help-17

[paper depot]

do they make you write it as m · s^-2 instead of m/s^2?

[river kettle]

no specific thing is asked

[paper depot]

ok then i think you should get in the habit of writing m/s and m/s^2 despite your book potentially writing it with a negative exponent

if only for visual memory because the slash can be read as 'per' in this case

[river kettle]

sure

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[coarse karma]

.repopen

=
5
9
(
�
−
32
)

The equation above shows how temperature
�
, measured in degrees Fahrenheit, relates to a temperature
�
, measured in degrees Celsius. Based on the equation, which of the following must be true?

A temperature increase of 1 degree Fahrenheit is equivalent to a temperature increase of
5
9
degree Celsius.
A temperature increase of 1 degree Celsius is equivalent to a temperature increase of 1.8 degrees Fahrenheit.
A temperature increase of
5
9
degree Fahrenheit is equivalent to a temperature increase of 1 degree Celsius.
A) I only
B) II only
C) III only
D) I and II only

[coarse karma]

<@&286206848099549185>

[vague root]

what do you need help with

[flat whale]

!15m

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[sage fjord]

Is there anything, I, a rising sophomore (HS), can do to get research in the math field, I understand it is already very late to get research this year. But, I am wondering if there are any research/internship opportunities I can get this summer, even if the chances are low...

pls @ or DM me if there is anything I can do/any opportunities there are

I would really appreciate your help!

If there are any math programs I can do that would also be nice!

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@sage fjord Has your question been resolved?

[sage fjord]

<@&286206848099549185>

can someone help ;/

[flat whale]

Help channels are for math questions. Go to #discussion

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[onyx meadow]

I keep getting 29%

Is this correct

[tulip nebula]

Did you get 28.57 before approximating it to 29 btw?

[onyx meadow]

No

[tulip nebula]

I did it like this, (18cent-14cent)/14cent and it comes out 4/14 whose percentage seems to be 28.57 so I was suspicious maybe that's how you got 29 as well

[onyx meadow]

The way I did it was

Took 800*(0.18) and got 144

800*0.14=112

And used the answer choices

To see how much greater it was

So I did 1.29*112

And that gave me a value closer to 144

I’m just wondering if B would be correct?

[tulip nebula]

How did you get 1.29

[tulip nebula]

I understood upto here

But what is using answer choices

[onyx meadow]

It’s just a short cut for percent increase

[tulip nebula]

Ok

[onyx meadow]

And so it’s asking for percent increase

[tulip nebula]

To find percent increase

I would do (144-112)/112 * 100 tbh

[onyx meadow]

So I did that to see which one would increase 112 to 144

Yeah so is the answer B?

That was my only question

[tulip nebula]

Seems to be

[onyx meadow]

Okay

Thanks

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Nw

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[vast sparrow]

Hello cute little slays

[vast sparrow]

I have a complicated problem; it involves integration with inverse hyperbolic functions

It seemed simple but I’m stuck on like the FIRST step..

[river minnow]

Would recommend multiplying top and bottom by e^-x

Then it becomes $\int\frac{9e^{-x}}{\sqrt{1 + e^{-2x}}}\dd{x}$

[twin meteor]

alonelybean

[river minnow]

Allowing you to make a pretty nice substitution, that is, let some u = e^-x

[vast sparrow]

that's what I was thinking, but I didn't realize that was possible

I guess it raises the question like; would you beable to put an x in an equation as long as you divide by it aswell?

so you can throw in variables into a integral just like you can with numbers? (2/2 vs x/x)

[mild flower]

sure, you can multiply the integrand by x/x if you want

[vast sparrow]

damn wow, I didn't realize it ran that deep

[mild flower]

it's just multiplying by 1

[vast sparrow]

integration is honestly more fun than derivatives idk, I'm pretty new to calc 2, but I got an A in calc 1 with an extremely foreign sounding teacher so tbh

literally that's crazy, first time I figured out you could do that my brain exploded legit

[mild flower]

differentiation is just mechanical, integrals require a brain

[vast sparrow]

lmao yeah literally, integration be slaying everybody in my class

but hopefully I can pull an A, at worst a B

okay well thanks yall!! I appreciate it <3

wait but @river minnow why would you do e^-x? when you could just do e^x

and also HOW does that e^-x get into the radical

[river minnow]

