sign of the product

[cosmic falcon]

Because if we have anything bigger than -2 the function will be either 0 or negative right?

[worn yoke]

if x is in the interval (-inf,-2) then we know that the product (-12x)*(x+2) will be the product of a positive number (-12x) and a negative number (x+2)

[cosmic falcon]

Because x can be up to 0 for 12 but only up to -2 in the x+2 so you just restrict it further

Instead of (-inf,0) it’s (-inf,-2)

[cosmic falcon]

This won’t click in my head it’s too confusing for me🥲

Is my logic wrong?

[worn yoke]

so we know that (-3) * (5) = (-15). in other words, the product of a negative number (-3) and a positive number (5) is a negative number (-15)

[cosmic falcon]

It still says it’s wrong🥲

[worn yoke]

wait to submit until we're done with the chart

[cosmic falcon]

That’s not the answer?!

I need a different method I don’t get the chart🥲

I prefer numbers than drawings

When they want me to find stuff on a graph I have problems with that too

[worn yoke]

can you perform the multiplication (-3) * (8)

[cosmic falcon]

That’s why I said here that I like that lol

[cosmic falcon]

Yes but here it’s different I don’t know it doesn’t click for me

[worn yoke]

what is the answer?

[cosmic falcon]

-24

[worn yoke]

what were the signs of the two things being multiplied?

[cosmic falcon]

-+

[worn yoke]

and what was the sign of the answer?

[cosmic falcon]

But I don’t get how that gives us the interval

[worn yoke]

so we know that (-)(+) = (-)

[worn yoke]

we already established that (-12x) is always positive on the interval (-inf,-2) and (x+2) is always negative on the interval (-inf,-2)

the chart is just a way of writing that down

[cosmic falcon]

Wait what

Oh yea

Wait that’s negative

I’m so confused

Why do we separate them in the first place

This is a nightmare of a question it’s been an hour🫠

Finding where it increases and decreases is so much simpler

How is this not the same

Concave is basically maxima and minima no??

[worn yoke]

because it's easier to find when each one is positive and negative individually, and we could apply this to as many factors as we wanted (e.g. if we had a polynomial with 5 factors)

[worn yoke]

we can use the same table for the first derivative to find that as well

[cosmic falcon]

Can we use the third derivative to find this then??

Just like how we can use the second derivative to find it for critical points

[worn yoke]

so we had that in (-inf,-2) the function is - because (+) * (-) = (-)

what about at x = -2? what sign is it there?

[cosmic falcon]

None

It’s 0

[worn yoke]

so we'll put a 0 there like we did above

[cosmic falcon]

This is way too much work, we have an hour and 40 minutes for 25 questions on the exam there must be an easier method

[worn yoke]

this will be much quicker if you practice it

it's only slow because i have to explain it

[cosmic falcon]

I don’t get where this is headed

So it’s - + -

Basically

[worn yoke]

yes

[cosmic falcon]

And so?

The function is negative then positive then negative?

[worn yoke]

yes. so from the chart we can construct the intervals where f''(x) is positive, and the intervals where f''(x) is negative

and remember that f is concave up when f'' is positive, and f is concave down when f'' is negative

[cosmic falcon]

So it’s the exact same thing as where it increases and where it decreases

..

[worn yoke]

it's the same strategy but with the second derivative instead of the first derivative, yes

[cosmic falcon]

So the much easier way is using the third derivative

Then plugging the values

And seeing if they are bigger or smaller than 0

24>0 means the plugged value -2 is a minima

Then drawing it accordingly

Yes it’s the exact same thing

Why all these clever ahh words it’s the exact same thing with different names

Concave and shit

It’s just where it increases where it decreases

One hour😭

[worn yoke]

f being concave up is where f' is increasing, yes

[cosmic falcon]

Why are they even asking it with different words this is so stupid

It’s the exact same question as doing it for critical points

[worn yoke]

with a different order of derivatives, which makes it look different on the original graph

[cosmic falcon]

That’s why I asked you this but you ignored it

[cosmic falcon]

It’s just another derivative

Instead of doing it with the second you do it with the third

It’s the same math

[worn yoke]

as long as the third derivative is nonzero, i suppose that works

[cosmic falcon]

If it’s 0 you just take the 4th derivative

[obtuse pebble]

@cosmic falcon Has your question been resolved?

