help-23

[safe radish]

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[wooden oyster]

question regarding limits

[wooden oyster]

when something is going towards pi

which way does it go from if it is pi^-

Does it go from 0 and out, or from the maximum value of pi

does it follow the green or black line

where the black starts at -2pi and pi and green starts at 0

[rapid igloo]

green

pi^- means you're approaching pi from the left

[wooden oyster]

so from 0 to -pi

[rapid igloo]

you're approaching pi

not -pi

[wooden oyster]

oh

so in the minus one you will be below the line

in the + one above the line

[rapid igloo]

nope

the other way around

a - in the exponent means you're approaching the value from the left

[wooden oyster]

if i come from the left

[rapid igloo]

you're above the x axis

[wooden oyster]

why is minus the left

[rapid igloo]

because it's related to the y values

[wooden oyster]

so approaching from the left means going right

[rapid igloo]

yes

[wooden oyster]

that shoud've been called right thne

[rapid igloo]

well no

[wooden oyster]

alright guess thats why it didnt make sense for me

[rapid igloo]

because you're going from left to right

[wooden oyster]

but i am going right

[rapid igloo]

so you're starting from the left

[wooden oyster]

i know

now

but it should still be called right

[rapid igloo]

uh no

[wooden oyster]

which way am i going?

i am going right

[rapid igloo]

but you're starting from the left

x -> pi^- means x -> pi, x < pi

[wooden oyster]

i'll accept that it is like that

but for me it meant the other thing

so

[rapid igloo]

yep

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[wintry sigil]

What are coincident points?

[lean otter]

2 points that are in the same place

So take this image I found

There is a point (x, y) on line red that is in the same exact place as (x, y)- they're pretty much the same point

Therefore they are coincident

Similarly coincident lines are just 2 lines that have the same exact values

[wintry sigil]

😮

Ok thanks

.close

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[nova junco]

hey

[nova junco]

y=(x-2)^3-5(x-2)^2+6(x-2)

how do I find the roots ?

with an easier way

[safe radish]

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

[nova junco]

I did

[untold topaz]

What did u try?

[nova junco]

I just expanded but Like

I want to find an easier way

[untold topaz]

What did u get after expanding

[nova junco]

y = x^3 - 11x^2 + 38x - 40

[untold topaz]

What abt the roots?

[nova junco]

x=2,x=5,x=4

[untold topaz]

I guess if u want to

U can reduce it to a quadratic eq

[nova junco]

how?

ohhh

now i remember

we make x-2=u

then we form a quadrati

ok thx

[untold topaz]

Uh not sure where that'd take u but good luck

[granite idol]

it wouldn't give you a quadratic but it would simplify it a bit I guess

[untold topaz]

What's the method called again?
For factorising cubic eq

Nvm I got it

Use Factor Theorem @nova junco

This is how u reduce a cubic eq to a quadratic eq

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[viscid grail]

Why does -5=5(cos(pi)+isin(pi))
Meaning how does pi happen? ik that -5=|-5|=5

[icy lance]

-5 doesnt = 5

but if you evaluate the trig, cos(pi)=-1 and sin(pi)=0

so you have -5=5(-1+0)

[icy lance]

if you visualise it in the complex plane, the argument is pi radians and the magnitude is 5 which is why you get this

[viscid grail]

I still don't understand, how did 5 make pi

pi is 180 degrees no?

[icy lance]

yeah

[viscid grail]

how did this happen

[icy lance]

because its -5

not 5

[viscid grail]

I uh

I still don't understand

[stray socket]

what is cos(π)

Or are you asking why π is showing up

[viscid grail]

it's this

[stray socket]

I'm asking you the value of cos(π)

[viscid grail]

yeah like how was it measured or how tf did it happen

I don't even know

[icy lance]

its a rotation anticlockwise from the positive real axis

[viscid grail]

-1

[stray socket]

If you visualize a complex number, z = a + bi, you can visualize a vector <a, b> on the complex plane that makes an angle with the positive x axis

[viscid grail]

yea

[stray socket]

And sin(pi) is 0 so that's how it's -5. The π is telling that z is antiparallel to the positive real axis

Not x* axis, complex planes don't have an x axis

[viscid grail]

man

idk man I still don't understand this

[stray socket]

Do you know what the positive real axis is

[viscid grail]

yeah

for x and y

right?

for y from 0 to ^

for x from 0 to >

[stray socket]

On the complex plane

The real axis is the horizontal, the imaginary is the vertical

[viscid grail]

that's clear yes

[stray socket]

So what's confusing you then?