$e^{-x} = \sqrt{e^{-2x}}$

[twin meteor]

alonelybean

[vast sparrow]

oh shit

oh shit the radical IS 1

like its 1/2

wait wtff so what ends up happening

you have du = -e^-x dx

[river minnow]

Yes, so e^-x dx = -du

And then you have $\int\frac9{\sqrt{1 + u^2}}(-\dd{u}) = -9\int\frac1{\sqrt{1 + u^2}}\dd{u}$

[twin meteor]

alonelybean

[river minnow]

If you are familiar with trigonometric substitution, this should be simple

[vast sparrow]

yeah it's new to me, like I learned it a few mins ago lmaoo

it makes sense but this idea of implementing e^x is tripping me oyt

idk why

[river minnow]

Now you can do another substituion, allow u = tan(theta)

What do you get?

[vast sparrow]

1 + tan^2theta wha

in the radical

shit

[river minnow]

Yeah, so 1/sqrt(1 + u^2) becomes just cos(theta)

And du becomes -dtheta/cos^2(theta)

[river minnow]

So this $= 9\int\frac{\dd\theta}{\cos\theta}$

[twin meteor]

alonelybean

[mild flower]

squared

[vast sparrow]

damn why is the theta there like

what does it do in integrals

[river minnow]

cos * 1/cos^2 ?

[mild flower]

oh reading

sorry am really tired and thought i saw a silly error

[vast sparrow]

wait is 1 - tan^2x = cos x

[mild flower]

there's nothing special about theta, it's just a variable

[river minnow]

Generally when you have something like x^2 - a^2 or x^2 + a^2 somewhere in your integrand, trig sub might be the way

[river minnow]

It is fine

[river minnow]

It's 1 + tan^2 = 1/cos^2

[vast sparrow]

damn how am I gonna memorize dozens of these random like

[river minnow]

So sqrt(1 + tan^2) = cos and 1/sqrt(1 + tan^2) = cos

[vast sparrow]

random substitutions

[river minnow]

Practice

[vast sparrow]

mannee

[river minnow]

They are not random

[mild flower]

this one comes from sin^2 + cos^2 = 1 and then you divide both sides by cos^2

[river minnow]

When you get to see how nicely they work, it will be simpler

[vast sparrow]

bruh I got it wrong

im pissed off im so dumb

How am I wrong

[river minnow]

What did you do after this?

[river minnow]

Btw the +C should be outside

[vast sparrow]

damn

wasnt paying attention

wait

it was still wrong :(

[river minnow]

.

[vast sparrow]

wait how

oh

wait you have to do the tangent thing?!

I thought it was just an option

[river minnow]

Hm, well

There could be an alternative way involving hyperbolic functions

[vast sparrow]

WAIT

LETS GO

I FORGTO THE NEGATIVE

WE MOVED OUT

I GOT IT RIGHT

[river minnow]

You got it?

Okay great

[vast sparrow]

thank you !!! yes I got it right, I just forgot the negative sign

thanks sm!

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[vast shale]

I have to do this line integral,i know i should swap x,y,dx,dy with t but i just dont know what to do with this condition

e^x-y=x+y

Do i express x or y

I have tried it

Doesnt work

Im frustrated

[flat whale]

Can you translate verbatim

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@vast shale Has your question been resolved?

[vast shale]

Calculacte the integral on the curve in xy-plain given the condition

for x>1/2

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[solid laurel]

this is a super simple question just forgot how to do it

[solid laurel]

some help would be appreciated

[regal bane]

Make a right triangle and apply sohcahtoa logic

[solid laurel]

our x is 3

and our y is 4

but whats our r?

[regal bane]

You have two sides of the triangle, and want the hypotenuse

[solid laurel]

yea?

[regal bane]

Pythagorean theorem is the tool you want

[solid laurel]

ok i got it

thank you!

[regal bane]

Np feel free to ask if you want anything else!

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[keen venture]

I know it's a straightforward answer but a little explanation if possible

and also with questions like these

[keen venture]

so 3

1 is alt to 3 like you said

really?

interesting

good to know for future problems

sad angle

haven't tried it

doin it rn

Is 65 the angle?

so the bottom

pythagorean?

alr

easy enough

66.7?

would it just be 67

yeah

doesn't say it explicitly but in videos it says 2 decimal places

or round

but considering the other numbers aren't decimals it's probably rounded

if its wrong I'll just do the practice again I got 2 attempts

alright

for this one do I just subtract 4 from 32 and approach it as the previous one

alright

32 and 80 are the sides and I'm finding the hyp

oh yeah

I did it but completely disregarded it in my head lol

yeah

84.8

I'm noticing that the hypotnuse isn't much bigger than the base

is that common?

ah

for this one is it sin cos tan procedure?

alright

would it be -tan or just normal

yeah what I got

so if im looking for a side it's -tan

yep

cause I've done some problems that call for -tan 1

29.99

is this 30?