[obtuse pebble]

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[glass rivet]

Me

[obtuse pebble]

Please don't occupy multiple help channels.

[gloomy vector]

notice how the first term is 4

so 1^6×a^2=4
im sure you can find a

for b i think you do binomial expansion for the first 3 terms of (1+3x)^6 and expand (a+x)^2 [you shouldve found a by now] then multiply them and take the constant for x^2

[obtuse pebble]

@glass rivet Has your question been resolved?

[glass rivet]

Yep

[obtuse pebble]

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[sinful kraken]

if my quadratic roots are 0 and 2, what is alpha and what is beta ?

[safe haven]

0 and 2...?

[sinful kraken]

x1=0 x2=2

[ruby path]

choose whichever one you want to be either

doesn't really matter

[sinful kraken]

beta = 0 sicne we are not in complex numbers

but what about alpha

[safe haven]

alpha and beta is just another representation for the roots

[sinful kraken]

how do we find them

[safe haven]

you literally just said its 0 and 2 no

[sinful kraken]

no

i'm missing something

if

[safe haven]

!xy

[obtuse pebble]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[sinful kraken]

(r-2)(r+2)=0, alpha = 2 and beta=0

what about for r^2-2r-3=0

[safe haven]

why is it 2 and 0?

[sinful kraken]

idk

what is alpha and beta ?

[safe haven]

its the roots of the equation no?

[sinful kraken]

idk

i know that beta is the complex part

and alpha the real part

[safe haven]

oh ur working with complex numbers?

[sinful kraken]

roots= alpha +- i*beta

yea

[safe haven]

should've said it earlier

[sinful kraken]

but what if we dont have complex numbers

beta = 0

[safe haven]

then the imaginary part is just 0

[sinful kraken]

what about alpha ?

[safe haven]

whatever the number is

2 = 2 + 0i

[sinful kraken]

i dont understand

[obtuse pebble]

@sinful kraken Has your question been resolved?

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[sonic umbra]

help me with this please?

[obtuse pebble]

@sonic umbra Has your question been resolved?

[sonic umbra]

no lil bro

wait

<@&286206848099549185>

[obtuse pebble]

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[sonic umbra]

AAHAHRGAGA I NEED HELP!

[sonic umbra]

could someone help me?

[late elbow]

I can only help with the first question because I honestly don’t know what’s going on with the 2nd graph 😭 @sonic umbra

[sonic umbra]

bet

YOO

ITS KANIN AGAIN

HI

[late elbow]

The first thing you have to keep in mind about the line of this graph is : it is a straight line, so the standard form of its equation is : y = mx + b

So the question is asking for m (slope) and b (y-intercept) so those are what you have to find first !

[sonic umbra]

okay

how do I find the slope

and y- intercept

[late elbow]

To find the slope, use the slope formula : y2-y1 / x2-x1

You must use x values and y values that are whole numbers because whole numbers are exact and they aren’t estimates

You have to determine 2 exact points that the line has

when plugging in the y’s and the x’s , only the y values go into the numerator and only the x values go in the denominator

[sonic umbra]

hmmm

wait so

[late elbow]

b (also known as the y-intercept) is a line’s point where x = 0

So you can just look at what the y value is when x value is 0 👍🏼

[sonic umbra]

y1 is = 20?

[sonic umbra]

okay

[late elbow]

Yes , that works 👍🏼
Make sure to also list x1

[sonic umbra]

bet

x1 is 1

right?

it couldnt be anything else

[late elbow]

yuppp

[sonic umbra]

bet

so y2 would be

40?

[late elbow]

yes, y2 can be 40
y2 can actually be any y value of the line , as long as it is not y1 and it is a whole number 👍🏼

[sonic umbra]

alr

so x2 would be

4

4-2 is 2

40 - 20 is 20

then 20/2

?

@late elbow

[late elbow]

You can only use 4 as your x2 value if 80 is your y2 value !