Like the overall concept?

[viscid grail]

the calculation of how pi happened

like idk why for example for 4 it's cos0, isin0

for 6i is cospi/2 isinpi/2

[stray socket]

because of the trig

like for -5 = 5 cis(pi)

cos(pi) = -1
i sin(pi) = 0i

5(-1 + 0i) = -5

its legit just trig

[viscid grail]

I guess this is from z=r(cos(x)+isin(x))

[stray socket]

it is

[viscid grail]

is it safe to say that I can distribute r like r(cos(x)+isin(x))=rcos(x)+r(isin(x))?

[stray socket]

yes

[viscid grail]

for z can be a+bi

[stray socket]

yes

in fact, a = rcos(x), b = rsin(x)

[viscid grail]

so, say z=a+bi=6i

oh also |z|=r

right?

[stray socket]

yes

[viscid grail]

and |z|=|a+bi|

[stray socket]

well uh

i mean loose notation

but if the verical bars refer to magnitude, then sure

[viscid grail]

right so |6i|=6

[stray socket]

yes

[viscid grail]

but why cos of 90 degrees

wait

[stray socket]

becase 6i lies directly on the positive imaginary axism whose vector is <0,6>, which is 90 degrees from the positive real axis

[viscid grail]

cos(x)=a/r

where a in this is 6?

[stray socket]

a = 0

[viscid grail]

oh

yeah

[stray socket]

6i = 0 + 6i

this implies a = 0 and b = 6

[viscid grail]

ohh yeah

[stray socket]

but you can legit think of complex numbers as position vectors

[viscid grail]

ngl idk how to measure the angle if it's something random

[stray socket]

i mean

[viscid grail]

if it's for 3 I'm flopping

100%

[stray socket]

protractor lol

theyll usually give you the ang;e

[viscid grail]

then wtf does it have to do with z and r

man

idk

[stray socket]

the literal relationship is right there

in the equation

[viscid grail]

yeah I get that

[stray socket]

like unless they give you z

and not the ang;e

then thats just trig

because lets say z = 4 + 3i

then find the angle it makes with the positive real axis

so you'd say:

5 cos(t) = 4
5 sin(t) = 3

[viscid grail]

oh

[stray socket]

like just compare terms

if you wanna go the vector route

<1,0> . <4,3> = (1)(5) cos(t)

[viscid grail]

yeah I got what I asked for

ty @stray socket

.close

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[stray socket]

np

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[ashen swan]

I don’t get how they simplified it

please help

oh

nvm

.close

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[ashen swan]

.reopen

[safe radish]

✅

[ashen swan]

why is it that

In this question

x/x^2 = 1

but for this qs

x/x^2= 1/x

??

[quiet juniper]

It's not tho

1/\sqrt{x^2} = 1/x

2x/\sqrt{x^2} = 2

[ashen swan]

like for here

[south epoch]

there is a square root on the x^2 so it becomes x and x/x is 1.

[ashen swan]

the denominator

[quiet juniper]

Ye ur reading it wrong

[ashen swan]

is 1-1/x^2

[ashen swan]

please explain :’>

[quiet juniper]

I'm referring to the second denominator here

[quiet juniper]

And here

[ashen swan]

?

[quiet juniper]

The denominator in ur second picture

U think it says x/\sqrt{x^2} = 1/x but it's not saying that at all

I'm pointing out what it actually says (and it's right)

[ashen swan]

ohhh

that just made things more complicated

[quiet juniper]

oh ye they might've screwed up in the first Pic if they meant to put 1/x^2 in the denominator

[ashen swan]

no yea I’m confused asf

😭

[quiet juniper]

It should just say 1 - 1/x

[ashen swan]

[quiet juniper]

In this instance putting 1 - 1/x^2 still gets u the right answer tho

[ashen swan]

could we like maybe

[quiet juniper]

But the logic isn't right

[ashen swan]

get on a call and like you could like screen share

and explain

[quiet juniper]

can't srry

[ashen swan]

💔

so does that mean

sqrt/\x / x^2 = 1/x

and x/ sqrt x^2 = 1?

[quiet juniper]

$\sqrt{x}/\sqrt{x^2} = \sqrt{1/x}$

[flat frigate]

992qqoloy

[quiet juniper]

It's just how radicals work

$\sqrt{ab} = \sqrt{a}\sqrt{b}$ if a and b are any two numbers

[flat frigate]

992qqoloy

[ashen swan]

what about

x/ sqrt x^2

?