30.0

gotcha

i mean like

rounded

up

30.00?

alright

how would I approach

ohhh

i just saw numbers

and an agle

and yeah

angle*

its sin

sinx = 8/30

?

how do i divide 8 = sin x

ahh

that's what i was talking about earlier

ye

15.5

so when missing a side and angle?

its

negative

gotcah

gotcha

ye

with no angle given its -?

uh

wait

so is

angle is 12

80 is base

im missing two sides

how do i figure this out

how

i only have the base

is the opp x?

tan 12 = x/80

x = 17

ez

is this one word

or just a phrase

alright

ok so for this one

75 is the base

35 is the angle

is the 22ft even useful

ah

is the answer minus 22?

wait wait nvm im overthinking

x = 52.5?

ah

that's what i was talking about earlier

yeah

so 74.5

took a little long but

atleast I understand it now

yeah

if I had asked my teacher she'd be like just try it by urself

like mf why u think i emailed u

yes

indeed

that's why i have come here

yeah for now

i'll be back later

thanks for the help

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[keen venture]

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[trim walrus]

Hello

[trim walrus]

How can do this??

[nova fulcrum]

!show

[vocal sleet]

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

[trim walrus]

Well what I did was open it up

And did some math and It turned out to be very dirty

[nova fulcrum]

show your expansion

[trim walrus]

Wait a minx

,w expand (x+1)(x+2)(x+3)(x+4)-24

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by _basudev#0

[nova fulcrum]

ok, now factor that

[autumn cedar]

is factoring not just what he had initially

[trim walrus]

Alr so one sol is 0

[autumn cedar]

nevermind

[nova fulcrum]

he had 24 in the other side

[trim walrus]

Yes

[autumn cedar]

yeah I forgot mb

[nova fulcrum]

so factor that

[trim walrus]

Should I go with like hit and trial methos

*method

[autumn cedar]

look at factors of the constant

[nova fulcrum]

start factoring what's in all of them

[autumn trail]

Do you know rational root theorem?

[autumn cedar]

and use the factor rule

[soft walrus]

first notice each term shares a common x, get that out first

[trim walrus]

Ye ik one sol is 0

[nova fulcrum]

factor

what do u have

[trim walrus]

Now I'm left with

[autumn cedar]

then factor out the polynomial excluding the x

[trim walrus]

X³+10x²+35x+50=0

[nova fulcrum]

ok, u know to factor that one?

[autumn cedar]

find factors of the constant and use the factor rule to determine which is a factor

[trim walrus]

Ye ik hit and trial one

[autumn cedar]

and then do polynomial division

[trim walrus]

Method

[nova fulcrum]

rational roots

[trim walrus]

No

Idk that

[nova fulcrum]

+-1 +-2 +-5 +-10 +-25 +-50

[soft walrus]

check each of those

if one returns zero it must a root

so you can divide it without remainder

[trim walrus]

Alr

[autumn cedar]

yeah and then you divide

[trim walrus]

Let me

[soft walrus]

by remdainder theorem

[autumn cedar]

and then repeat the process

until everything is in brackets

[soft walrus]

well then you get a quadratic after running division once

quadratics are pretty easy

[autumn cedar]

[nova fulcrum]

u sure?

xd

[trim walrus]

Alr non

Non of those work

Hold up

Ale

-6

*5

[autumn cedar]

now if plugging -6 = 0

that means (x+6) is a factor of the polynomial

and nowu can divide

[trim walrus]

Ye

Alr

(X+5)(x^2+5x+10x)=0

,w x^2+5x+10=0

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by _basudev#0

[trim walrus]

Just 1 real root

[soft walrus]

be careful you still have one more you got rid of earlier 👀

[trim walrus]

Oh ye x=0

Yep

Thanks

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[dusky trout]

hope its not against the rules...
which of the following is not based on earth`s acceleration?