[sonic umbra]

oh

so

x2 is 2?

cuz half of 80 is 40?

and thats y2

[late elbow]

yesss
‘Cause you chose 40 as your y2 👍🏼

[sonic umbra]

okay

soo

2-1

is 1

40 - 20 is 20

20/1

what is that as a decimal?

if I need to

[late elbow]

yup 👍🏼
20/1 is just the same as 20

‘Cause any number you divide by 1 does not change at all

[sonic umbra]

ooo

okay I understand

now what

[late elbow]

So 20 is your slope

20 will be your m value in the equation y = mx+b

But now you need to know too what b is

just look at the y axis and see if the line crosses there and what point is there

[sonic umbra]

hmm

[sonic umbra]

okay

also mb I was afk

wdym cross

like here?

[late elbow]

I mean, if the line is exactly on the y axis 😅

My English is not that good since it’s not my first language , sorry 🥲

The y axis is the line where the y values are written

And the x axis is the line where x values are written

[sonic umbra]

hmm

[sonic umbra]

nah your good

[sonic umbra]

so how do you tell

if it's crossing the y

[late elbow]

For question 1,

when you look at the line, one of its points is in the origin (origin of any graph is known as (0,0)

From (0,0) , what do you think is the y intercept ?

The line has a point on both x and y axis , not just the y axis

[sonic umbra]

SORRY

I WAS AFK

[sonic umbra]

ohhhh

[late elbow]

We can take this graph just as an example 👍🏼

In your observation, what do you think is the y intercept ?

[sonic umbra]

uhh

[sonic umbra]

the line

[late elbow]

To know what the y intercept is, you need to look at the y axis. if the line is on the y axis, then look at what the y value is for that 👍🏼

[sonic umbra]

so

y is 0

[late elbow]

For the question you sent , yes
The y intercept is 0

So that is your b value 👍🏼

[sonic umbra]

0

ohh

20 is m

0 is b

WAIT

[late elbow]

That is how to know the y-intercept

but since the y-intercept for the graph is 0 and it’s not another number , you don’t write the equation as y = 20x + 0

It will be written as y = 20x

Because nothing happens when you add 0 to 20x because it’s zero and 0 doesn’t change anything when it’s added or subtracted 👍🏼

[sonic umbra]

I solved it

[late elbow]

👍🏼🫶🏼

[sonic umbra]

thank you so much kanin

[obtuse pebble]

@sonic umbra Has your question been resolved?

[sonic umbra]

no

wait lil bro

[obtuse pebble]

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[heavy anvil]

Da problem mate?

[polar fossil]

what

[opaque dome]

Its probably a miss-open

[polar fossil]

.close

[obtuse pebble]

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[obtuse pebble]

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[wanton dune]

How to solve lim x to inf x(cosx +a) for a real

[wanton dune]

I know that it’s infty(a+(-1 to 1))

Is that undefined?

[grand yoke]

i think its undefined

since lim x -> inf (cos x ) is undefined

[timid silo]

I concur

[wanton dune]

How can I prove how that’s its undefined?

What theorem should I use?

[grand yoke]

there is no theorem

i guess you can just state that it is undefined

[timid silo]

,w evaluate lim x-> inf x(cos(x))

[warm shale]

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[obtuse pebble]

@wanton dune Has your question been resolved?

[obtuse pebble]

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[tawny tree]

when does a_n->g imply f(a_n) -> f(g)? where -> means the limit as n-> inf and why

[fossil crag]

continuous f is sufficient

but without any more information about (a_n), if f is not continuous then the statement is false in general

In fact, f:R->R is continuous in g if and only if for any (a_n) such that a_n->g, then f(a_n) -> f(g)

[opaque dome]

Si et seulement si

[tawny tree]

wait thats literally the definition

omg

[fossil crag]

it's not the definition for continuity

but it certainly is a characterization of continuity

for real functions

[tawny tree]

continuity means a function equal to its limit at every point right?

[alpine bison]

the literal definition doesn't use sequences

[opaque dome]

From right and left equal to f(the point)

[tawny tree]

ah yes

[fossil crag]

a function is equal to its function limit at every point

[tawny tree]

oh yes sorry thats what I meant

[fossil crag]

yeah and in some spaces (non sequential spaces) what we call sequential continuity isn't the same as continuity

[fossil crag]

"for any (a_n) such that a_n->g, then f(a_n) -> f(g)" is sequential continuity

"f(x) -> f(a) when x-> a" is continuity

[tawny tree]

alright

thank you

[fossil crag]

happily for you, the set of real numbers is a sequential space

[fossil crag]

so this

[fossil crag]

and this

are the same

[opaque dome]

Which one isin't ? Rational ?