[quiet juniper]

Well that's just properties of powers

$(x^a)^b= x^{ab}$

[flat frigate]

992qqoloy

[quiet juniper]

And $x^{a - b} = \frac{x^a}{x^b}$

[flat frigate]

992qqoloy

[ashen swan]

[quiet juniper]

$\sqrt{x^2} = (x^2)^\frac{1}{2} = x^{2*\frac{1}{2}} = x$

[flat frigate]

992qqoloy

[quiet juniper]

then $x/x =... 1$

[flat frigate]

992qqoloy

[ashen swan]

but

it’s

a sqrt

why is the

OHHHHH

[quiet juniper]

This is algebra 2 stuff my dude

[ashen swan]

sqrt of x^2

the sqrt

[quiet juniper]

ye

[ashen swan]

gets canceled cuz of

ye

ok

ye

I get it

[quiet juniper]

[ashen swan]

ty amigo

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[wise schooner]

does it matter if I find the inverse first and then multiply by -5

or multiply by -5 then find the inverse

or is it all the same

[peak estuary]

there is a difference between 5M^-1 and (5M)^-1 =5^-1 M^-1 = 1/5 M^-1

first find inverse

[wise schooner]

right, thanks.

.close

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[indigo tundra]

Is this correct

Oh wait

I need to switch the direction

.close

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[hardy sky]

I am watching a video on how to convert into vertex form, and I am confused on one of the steps he took, where he multiplies one side, 5/6, by 2 each, and 7/1, by 12 each, why does he do that?

[hardy sky]

he doesnt explain in the video

<@&286206848099549185>

[safe radish]

@hardy sky Has your question been resolved?

[hardy sky]

someone help....

[karmic falcon]

Not a helper but I could help explain if you want

[hardy sky]

okay

thank u

[karmic falcon]

He's multiplying the fractions by 1 each to keep their values the same, but their denominators change

When their denominators are the same (12 in this case) he can subtract the two fractions from each other

[hardy sky]

why doesnt he multiply the right fraction by 6?

or does he have to change the left fraction as well

[karmic falcon]

Its because of the leftmost fraction

the leftmost fraction simplifies into 25/12

so we already have a 12 in one of the denominators

we might as well change the other two fractions to also have 12s in the denominators to save time

then we can subtract all three fractions more easily

[hardy sky]

oh i see, so 3(25/36), just simplified into -25/12, and he mutliplied the other fraction each by 12 to have the same denominator?

[karmic falcon]

yep

[hardy sky]

no, he did 5(-5/6), and got -25/6, im still confused lol

[karmic falcon]

its (-5/6)^2 then multiplied by 3

[hardy sky]

so hes trying to make the -25/6, a 12 denominator so that it matches the 3(25/36)?

[karmic falcon]

So (-5/6)^2 becomes (5^2/6^2) which becomes (25/36) [no negative 25 as (-5)^2=25]

so we have a positive (25/36) which is multiplied by 3

[karmic falcon]

Yes- we need the denominators to match for simple fractional addition/subtraction

[hardy sky]

okay i get it i think, so he has 3 fractions, and the leftmost has a denominator of 12, and hes trying to make the rest 12 denomator as well?

[karmic falcon]

Yes

[hardy sky]

that way its easier to evaluate

[karmic falcon]

Yeah

[hardy sky]

okay i get it now, thank you so much

[karmic falcon]

If we dont change the denominators, doing the subtraction would be very messy

[hardy sky]

yea i see

[karmic falcon]

no problem

[hardy sky]

i think i just overlooked there being 3 fractions, as i always work with just 2

so it looked weird

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[lean otter]

f(x)=(x+1)/(2x-5)

[lean otter]

trying to find the inverse of the above function

got to y=2xy-5x-1 but that was wrong

here are the steps I followed:

x=y+1/2y-5

[quasi bison]

yeah cause it is not the final answer lol

[lean otter]

It's not?

[quasi bison]

it's not

you have not isolated y

[lean otter]

y is isolated though..

[quasi bison]

no it isn't

it appears once more on the right, which it isn't supposed to

your work up until that step has an even number of mistakes (possibly 0) that cancel each other out

[lean otter]

oh wait, there is a merge

alright, here is how I'm trying to work it out

(y+1)=x(2y-5)

If I could somehow isolate the x from either value

could I shift y+1 and set the equation to 0?

Yeah I'm not getting it

<@&286206848099549185>

having some trouble finding f^-1 of: f(x)=(x+1)/(2x-5)

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[lean otter]

I’m confused on where to start

[upbeat ridge]

What type of equation should you be using?