[dusky trout]

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[stoic knoll]

Hey, i got the task to write down all "idempotent 2x2 matrix", first i thought about the null matrix but arent there way to many possible idempotent matrix to write down ?

[cyan talon]

"2x2 idempotent"

[paper depot]

you will most likely have a family of them

[cyan talon]

I mean sure there's prolly infinitely many of them

[paper depot]

described by one or more real parameters

[stoic knoll]

so overall it would be more smart to just write down the formular to build a idempotent matrix instead of trying to write down infinitely variants ?

[cyan talon]

depends if you want to die doing this task or not

[stoic knoll]

i dont really wanna write down matrices for hours tbh

[cyan talon]

basic idea is

take some arbitrary 2x2 matrix ((a,b) (c,d))

write what it means for that matrix to be idempotent

that should give you conditions on a, b, c, d for the matrix to be idempotent

[stoic knoll]

Alright. Thank you! Was just wondering bc my theory was that there should be an infinite amount of idempotent matrices and the task to "write all idempotent 2x2 matrices" seemed a bit off

[cyan talon]

either the person who wrote that wants you dead

or they just had the same thought process as mine while writing the q

I hope it's the 2nd one

[stoic knoll]

+1

Thanks!

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[strong cedar]

Hey, I am unsure how to do the second part of this question

[paper depot]

the way this is worded is mega sus

as written they are just asking for u_1?

[strong cedar]

It doesn't have a limit

[paper depot]

like what does the convergence have to do with it

or what

or wait

oh i get it now

[strong cedar]

Find a u1 when the sequence converges?

[paper depot]

yes

there will be only one such point

[strong cedar]

Is there a method for that or is it just something like 0?

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[fleet vessel]

Can some help me through this

[fleet vessel]

Part b

[outer warren]

what was your decomposition?

[fleet vessel]

Wdym?

[outer warren]

what was your partial fraction decomposition from part a

[fleet vessel]

A 3 b 4 c -2

[outer warren]

did you find f'(x)

[fleet vessel]

No

Do I need to do that

[outer warren]

find f'(x) then

that's the most efficient way to determine if the function is decreasing

[fleet vessel]

What would I do after that then?

[outer warren]

what values of f'(k) would indicate the function is decreasing at k

[fleet vessel]

I’m gonna be honest I have no idea

[outer warren]

do you know what the derivative indicates

[fleet vessel]

Yes the rate of change of gradient

[outer warren]

so what does f'(k) give

[fleet vessel]

1

[outer warren]

no

[fleet vessel]

When you differentiate k it’s not one

[outer warren]

not asking for (f(k))' nor k'

[fleet vessel]

Oh

[outer warren]

but the derivative evaluated at k

[fleet vessel]

Oh ok

[outer warren]

f'(k) gives the slope/gradient at x=k

if given the numerical value of that, how would you determine if the function is increase/decreasing at that location

[fleet vessel]

Well if you put in a bigger value. Which would be smaller than the one before does that show it is decreasing?

[outer warren]

just from the derivative itself, nothing else

e.g. if i just gave you that
f'(k) = 5

would you be able to determine from that whether the function is increasing/decreasing at k

[fleet vessel]

Not sure

[outer warren]

recall what the derivative does

[fleet vessel]

Rate of Change of gradient

[outer warren]

didn't catch what you said earlier

no

its not

rate of change OR
gradient not of

rate of change of gradient would be the second derivative

[nova fulcrum]

If this is too hard for you, you could also use the definition of increasing function.

although i think using derivative is easier anyways

[outer warren]

consider a random curve like that

[vast shale]

(off topic) what is 8.539734222673567

[outer warren]

is the slope/gradient at the green point positive/negative

[vast shale]

in your about

[outer warren]

oh

[vast shale]

im not

its a legit quesiton

[outer warren]

its pie

[vast shale]

OH

[outer warren]

that's better asked in chill

[vast shale]

alr

[outer warren]

not in someone elses channel

[vast shale]

sorry about that

[fleet vessel]

Negative

[outer warren]

and is the function increasing or decreasing there

[fleet vessel]

Decreasing

[outer warren]

what about here, same two questions

[fleet vessel]

Would it negative as well

And decreasing

[outer warren]

yes

and for this

[fleet vessel]

Increasing

And positive

[outer warren]

yes

so can you now see that positive slope/gradient indicates function is increasing
so can you now see that negative slope/gradient indicates function is decreasing

[fleet vessel]

Oh ok

I get it now thanks

.close

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[elfin moon]

How to solve this question?