[fossil crag]

you can only hope to find non sequential spaces with non-metric spaces

Q has the absolute value that induces metric

[opaque dome]

Ah

[tawny tree]

its for real analysis so thankfully im working with R

much thanks

.close

[obtuse pebble]

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[obtuse pebble]

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[winter olive]

What flag is set when a larger positive signed number is subtracted from a smaller positive signed number?
Answer: Carry Flag or Sign Flag
I feel like both are correct but I'm not sure help

[last pilot]

"flag"?

[winter olive]

Ya

[last pilot]

what is a "flag"

[fossil crag]

this is more computer science than math related

I don't know if there's a more suited channel

[winter olive]

Your right i will be on my way

.close

[obtuse pebble]

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[last pilot]

wat

[obtuse pebble]

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[subtle kayak]

hi i need help

[obtuse pebble]

@subtle kayak Has your question been resolved?

[subtle kayak]

<@&286206848099549185>

[obtuse pebble]

@subtle kayak Has your question been resolved?

[sharp ivy]

draw it

[obtuse pebble]

@subtle kayak Has your question been resolved?

[last pilot]

<@&268886789983436800>

[oak escarp]

why are you pinging mods

[last pilot]

...?

[high lily]

N slur using reacts

[upbeat plinth]

.close

[obtuse pebble]

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[oak escarp]

oh I see

[last pilot]

thanks roketto+ramonov+derivada

[high lily]

it helps to describe the issue as it may not be immediately obvious

[obtuse pebble]

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[celest karma]

can anyone help me with this

[celest karma]

can anyone help me how the dP/dtheta get?

[crimson gust]

Do you know the quotient rule?

[celest karma]

f(x)= vu' - uv'/v^2

is this one?

[crimson gust]

yes

[obtuse pebble]

@celest karma Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[high ermine]

yoo

[high ermine]

5+(8y-14)=11y-9 is not possible righr?

[arctic bloom]

,w 5+(8y-14)=11y-9

[warm shale]

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[arctic bloom]

looks possible to me

[high ermine]

what

bah

no

[high ermine]

explain how 8y=11y

[arctic bloom]

nah

[high ermine]

no posible

[thorny idol]

anything times 0 is 0

[high ermine]

yes

why

quite not how i was suposed to calculate that but ok 👌🏻

thanky you

[thorny idol]

you can isolate y on your first equation and find y = 0

[high ermine]

make sense

[thorny idol]

and with 8y = 11y, sub 11y both sides youll have 8y - 11y = 0, so -3y = 0 finally y = 0

[high ermine]

ok

how close

this problem

.close

[obtuse pebble]

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[obtuse pebble]

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[drifting sentinel]

Preparing for an university entrance exam, there's some stuff i've never even done in hs
need help on this particular question

[timid silo]

|(x-2)(x-1)| * x = k, right?

[drifting sentinel]

oh wait.

yeah...that should be it

but how do I solve for k?
do I just solve it like a normal equation for x and calculate the range of k?
english is my second language, so the question threw me off a little

.close

[obtuse pebble]

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[topaz warren]

Im not sure how to solve this

[warm shale]

syecko

[sterile kayak]

oops

[ionic oar]

triangles are similar

implies angle of small triangle = angle of big triangle

[topaz warren]

But I need the x and y

[ionic oar]

i think they want you to solve each one seperately

[topaz warren]

How?

[ionic oar]

we have y + 16 = 3y - 34

[topaz warren]

Oh, like that

Ok

y=-25

?

[ionic oar]

i dont think y is negative

[topaz warren]

No its positive

Sorry

[ionic oar]

yeah its correct then

anyways, do you know how to solve for x?

[topaz warren]

Yeah

Thanks for your help

[ionic oar]

yw

[topaz warren]

.close

[obtuse pebble]

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[obtuse pebble]

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[red cipher]

What steps to do to get 1/2 ln5

[digital kite]

But why did you do all these steps?