[lean otter]

Ce^kt

[upbeat ridge]

Right

What does C mean

Or what should you put in C?

[lean otter]

C is the 900 bacteria?

That’s what I got and I got stuck

[upbeat ridge]

Why did you divide 900?

Ohh wait

Hold on

You sure it's 15k?

[lean otter]

I’m not sure

Idk if i did any of it right

How should I initially set up this problem

Hello?

.close

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[fleet pine]

why is there a minus there?

[tall bough]

Coz of the usub

[fleet pine]

ah i see

thank you

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[granite cape]

$$\int _{ }^{ }\left(1+4x\right)\sqrt{1+2x+4x^2}dx$$

[flat frigate]

Lex1729

[granite cape]

u = 1 + 2x + 4x^2

du = 2 + 8x dx

dx = (du)/(2+8x)

$$\int _{ }^{ }\left(1+4x\right)u^{\frac{1}{2}}\cdot \frac{du}{2+8x}$$

[flat frigate]

Lex1729

[granite cape]

Can someone explain to be why the x in the "4x" term and the x in the "8x" term don't cancel out?

[fluid token]

you can, but you ahve to a bit of algebra first

take out the common factor from 2 + 8x

[granite cape]

Oh ok, thanks.

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[stray palm]

Here is the second statement p?

Like do I assume x and y are even?

[spring wasp]

let x = 2k and y = 2m

[quiet juniper]

maybe could prove it by splitting up into cases yeah

[stray palm]

So you can assume then?

Ya

[tidal imp]

You can do casework, but you don’t need to for this one

[stray palm]

How else would you do it?

[tidal imp]

Try squaring (x+y)

(Of course you start by stating “assume x+y is even, let x+y=2k for some integer k”)

[stray palm]

You get xy though?

So how does the xy affect it

[tidal imp]

What’s the square of x+y first?

[stray palm]

X^2 +2xy +y^2

[tidal imp]

What’s the parity of 2xy then?

[stray palm]

Oh xy is an integer

So it has the form even

[tidal imp]

Yeah

Done with one side

[stray palm]

So x^2 + y^2 = -2xy?

[tidal imp]

No

[stray palm]

Why not

[tidal imp]

I forgot to add initially, but you do have to represent x+y by something

It’s not necessarily zero

[stray palm]

Wdym

[tidal imp]

You have to use the assumption that x+y is even

[stray palm]

Hm

0 is even though no?

But I guess it can be any even

So

I can do x + y = 2j for some integer j

[tidal imp]

Yes

[stray palm]

Then square both sides?

[steep frigate]

Yes

[stray palm]

How does this look

[tidal imp]

We don’t need x and y to be even here

All the proof needs is for x+y to be even

It doesn’t matter if x and y are even or x and y are odd, still works the same

However

This is only one side

We proved <-

Now we need ->

[stray palm]

I’m confused

Isn’t this just cases?

[tidal imp]

About what

[stray palm]

I thought by doing it this way it is t cases and we only need to show this

Cause we squared x + y = 2j while assuming x + y is even

So then we can solve for x^2 + y^2

And we showed that it’s subtraction of two even integers so it’s even?

[stray palm]

What else do we need to do

[tidal imp]

This is sufficient already (with some polishing)

But again, note this is an if and only if problem

Wait actually you forgot to put the conclusion that x^2+y^2 is even

Probably should put that

[stray palm]

Yeah I did

[stray palm]

So that makes this insufficient?

Now I need to show that x^2 + y^2 is not even if x + y is odd, right?

[steep frigate]

Yes

[tidal imp]

Yes

This is the best way to do this

So essentially you’re going to repeat the argument, except x+y is odd this time

[stray palm]

Ya

[tidal imp]

And you know that’s fine because the contrapositive is equivalent to the original statement

[stray palm]

How does this look?

[tidal imp]

Hold on why is it 4j+1

[stray palm]

Oh I meant 2j+ 1

[stray palm]

Lemme fix

Doesn’t this show even?

So it proves false?

[tidal imp]

+2?

[stray palm]

Ya

Unless I failed a

O

Lol

I failed algebra

[tidal imp]

1x1=2 folks

[stray palm]

Whoops

[stray palm]

Real

[steep frigate]

https://tenor.com/view/its-completely-foolproof-joshua-weissman-foolproof-infallible-idiot-proof-gif-18351329

W

[stray palm]

@tidal imp

This better?

[steep frigate]

Its better

good job young man

[tidal imp]

8j^2?