I didn't understand their solution too which provided

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[placid plover]

What is the difference between du/dx and du? I understand the dx in the denominator means 'with respect to x', so what does just du stand for?

[steep sand]

du/dx means rate of change, du means small change in u

[celest tartan]

you just replace (3x**2+2x)dx with du
it’s handy for substitution

[celest tartan]

[mild flower]

yep so for an infinitesimal change in u (we call the change du, or "difference of u"), x changes by that amount times (3x^2 +2x), or (3x^2 + 2x)du

so, dx = (3x^2 + 2x)du

and yes you can treat those like variables that you can substitute, just treat dx and du as one symbol each (don't cancel the d's)

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[placid plover]

When performing integral substitution, why don't you need to apply the chain rule on e^u dx? E.g. integral(e^u) dx would be (1/u)(e^u)

[faint fog]

youre integrating with respect to u

[obtuse radish]

in the highlighted bit its du

[faint fog]

that's the point of the whole getting du/dx thing, is that you can do a substitution for dx in terms of du

in fact u-substitution is exactly the chain rule in reverse

if you take the derivative of what you got as an answer, you'll see that the "inner" piece is exactly u

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[vast shale]

What did i do wrong here ? i got a really weird answer when using quadratic formula

i did a common factor of (1+y) (1+y) (1-y) (1-y)

[urban edge]

3(1+y)(1-y)=?

[vast shale]

ohh

wait

my bad

3 -3y2

thank you

[vocal sleet]

@vast shale Has your question been resolved?

[vast shale]

so i got -7/3 instead of -3/7

@urban edgeany idea why ?

[urban edge]

14y=-6

How do you isolate y?

[vast shale]

i do /14

oh

i did the reverse

[urban edge]

Yeah

[vast shale]

ty

[vast shale]

.flip

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[trim walrus]

Hi

[trim walrus]

Any idea how to solve this

??

[astral shadow]

Rewrite

log_{bx}b = log(b)/(log(b) + log(x) etc

You'll get a quadratic.

You should be able to comment about its roots.

[trim walrus]

Which property of log are you using over herer?

[astral shadow]

Probably called base changing property

[twin meteor]

What the hell am I doing here?

[astral shadow]

Then also,

[trim walrus]

$\log_{bx} b = \frac{\log {b}}{\log {bx}}$

[twin meteor]

What the hell am I doing here?

_basudev

[trim walrus]

Smth like this??

[astral shadow]

Yes, then this for your denominator.

And do this for every term.

[trim walrus]

Alr

Smth like this

Should I proceed by taking lcm or wot??

[patent nymph]

Yes, then factor it

[trim walrus]

Alr

Man, its so lengthy

,w simplify (2 log (b))/(logx) + (logb)/(logb+logx) + (3logb)/(2logb+logx)

[twin meteor]

Results provided by WolframAlpha

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Requested by _basudev#0

[trim walrus]

Alr

I don't see anything

Helpful

After doing that

[merry python]

After this take logb as common

in the numerator

[trim walrus]

Alr

Then

[merry python]

shwo what you get

[trim walrus]

Alr

[twin meteor]

Couldn't find an attached image in the last 10 messages.

Jun 26, 2023, 7:23 AM

[trim walrus]

,rotate

Wait wait

It has already been taken common

[merry python]

Okay

Since that looks ugly

proceed with substituion

logx=p
logb=q

[trim walrus]

Alr

,rotate

[merry python]

Okay

RHS=0 so can you find how many possible values of p are there?

[trim walrus]

Yes

Wait let me write it

Infinite

??

[merry python]

How infinite?

[trim walrus]

,rotate

[trim walrus]

Zero

Since logb=0 is not defined

[merry python]

logb=0 is defined

log0 is not defined

and also you don't need to look onto that

[trim walrus]

Yep 2

[merry python]

since we are only dealing with p

q+2p=0 and 4q+3p=0

This should give two possible values of p

[trim walrus]

Yes

So 2 real roots

[merry python]

Yes

[trim walrus]

Alr mate thanks

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[paper rampart]

.

[vast shale]

Hello

[vast shale]

How do I do it??