[red cipher]

Idk what's the right approach

Just showing my work

[digital kite]

You noticed that below the derivative was 2y

Multiply both the numerator and denominator by 2 and you were done

[red cipher]

Um

[digital kite]

Anyway

Mmm

But since it was y^2 shouldn't you have done (2x)^2?

[red cipher]

Right

Um I can't find the reason why I have an extra x

[digital kite]

What Is your result?

[red cipher]

Wait not this

1 sec

[digital kite]

Why do you think this is wrong?

,w $\int_{0}^{2x} \frac{xy}{x^2+y^2}dy$

[warm shale]

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[red cipher]

Nvm its right haha

.close

[obtuse pebble]

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[celest ether]

I don’t know where to start

[full elm]

join OS, OQ
Tangents are perpendicular to radii
find angle SOQ
It will be 92
so its not a diameter as angle subtended by an arc from the diameter is 90

[celest ether]

How did you work it out to be 92

I’m a bit slow

[celest ether]

Oh yeah tangent & isosceles

[celest ether]

But I don’t get the last line

I’ve never seen the word “subtended”

[full elm]

this angle is always 90

[celest ether]

WHAT

How on earth did you think of that

You literally picked a needle out of a hay bale

Thank you so much bro

I couldn’t see the right angle thingy

!close

How do I close channel

.close

[obtuse pebble]

Channel closed

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[obtuse pebble]

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[timid silo]

Hello, I do International A level Math Pure Mathematics 1 in Y12, I am really struggling with everything my grades have dropped drastically. What do I do, I do the questions in the textbooks, My teachers are of no help, I take tutoring as well they are fine. What else can I do?

[chrome pond]

plz any one explain how to solve these questions, and plz do one of them for me as a sample

[mortal mason]

@chrome pond open a help channel for yourself

[chrome pond]

?

[mortal mason]

this chat is already taken by omar go to the help chats which aren't occupied and ask there

[restive gorge]

What is it specifically that you struggle with?

[timid silo]

Literally everything, Mostly the graph and algebra part.

[restive gorge]

Is there an exercise for example?

[mortal mason]

when solving the questions from the textbook are you solving them and understanding ?

[fierce maple]

@mortal mason I want to ask a question but #help-0 is closed, what should I do?

[chrome pond]

<@&286206848099549185>

[mortal mason]

theres help-41, help-36,help-34

these 3 chats are free

[timid silo]

Yeah

[restive gorge]

you can post it

[fierce maple]

@mortal mason Those do not show for me, I only see 3 channels under the "suggested" section, how can I see the ones you are talking about?

[mortal mason]

here press this #help-41

[chrome pond]

plz help anyone

[fierce maple]

@mortal mason Thank you

[mortal mason]

@chrome pond go to a different chat and ask your question, this chat is already occupied.

[restive gorge]

Why are you not listening?

[mortal mason]

#help-36 go here @chrome pond and post your question

[chrome pond]

Okay

[chrome pond]

i hv posted there

[mortal mason]

yes i saw

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[last pilot]

!15m

[obtuse pebble]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[last pilot]

jeez man give us a bit

;-;

[timid silo]

Ok

sorry

[pallid aurora]

have you drawn it out?

[timid silo]

YEa

[pallid aurora]

can i see your drawing?

[timid silo]

1 min bro

it's uploading

for a long time

[pallid aurora]

your drawinf is a little off, the 6 and 12cm chords are switched

[timid silo]

Ab = 6

cd = 12

[pallid aurora]

why would a chord get smaller as it approaches the center?

the diameter*

[timid silo]

the chord gets bigger as it approaches the center

[pallid aurora]

okay so what is wrong in the diagram?