[stray palm]

Forgot to change that

[tidal imp]

Yeah just polish the two proofs but the idea is there

Just minor details at this point

[stray palm]

Alright, thank you

[tidal imp]

gl professor frankly

[stray palm]

<3

.close

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[amber orchid]

Im stuck on 9c, ive tried using a tree diagram to find all the possibilities but it looks like im missing something

[amber orchid]

I think im on the right track, but not sure what to do

[gentle herald]

Find how many total choices r there

[amber orchid]

That would be 2 + 9 + 18= 36.
Is it that simple?

[gentle herald]

How r u getting 2+9+18

[amber orchid]

2 options, soup and salad. Then there are 6 options of mains after that. Then 18 different path for deserts

If its not 36 then most likely 18

[safe radish]

@amber orchid Has your question been resolved?

[amber orchid]

Ive understood that there are 36 options, but what does it mean by all three courses

.closeudu

Clos

.close

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[steep gust]

how i can solve this ?

[thin bridge]

do you know your exponent laws

[steep gust]

ye

[steep gust]

but maybe idk

[thin bridge]

which one is it,
yes or don't know

[steep gust]

sum?

[thin bridge]

wdym

[thin bridge]

that law will be useful here, yes

[steep gust]

i see thx

.close

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[granite idol]

looks like it might be asking you to find elementary matrices?

[wise schooner]

I tried getting A to reduced row echelon form then doing those same operations on a 3x3 identity matrix but its wrong so idk

[granite idol]

each step in row reduction would give you a single elementary matrix. then you take the product of all of them

[safe radish]

@wise schooner Has your question been resolved?

[safe radish]

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[languid plover]

hello~

[languid plover]

can anyone help me

[quiet juniper]

no one can ur on ur own , it's a doggy dog world

[languid plover]

this is a little bit about physics and math at the same time

are you ready?

[quiet juniper]

I never am

[languid plover]

it might be easy for you

but I just don't understand this question

[quiet juniper]

okie

[languid plover]

so what they are saying is

I need to get the acceleration

so what I did is to calculate it to get final velocity

and acceleration = delta V/T

right?

[quiet juniper]

Assuming velocity increases linearly with respect to time that should work

[severe pond]

acceleration should be zero tho

[languid plover]

that is what I thought

[severe pond]

since the velocity is constant

[languid plover]

since it is a linear

[severe pond]

yea x vs t is linear so velocity is constant which means it isn’t accelerating

[languid plover]

but for some reason the book shows the answer is 0.25 cm/s2

[severe pond]

the derivative(slope) is 2.5

[languid plover]

that is velocity tho

right?

[severe pond]

yea

the acceleration is zero

[languid plover]

but again but book shows the answer is 0.25cm/s2

for some reason

[quiet juniper]

book dumb

maybe check the errata

[languid plover]

is it somehow related to the time?

like

it is not constant

[severe pond]

if u have some linear distance function it will be of the form y=mx+b where m is the velocity so if u find the second derivative it’s always zero

the book is wrong

if u use the data it’s increasing at a constant rate too

[languid plover]

I know but

please see the time

X axis

it is not constant

[severe pond]

wdym?

[languid plover]

I just wanna know if it makes a difference

[quiet juniper]

That just means the rate at which they sampled points is not constant

[languid plover]

like the inverval

between each time

[severe pond]

the graph models the points accurately

the books answer is just wrong

[languid plover]

it doesn't go up like 0.1, 0.2, 0.3

[severe pond]

yea it goes up by 0.05

it’s irrelevant

[languid plover]

oh shoot

guys sorry

[severe pond]

u can scale the x axis however u would like

[languid plover]

the x axies is wrong

[severe pond]

is it t^2

[languid plover]

you gotta see the time in the table

[severe pond]

lol

[languid plover]

yes

[quiet juniper]

t axis is fine

[languid plover]

it is t^2

[quiet juniper]

Yeah

But the axis has them spaced apart properly

[severe pond]

but the table says t in seconds

[languid plover]

ohhmm

[quiet juniper]

it's irrelevant to the acceleration tho

[languid plover]

yes

seconds

okay

so basically

acceleration should be 0?

[severe pond]

yes it’s zero

[languid plover]

hold on imma capture the question from the book

and show you guys

[severe pond]

ok

[languid plover]

this is it

[severe pond]

the graph is wrong

this data is different

[languid plover]

I made the d vs t graph based on d.

[severe pond]

there is an acceleration for this data

velocity is changing

[languid plover]

but if you see the question e.

use the second graph

which I made for the question d.