[sudden compass]

factorize the expression

notice that you can take something common out

from all three terms

[vast shale]

Yes

A

a(ax²+bx+c)

[sudden compass]

ax^2 + bx + c is given to you

[vast shale]

Is its imaginary

[sudden compass]

bruh 😂

[vast shale]

It's roots are

Fuck

[sudden compass]

look a little harder

[vast shale]

No

0

[sudden compass]

exactly

[vast shale]

But no opt such as 0

[sudden compass]

0 is non-negative

[sudden compass]

it should be B

u have an answer key?

[vast shale]

Hold up let me check

[empty frigate]

...i don't get it, how is it B?

[vast shale]

Yes answer key

[sudden compass]

its B?

[vast shale]

It says A

[sudden compass]

[empty frigate]

how would a(ax^2 + bx + c) be zero?

for x a real number

[sudden compass]

ax^2 + bx + c is 0

its fudging given

[empty frigate]

we're given the exact opposite

for real x, ax^2 + bx + c is never 0

[vast shale]

That's is for complex ig

[sudden compass]

huh?

[empty frigate]

"the roots of ax^2 + bx + c are imaginary"

[vast shale]

Ye

See I told ya

[sudden compass]

wait oh

am dumb

its 0 only for x = a,b where a and b are the imaginary roots

but they are saying x is real

[empty frigate]

(using a,b to write those is not a good choice of letters lol, we already have an a and b)

[sudden compass]

(true, but i was lazy to use alpha and beta)

[vast shale]

Alr

Do

So

Ye

How would I proceed

[sudden compass]

he isnt there some rule that says if one root is imaginary the other is the conjugate

[vast shale]

Yes

There is

I g

Case - II

[sudden compass]

its for D>0

i dont think it applies for D < 0

[vast shale]

Oh ye I didn't see that it's fir D>0

[sudden compass]

we dont know if the roots are unequal or conjugates

[vast shale]

Hmm

[sudden compass]

oh nvm

[vast shale]

[vast shale]

Yep

[sudden compass]

if they are conjugates they must be of the the form a+bi and a-bi

[vast shale]

If the one root is x+i then other is x-i

?

[sudden compass]

yeah

[vast shale]

Alr

[sudden compass]

a quadratic can be written as $(x-\alpha)(x-\beta)$

[twin meteor]

itzkraken.

[vast shale]

Yes

[sudden compass]

where $\alpha and \beta are the roots$

[twin meteor]

itzkraken.

[vast shale]

Alr

OK

[sudden compass]

so substituting

[vast shale]

$(x-a-bi)(x-a+bi)$

[twin meteor]

.fire007

[sudden compass]

yeah thanks

[sudden compass]

you can use identities to simplify

ur answer will be a rational expression

with no is

[vast shale]

Like what identity??

Algebraic identity??

[sudden compass]

yes

[vast shale]

,w expand (x-a-bi)(x-a+bi)

[twin meteor]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by .fire007#0

[thick adder]

What is going on here

[woeful cloak]

ax^2+bx+c could be strictly negative btw, can't believe the question missed that

[thick adder]

Was this the orignal problem

It's much simple then whatever you are doing

we not the for no roots

[thick adder]

b^2-4ac must be negative

Now we note 2 things in the second expression

[vast shale]

Yes

?

[thick adder]

squares are always pasitive

Hence it's leading term is +

And it's an upward parabola

[vast shale]

Yes

Agreed

[thick adder]

Compute the discrimination of 2

its

[vast shale]

(ab)²-4a³c

[thick adder]

a^2b^2-4a^3c

Now take a^2 out

[vast shale]

a²(b²-4aç)

[thick adder]

So we have

[vast shale]

Oh it's neg

[thick adder]

Yup

No idea what was going on here before

[vast shale]

But the answer key says its A)

[thick adder]

Yes

The Discriminant is negative

Hence there are no roots

Hence it never crosses the x axis

And since it's upward facing

...

QED

[woeful cloak]

that proof is not quite incorrect, having no real roots does not imply it is positive

for example -x^2-1 is negative for all real x

[thick adder]

I mentioned

a^2>0

[woeful cloak]

there is something else that you have to considered here

[thick adder]

Hence upward facing

[woeful cloak]

ah, right then

[vast shale]

a²(b²-4ac) this exp is neg right, hence the Discriminant is img

[thick adder]

Yup

[vast shale]

So roots are img

So now how we know if the exp is positive or neg??