[timid silo]

idk

[pallid aurora]

the chords should be switched

[timid silo]

Ok

[pallid aurora]

i’d recommend redrawing and then i can guide you through a solution

[timid silo]

i am constructing

[pallid aurora]

alright nice

can you draw in lines OC and OA

[timid silo]

ok

[pallid aurora]

these will be radii of the circle and we can label their length with the variable r

[timid silo]

is it understandable

[pallid aurora]

yep

can i label there midpoint of CD as M and the midpoint of AB as N

[timid silo]

ok

[pallid aurora]

then we can say ON - OM = 3

let’s me know if you don’t understand why

[pallid aurora]

good

[pallid aurora]

.

do you understand this

[timid silo]

OM -On = 3

how

[pallid aurora]

MN is equal to 3

[timid silo]

yeah

[pallid aurora]

so we just trim off OM

from ON

[timid silo]

ok

[pallid aurora]

now can you use pythagorean theorem with our radii to get ON - OM = 3?

like what is ON in terms of r and would is OM in terms of r

[timid silo]

I kinda got confused in ON -OM = 3

[pallid aurora]

we want to find a way to work with the 3 cm

[timid silo]

Shall i keep Om = x

[pallid aurora]

nope

are you confused as to why ON - OM = 3 or as to how to get ON and OM in terms of r

[timid silo]

OM = x and ON = 3-x

does that make sense

[timid silo]

yeah

[pallid aurora]

ON should be 3 + x

then 3 + x - x = 3

[timid silo]

yeah

sorry

[pallid aurora]

understand now why ON - OM = 3

[timid silo]

yes

[pallid aurora]

okay

can you find ON in terms of r?

[timid silo]

Ok

AN = 3 cm right?

[pallid aurora]

no

wait yes

sorry

AN = 3

[timid silo]

A perpendicular bisector from center bisects the chord

that theorem

[pallid aurora]

yep ur all good

my mistake

keep going

[timid silo]

Then waht can i ddo

Ok

[pallid aurora]

use pythagorean theorem on triangle OAN

[timid silo]

i got it

[pallid aurora]

okay then repeat that process for length OM and solve for r

[timid silo]

$$ 3^2 + 3^2 = R^2 $$

[warm shale]

Cristiano Ronaldo Dos Santos

[timid silo]

$$ 18 = R^2 $$

[warm shale]

Cristiano Ronaldo Dos Santos

[timid silo]

$$ R = \sqrt(18) $$

[warm shale]

Cristiano Ronaldo Dos Santos

[timid silo]

These things are correct or not ?

OM = 0 cm

IS this right

[pallid aurora]

doesn’t seem right

[timid silo]

OK

[pallid aurora]

i believe r would be sqrt(45) not sqrt(18)

[timid silo]

how

[pallid aurora]

sqrt(r^2 - 9) - sqrt(r^2 - 36) = 3

i got this equation

is this what you had?

[timid silo]

$$ 6^2 + x^2 = R^2 $$

[warm shale]

Cristiano Ronaldo Dos Santos

[pallid aurora]

so x = sqrt(r^2 - 36)

[timid silo]

yeah

[pallid aurora]

which is side OM

[timid silo]

yeah

[pallid aurora]

what about ON

what was ur equation

[timid silo]

3-x

[pallid aurora]

3 - x makes everything a little complicated

also it was 3 + x

[timid silo]

{sqrt(r^2 - 36)+ 3-x} + 3^2 =

[timid silo]

THis thing is so confusing

man

[pallid aurora]

😭😭

gimme like 2 minutes and i’ll draw you a diagram

[timid silo]

Okk bro

[pallid aurora]

sorry for the atrocious handwriting

[timid silo]

No it's alright

[pallid aurora]

do you understand what i have done

[timid silo]

i dont have phone so it's hard to draw a diagram for me

[timid silo]

1 min

[pallid aurora]

okay

[timid silo]

I understood

[pallid aurora]

alright good

[timid silo]

Then 3 is the final answer bro

[pallid aurora]

huh?

[timid silo]

*then

[pallid aurora]

no

you need to solve for r

[timid silo]

ok

2 mins

[pallid aurora]

alright

[timid silo]

i am getting

Root 3sqrt(2

[pallid aurora]

you are doing something wrong in your arithmetic then

[timid silo]

what did u get man

[pallid aurora]

sqrt(45)

[timid silo]

only ans

is enough

Ok

thanks

Thank you so much

[pallid aurora]

go through the math

incase i’m wrong

[timid silo]

ok

.close

[obtuse pebble]

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[obtuse pebble]

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[hearty terrace]

If we have that $\bf{u}=\bf{v}/||\bf{v}||_1$ for one norm, why is it that $||\bf{u}||_2=||\bf{v}||_2/||\bf{v}||_1$ for another norm?