[severe pond]

oh u started from part d

[languid plover]

it says make a second one

with

t^2

so it has a linear line

what is what I did

did I do something wrong??

miss something?

[severe pond]

yea yea ur graph had it as t

when the x axis should be t^2

[languid plover]

yeah

[severe pond]

u just labeled it wrong

[languid plover]

oops

imma change it

[severe pond]

this means there’s constant acceleration

[languid plover]

really?

[severe pond]

acceleration should 2.5 cm/s^2

which is 0.025 m/s^2

[severe pond]

yes notice that acceleration is defined as change in velocity over time which translates to (m/s)/s which is just m/s^2

[languid plover]

waittttt

[severe pond]

since the graph is linear

[languid plover]

isn't 2.5 velocity???

[severe pond]

no

2.5 cm/s^2

is acceleration

velocity changes

depending on the time measured

[languid plover]

omg

[severe pond]

yea

[languid plover]

wait

what equation did you use?

[severe pond]

to find the acceleration?

[languid plover]

yes

I got 2 options

[severe pond]

just do the slope using the points

[languid plover]

and I wanna know if they both have the same answer

[severe pond]

change in position divided by change in time squared

[languid plover]

but if I do that

[severe pond]

u can use any of the points

take the first two

[languid plover]

d/t

is v

isn't it?

[severe pond]

(0.025-0)/(0.100-0)=2.5 cm/s^2

[severe pond]

the table is d/t^2

not seconds

[languid plover]

huh???

[severe pond]

seconds squared

[languid plover]

hold on

[severe pond]

look at the units

[languid plover]

are you saying I should use the original table?

[severe pond]

or wait nevermind

i was looking at that table

hold on

[languid plover]

mhm

[severe pond]

i’ll use the time squared table

[languid plover]

yes

that is the second graph

as the question e says

[severe pond]

no yea i was right

it’s 2.5

use ur graph

[languid plover]

so

[severe pond]

(0.025)/(0.01)

2.5 cm/s^2

[languid plover]

Vf-Vi/Ttotal = a

right?

[severe pond]

=0.025 m/s^2

yea

change in velocity over time

u can use any of the points

[languid plover]

0.625/0.25 = v

[severe pond]

since the graph is linear

yea

and u get 2.5

[languid plover]

2.5cm/s

[severe pond]

2.5cm/s^2

not/a

[languid plover]

2.5-0/0.25

[severe pond]

it’s seconds squared

[languid plover]

I don't get "seconds squared"

what does that mean?

[severe pond]

it means

u divide by time twice

look

velocity is

[languid plover]

d/t

[severe pond]

distance/ time right

yes

[languid plover]

yes

[severe pond]

and u know

acceleration is v/t

right

well we know

[languid plover]

yes

[severe pond]

v=d/t

so plugging d/t in for v

[languid plover]

mhm

2.5cm/s

[severe pond]

we get a= (d/t)/t

which is d/t^2

[languid plover]

ah~

so (d/t)/t = (2.5cm/s)/t

right?

[severe pond]

no.

[languid plover]

whyyyyyyyy

[severe pond]

because look at the question

[languid plover]

yes

[severe pond]

the question asks u to graph distance vs time squares

ao

so

u made that

but u didn’t label the x axis as time squared

[languid plover]

waitttttttt

so whichmeans

[severe pond]

look at part d lol

[languid plover]

it is already calculated

?

[severe pond]

it says t^2

u already did the calculations for t^2

[languid plover]

yess

[severe pond]

0.1^2= 0.01

[languid plover]

mhm

[severe pond]

0.2^2= 0.04

see that’s how u got the data on the x axis

in ur new table

it’s time squared

[languid plover]

so d/t is already a

cuz the t is actually t^2

[severe pond]

yes

[languid plover]

omgggggg

[severe pond]

u just labeled wrong

[languid plover]

you are a genius

[severe pond]

all u brother

[languid plover]

https://tenor.com/view/crycat-crying-cat-crying-cat-thumbs-up-thumbs-up-ok-gif-22851318

but I still don't get why the answer is 0.25 cm/s2 on the book

[safe radish]

@languid plover Has your question been resolved?

[safe radish]

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[lean otter]

How do I do part ii?

I attempted it

[plucky elk]

Solve for v(t)

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

Can you please look at my working out

I found an expression for time in terms of v

Is it right?

<@&286206848099549185>

[dry agate]

Hm...