[thick adder]

Yup

[vast shale]

Or.zero

??

[thick adder]

Draw the graph

It's a parabola

Now

[vast shale]

It doesnt touches x axies

[thick adder]

Can it ever touch the x axis?

Yup

So

It's either all + or all -

Cause it never crosses

Got that?

[vast shale]

Not quite

[thick adder]

And since it's facing upwards

Ok

You understand that is never crosses the x axis

?

[vast shale]

Ye

[thick adder]

So if suppose

It was positive at one point

But negative at another

Well the we have 2 points on opposite sides of the axis

[vast shale]

Yes

[thick adder]

And since parabola are continuous curves

It's implies the curve must also cross the axis

Which we just showed is false

[vast shale]

Any example you got??

[thick adder]

Umm

Think about it

Suppose you wanted to draw the graph

[vast shale]

Yeah

[thick adder]

You know you can do that without lifting your pen right?

Just start at one arm of the curve

The draw the hump

Then the other arm

And at no time lift the pencil

[vast shale]

Alr

[thick adder]

Now you have 2 points on the curve

One at the bottom

One at the top

[vast shale]

Wlr

[thick adder]

Since they are on the curve

You can connect them

[vast shale]

Yes

[thick adder]

Without lifting your pencil

Through the same curve

But think about it

If you start at a point at the bottom

And go to the top

Surely you must cross the middle

Got that?

[vast shale]

I go through the curve or a straight line??

[thick adder]

Any way you go

You must cross the middle

Go in a zig zag, a line, an arc

[vast shale]

A visual representation would have been better....

Like a drawing

[thick adder]

There is no way to go from red to blue

Without crossing green

[vast shale]

Yes

[thick adder]

We know the parabola never crosses green

As no roots

So it can't have points on both sides

[vast shale]

A parabola crosses the green if it has real roots right??

If its img it stays up

Or down

Depending Upon the sign of (a)

[thick adder]

YUP

And since the sign of a here is positive

(It's a square)

It always stays up

[vast shale]

It stays up

[thick adder]

And hence

[vast shale]

Yes

[thick adder]

Its always

...

[vast shale]

Hold up but the roots of the eqn had D < 0, it means it stays up and never croess x-axis

[thick adder]

EXACTLY

And what does it mean to "stay above the x-axis"

[vast shale]

It means there roots are img

[thick adder]

It means another thing

If a point is above the x axis

That means the function at that point is

[vast shale]

Oh ye it's positive

[thick adder]

Gg

Ez

[vast shale]

Oh yep

OHHHHH

I get it.

I get it

Thanks man

Thanks.for the help 🙂

I really appreciate it

[thick adder]

That's what I am here for

That and procrastinating doing my own work

[vast shale]

Alr.

Alr

.close

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[elfin moon]

If there is a matrix A(5×5) which entries are 0 and 1
Where A is upper triangular matrics

[elfin moon]

Then min| n(0)-n(1) | will be

N means number of

[paper depot]

so there are 25 cells each containing a zero and a one, and you want to know the smallest possible difference between the amount of zeros and the amount of ones?

[viscid ore]

What does this notation mean?

oh

got it

[hard atlas]

!nosols

[vocal sleet]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[viscid ore]

....my bad

[viscid ore]

this

[vocal sleet]

@elfin moon Has your question been resolved?

[paper depot]

@elfin moon

[elfin moon]

Yes Ann

[elfin moon]

The difference between number of zeros and number 1

With modulas

Please check again i edited the question @paper depot

[elfin moon]

Yes

Exactly

[paper depot]

so A is upper-triangular

can you write down what A looks like, using asterisks for entries you don't know yet?

[elfin moon]

Huh what? Where is solution

[paper depot]

riku must have deleted it already

[elfin moon]

If riku has written then they shouldn't tell her what to do what to not

[elfin moon]

I don't know

Do you want me to write down matrix?

Should I send picture?

[paper depot]

yes

[paper depot]

great

$A = \bmqty{&&&&* \ 0&&&& \ 0&0&&&* \ 0&0&0&& \ 0&0&0&0&*}$

[twin meteor]

Ann

[paper depot]

ok

[elfin moon]

Yes

[paper depot]

if you didnt know that A was upper-triangular, what would be the minimum of |n(0) - n(1)|, and how could you achieve it?