[warm shale]

Vѳrtєx-

[wraith wing]

if 2 circles have the same circumradius and one angle is the same can we say the 2 circles are congruent??

[hearty terrace]

I'm told it's by the homogeneity property that $||c\bf{w}||=|c|||\bf{w}||$ but i'm not sure how we use it

[worn yoke]

take the norm of both sides

[wraith wing]

specifically if 2 right triangles have the same circumradius

[warm shale]

Vѳrtєx-

[hearty terrace]

oh, and you can take out $\frac{1}{||v||_1}$ from the norm by homogeneity?

[warm shale]

Vѳrtєx-

[worn yoke]

yes

[hearty terrace]

ok coolio thanks!

.close

[obtuse pebble]

Channel closed

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[hearty terrace]

.reopen

[obtuse pebble]

✅

[hearty terrace]

So this is the unit sphere for the 2-norm, $S_1={v\in V | \quad||u||_{\text{2 norm}}=1}$

[warm shale]

Vѳrtєx-

[hearty terrace]

nvm ill use my noggin a lil more

.close

[obtuse pebble]

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[lavish sand]

Theorem:
Let f : V = Rn →Rm =W be a linear map.
1 f is injective if and only if
f (v) = oW ∈ Rm only for v =oV ∈Rn.
2 f is surjective if and only if
rangf = m =dimW

can someone explain this to me especially the first one

[unreal musk]

Injectivity of linear maps can be boiled down to checking if the only thing that gets sent to the zero vector is zero

[lavish sand]

can you give me a simple example

[unreal musk]

(note that the image of zero under any linear map is always zero, injectivity would mean that it's the only thing sent to zero, but that's not only necessary for injectivity, but sufficient: as soon as you know that, you know it's injective)

And e.g. take multiplication by an invertible matrix, send v to Av, with A invertible

Then the only solution to Av = 0 is v = 0 (multiply both sides by the inverse of A)

[lavish sand]

.closed

.solved

[obtuse pebble]

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[silent dirge]

I cant wrap my head around how to represent the function for every week. If someone could explain that would be amazingg!!

[thorny idol]

Notice that increasing by 7% is the same as multiplying by 1.07

It starts with 150, after one day, it becomes $150 \cdot 1.07$, after two days $150 \cdot 1.07 \cdot 1.07$, after theee days $150 \cdot 1.07 \cdot 1.07 \cdot 1.07$. So in general, after d days you will have $150 \cdot 1.07^d$

if you know $w = 7d$ can you write that function in terms of weeks?

[warm shale]

kaue

[silent dirge]

the rate^7?

[thorny idol]

You can solve for d in $w = 7d$ and replace it into the function

[warm shale]

kaue

[silent dirge]

w/7

one week

so 1/7

[thorny idol]

If a week has 7 days, then a day is one seventh of a week. si

$150 \cdot (1.07^d) = 150 \cdot (1.07^{w/7})$

[warm shale]

kaue

[silent dirge]

so if it grows 7% a day, then it grows 7%^7 in a week right

[thorny idol]

if it is multiplied by 1.07 every day, then it is multiplied by 1.07^7 in a week

but you cant really say it grows 7%^7 a week because a grownth of 7%^7 is actually smaller than a grownth of 7%

[silent dirge]

so the function would be

150(1.07^7)^t

[thorny idol]

No, $1.07^{w/7} = (1.07^{1/7})^{w}$

[warm shale]

kaue

[thorny idol]

a grownth of 7% in a day means a grownth of 60.578147647843...% in a week

because 1.07^(1/7) = 1,6057...

cuz a day is a seventh of a week

[silent dirge]

so

150(1+0.07^7)

1+0.07^7 is the rate

no

could it be 1+0.07^7

[thorny idol]

a day is a seventh of a week, so its to the power of a seventh not to the power of 7

[silent dirge]

does the 1/7 apply to the initial value as well

[thorny idol]

wym

[silent dirge]

would it b 1.07^1/7 or 1+0.07^1/7

[thorny idol]

and $1,07^{1/7} \ne 1 + 0,07^{1/7}$ btw

[warm shale]

kaue

[thorny idol]

its $1.07^{1/7}$ which is $1 + 0,605...$

[warm shale]

kaue

[thorny idol]

$f(t) = 150 \cdot (1.07^{1/7})^t$

[warm shale]

kaue

[thorny idol]

because $(1.07)^d = (1.07)^{w/7} = ((1.07)^{1/7})^{w}$

[warm shale]

kaue

[silent dirge]

ohh

like the half life formula

t/2

okok

so it would be b

thanks!!