I'm in 8TH class bruh

[little mesa]

nice

[dry agate]

[little mesa]

yeah it looks ok, now just find the time taken

[dry agate]

Interesting

[lean otter]

Nvm

I got it

.close

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[wispy grail]

yo boys

[wispy grail]

36 x 6^-2

Do I only multiply 36 with the numenrator or whatevr it is called?

Because to solve it I make the question into: 36 x 1/6^2

[white swallow]

Denominator

[devout shale]

That is how you multiply it

[white swallow]

Yes

[devout shale]

What you wrote it correct

36 is 36/1

So multiply numerator with numerator and denominator with denominator

[wispy grail]

what bro

multiply 36 with 1 and multiple it with 6^2?

[devout shale]

36/1 * 1/6^2

Multiply the numerator of both fractions

[wispy grail]

oh

makes sense

[devout shale]

And multiply the denominator of both fractions

That is the result

[wispy grail]

ty i wanna be a nerd

I need to learn much math

[devout shale]

You can be good at math without being a nerd

Good luck

[wispy grail]

🙂

[wispy grail]

Being nerd is good tho

ok I will close ty guys

.close

[safe radish]

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[coral bridge]

Using a function that you’ll chose, prove that for all x in ]0;1[, x^x * (1-x)^(1-x) >= 1/2.
I know that we have to use f(x) = x^x so we have f(x) * f(1-x), but I don’t know how to proceed

[stable inlet]

differentiate x^x * (1-x)^(1-x)

[safe radish]

@coral bridge Has your question been resolved?

[coral bridge]

Thx I juste messed up the first time I did it lol

.close

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[prisma hamlet]

can someone prove a) for cases (i) and (ii) for me

[prisma hamlet]

I think I understand what it means, I've just got no nidea how to prove it

<@&286206848099549185> sorry to disturb, but I really need help with this exercise

I can't seem to be able to do any of them

[obtuse prawn]

Hello in which metric you are working

[prisma hamlet]

what do you mean metric?

[obtuse prawn]

The segment is in what space

[prisma hamlet]

I think I'm supposed to say two dimensional?

[obtuse prawn]

Let AB Belong to your space where A and B are the endpoints of the segment. AB={(x,y)/x=(1-t)x1+tx2 y=(1-t)y1 + ty2 , t belong to [0,1]

We apply y to F by translating we mean moving all point of this segment then A’B’ = {(x’,y’)/x’=x+a y’=y+b , x y belong AB

Substitute x and y

Arrange the variable and t to get the Shape of a line which is the same as AB

[prisma hamlet]

wait hold on

can you explain to me what you did with (1-t)?

[obtuse prawn]

Try to plot it on a paper it will make more sense

[prisma hamlet]

in this book, it is stated that any point on the segment A, B is (1 -t)A + tB

but with x's and y's, it got very confusing

[obtuse prawn]

Perfect

A segment have two point at the edges

Yes ?

[prisma hamlet]

yes

let's call them P, Q instead?

[obtuse prawn]

Ok so we set x and y that are the edges of the segment

Ok PQ

[prisma hamlet]

x and y's are kind of confusing me

[obtuse prawn]

I’m gonna write it on a paper

Hold on

[prisma hamlet]

alright

[prisma hamlet]

I'm checking it

a minute please

so alright, could you explain to me what the x and y's mean exactly?

I understand they are coordinates, of what?

A, B?

or is it x1 and x2 I'm seeing?

[obtuse prawn]

So x could be anywhere in your space

Yes ?

By x I mean a point

Any point

[prisma hamlet]

alright

[obtuse prawn]

How can you write this

[prisma hamlet]

so x1 and y1 for A, x2 and y2 for B?

[obtuse prawn]

No

X1 and x2 are for x and same for y ….. x and y are coordinate of each point between A and B

[prisma hamlet]

I think I understand

[obtuse prawn]

Can you send me the book you are studying in ?

[prisma hamlet]

the whole book?

[obtuse prawn]

Or tell me it’s name

Yes

[prisma hamlet]

it's Serge Lang's Basic Mathematics

[obtuse prawn]

Okay thank you

[prisma hamlet]

thanks for the help

.close

[safe radish]

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[crude bear]

If $I_n = \left{ x \in \mathbb{N} | 1\leq x \leq n \right}$

[flat frigate]

tales

[crude bear]

Theorem :
If there is a bijection between $A\subset I_n$ and $I_n$, then $A = I_n$

[flat frigate]

tales

[crude bear]

The book does induction over $n$

[flat frigate]

tales

[crude bear]

If $n=1$, then $I_n = \left{1\right}$, so the only possible subsets are the empty set and the set $\left{1\right}$