[elfin moon]

Ohh wait

Surely 0

[paper depot]

and how could you achieve it?

give me a 5 by 5 matrix with an equal number of zeros and ones.

[elfin moon]

Max entries 25 for 0
25 for 1

Substrate minimum 0

[paper depot]

...

no, you cannot just subtract their maximums like this...

i mean, you can, but it won't help you at all

also "substrate" is a different word and has nothing in common with "subtract" (which only has one S)

[elfin moon]

U mean both entries present?

[paper depot]

$|n(1) - n(0)| = 0$ means $n(0) = n(1)$

[twin meteor]

Ann

[paper depot]

so if you claim the minimum value of |n(1)-n(0)| is 0 it means you know a matrix for which |n(1) - n(0)| = 0

and i am asking you to give me such a matrix

[elfin moon]

Then it's not possible 25

Not divisible by 2

[paper depot]

ok

so we come back to the same question:

if you didnt know that A was upper-triangular, what would be the minimum of |n(0) - n(1)|, and how could you achieve it?

you now say the minimum isn't 0. what is it instead?

[elfin moon]

Wait a minute

I need to think more

[paper depot]

take as much time as you need.

[elfin moon]

Then answer is 1

[paper depot]

ok

and how could you achieve it?

i.e. give me a matrix with |n(0)-n(1)| = 1

[elfin moon]

I put 13 0 and 12 1

[paper depot]

13 zeros and 12 ones, great.

can you make that happen when A is upper-triangular?

[elfin moon]

Ahhhhhhhhhhh

I need to understand what the actual definition of upper triangular first

So a zero matrix is upper triangular too?

I guess yes

[paper depot]

yes it is

but also you wrote out what A looks like with the "upper triangular" condition

and i copied it

right here

[elfin moon]

So when it is upper triangular then 10 zero and other side 15
Then minimum 5

And if i give zero to more diagonal

Then 13,12

So 1

[paper depot]

you realize that these asterisks don't have to be 1's, right?

some of them could be zero. all of them could be zero if you wanted.

you can make a "13 zeros, 12 ones" matrix from this.

you made it more difficult for yourself just now

[elfin moon]

What?

My answer is 1

Where is it difficult?

Sometimes I don't get what point you want me to focus

[paper depot]

So when it is upper triangular then 10 zero and other side 15
Then minimum 5
And if i give zero to more diagonal
Then 13,12
So 1

all of this was unnecessary

you confused yourself into thinking the minimum was 5 for a moment

13 zeros and 12 ones, great.
can you make that happen when A is upper-triangular?

the correct answer to this was "yes"

maybe also with an example like

1 1 1 1 1
0 1 0 0 1
0 0 1 1 1
0 0 0 1 0
0 0 0 0 1

or whatever

[elfin moon]

Yes. It doesn't matter if diagonal is zero or whatever

We need to make it max and substrate

Why wrong?

I am right

[paper depot]

nobody will understand "make it max"

[elfin moon]

Ahhh i am not native to english

Maybe riku will understand my words

I meant we are here wanting to make the whole thing minimum

And it is possible when we make zeroes and 1 maximum and close to each other

[paper depot]

"maximum" is a bit weird still

[elfin moon]

So their subtraction tends to minimum 0

You can't maximize one of them exceeding 13 this is what I mean

Maybe my words of selection are weird

If we want to get values of a function minimum then we will make them possible max at size so their max part substract and gives us minimum value

[paper depot]

subtract not substrate

your choice of words is weird, yes. but i think that's because of your math skills, not because of your english skills.

[elfin moon]

Math skill??

Here I am telling you the fact why it is minimum

We maximize zeros/ones so that they subtract and give minimum values

Can you get minimum values by minimum?

Or by minimising?

[paper depot]

...

[elfin moon]

Tell me what math skill is wrong

It's a bit insulting

. close

[paper depot]

my apologies

i did not mean to insult you

[elfin moon]

I am just here to learn from you guys

[paper depot]

what i had in mind is: you struggle with expressing your ideas in mathematical language, but native english speakers struggle with it all the same

[elfin moon]

If i am somewhere wrong in any way, then you should have helped me or correct me

[paper depot]

so the fact that you aren't native plays a lesser role in this than the fact that math english is just different from everyday englsih

if i knew how to correct you i would have already done so

[elfin moon]

I shouldn't engage in controversy. Let me try to solve another question...sorry by my side