.close

[obtuse pebble]

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[daring zodiac]

how do i go on about solving this?

[worn yoke]

are there any values of x in particular where you would suspect it's discontinuous?

[daring zodiac]

am i correct for this:
f(2)=(2^3)-2c=8-2c

[worn yoke]

yes

[daring zodiac]

and:
lim as x approaches 2 from left side of f(x) = 4c+4

and from the right side is 8-2c

[worn yoke]

yes

[daring zodiac]

so what do i do after that

[worn yoke]

the limit must exist and equal the value of the function

[sage zinc]

and the limit exists if both left and right side equal to each other

[daring zodiac]

does that mean this is discontinuous

[worn yoke]

for the limit to exist, both sided limits must be equal

you have both sided limits. how can we make them equal?

[daring zodiac]

im not sure

[worn yoke]

just set them equal

then you have an equality which can be solved

[daring zodiac]

so we solve to find c?

c is 2

[worn yoke]

yes

[daring zodiac]

thats it?

[worn yoke]

well we need to make sure you actually solved for c correctly

[daring zodiac]

so we just plug c into those limits?

[worn yoke]

that would be a good idea

[daring zodiac]

alright

ok so i got 12 for left side and 4 for right side

so only true for right side @worn yoke?

[worn yoke]

that means you solved for c incorrectly

[daring zodiac]

c is actually 2/3

ok now i have 20/3 from both sides @worn yoke

[worn yoke]

ok now we have solved for c successfully

[daring zodiac]

so thats it?

[obtuse pebble]

@daring zodiac Has your question been resolved?

[worn yoke]

yes

[obtuse pebble]

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[swift solstice]

heyy

[swift solstice]

can anyone explain to me the method the finding the product of prime factors of a number

ik it's simple but i rlly dnt have a clue cl

[rugged lava]

that would be the number itself

every composite number can be written as the product of primes

I think you might mean something else

[swift solstice]

oh wait then how to write a number as a product of its prime factors?

[hasty vessel]

take the number and divide it by prime numbers until you get 1

[rugged lava]

divide it with prime numbers

yeah

[swift solstice]

can you show me an example plz

[rugged lava]

for example

12 can be divided by 2

which gives us 6

6 can be divided by 3, which gives us 2

[hasty vessel]

in my country we draw a long vertical line and put the number on the left and on the right the prime number we divide it by

[rugged lava]

2 can only be divided by 2, which gives us one

now we multiply all the quotients we got

2, 3, 2

2*3*2=12

[rugged lava]

yea

[swift solstice]

ahhh okay imma try

okay so for 98 for example

[rugged lava]

something like this

[hasty vessel]

yea like that

i didn't think of screenshotting and sending

[swift solstice]

OOHH i actually get thatt

[hasty vessel]

it's a neat way to do it

[swift solstice]

TYSM u both

[hasty vessel]

without occupying too much space

[swift solstice]

😊

[hasty vessel]

no worries

[swift solstice]

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[rugged lava]

useful technique lol

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[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[onyx belfry]

The eigenvalues of this matrix should be $\lambda=-2, \lambda = 9 $ and $\lambda = \alpha^2$

[warm shale]

cristorenzo99

[onyx belfry]

So you should put $\alpha^2 = -2$ and $\alpha^2 = 9$ to have a "double eigenvalue". This gives you $\alpha = -3$ and $\alpha = 3$.

[warm shale]

cristorenzo99

[onyx belfry]

Since the other eigenvalue has multiplicity = 1, the algebraic multiplicity for this one will be equal to his geometric multiplicity

So you have to study the subspace $V_3= \ker (A-3Id)$ and $V_{-3} = \ker(A+3Id)$

[warm shale]

cristorenzo99