[flat frigate]

tales

[crude bear]

There is no bijection between empty set and nonempty set

so we have $A = \left{1\right} = I_n$

[flat frigate]

tales

[crude bear]

Supposing it is valid for some natural n

We take $A \subset I_{n+1}$ with a bijection with $I_{n+1}$

[flat frigate]

tales

[crude bear]

Say $f: I_{n+1}\to A$

[flat frigate]

tales

[crude bear]

We restrict f to $I_n$, getting $f':I_n\to A-\left{f(n+1)\right}$

[flat frigate]

tales

[crude bear]

f' is a bijection

So we can use the induction step, getting $A - \left{f(n+1) \right} = I_n$

[flat frigate]

tales

[crude bear]

Now my book says $A = I_{n+1}$ and $f(n+1) = n+1$

[flat frigate]

tales

[crude bear]

Which is not obvious at all

Why is this true?

[safe radish]

@crude bear Has your question been resolved?

[crude bear]

<@&286206848099549185>

[crude bear]

<@&286206848099549185>

[safe radish]

@crude bear Has your question been resolved?

[crude bear]

<@&286206848099549185>

[safe radish]

@crude bear Has your question been resolved?

[crude bear]

.close

[safe radish]

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[safe radish]

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[dire topaz]

How do I find a?

In part b

[safe radish]

@dire topaz Has your question been resolved?

[safe radish]

@dire topaz Has your question been resolved?

[safe radish]

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[buoyant sphinx]

Let $(s_{n})$ be a convergent sequence. Let $s=lims_{n}$
\If $s>a$, then there exists a number $N$ for which $n>N$ implies that $s_{n}>a$.

[flat frigate]

Matt.Rey

[buoyant sphinx]

I'm confused on this proof. I know have an overall definition that the sequence sn converges to s for every epsilon > 0, but I don't really know how to include it.

[split ether]

Try perhaps taking the definition of convergence and letting epsilon be equal to s - a

[buoyant sphinx]

alright. I'll give it a crack

I guess another dumb question would be how do you know to do that just by looking at the problem?

[split ether]

Intuitively the implication is true as s_n should get arbitrarily close to a, so at some point the terms must be greater than a

Since the choice of epsilon gives you the terms which are as close to the limit as you want, you could choose epsilon to be the distance between s and a

So that each s_n is closed to s than a, therefore s_n > a must be concluded

[buoyant sphinx]

so I let epsilon=s-a, so then the new inequality I would get is $|s_{n}-s|<s-a$ but we should make this also $s+a<|s_{n}-s|<s-a$ ?

[flat frigate]

Matt.Rey

[split ether]

Right, now you use the fact that |x| < y implies x < y and -x < y

Meaning here you have s - s_n < s - a

[buoyant sphinx]

awesome. I understand now. Thank you for your patience! Truly appreciate it

.close

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[undone pecan]

Fastest way to simplify this

?*

The end result:

[fiery merlin]

Handle when n is even and when it's odd separately.

Then combine them.

[undone pecan]

I want to simplify the equation before this step.

Or is it considered faster when doing the seperating step first?

[fiery merlin]

I'm not sure which way is faster, it's just that it should simplify quite a bit.

[undone pecan]

Not what I asked for but thanks a lot.

<@&286206848099549185>

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[weary stream]

U know how to calculate
MIN of (X+3)^2/2x ?

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[hard crest]

calculus, find critical points

[weary stream]

I don't know calculus

Am in 9 th grade

[hard crest]

hmm well what domain are you looking at? what are some properties of that expression? does it have any asymptotes?

[weary stream]

X>0

We don't know nothing more

Only x > 0

[hard crest]

i'm not sure how i'd do this without using calculus

[lunar rampart]

it doesn’t have a minimum right?

[hard crest]

i didn't think so at first but it does

[lunar rampart]

its asymptotic to y = 0

[lunar rampart]

itll have turning points but not a minimum

i think

[hard crest]

on x<0 it won't have a min or max, but on x>0 it has a minimum

[lunar rampart]

ah right yeah

[lunar rampart]

youd need to find the derivative

not sure how else to do it

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@weary stream Has your question been resolved?

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[polar birch]

ok so for this one

u want to differiantiate the circumfference of the sphere

radius is circumference divided by 2 pi

so the surface are is 4pi(c/2pi)^2

which reduced to c^2/pi

differentle of ds is 2/pi* C * dc

So when C is 84

and when dc is 0.5

ds is 2pi/pi(84)(0.5)

which comes out to 84/pi