help-28

[coarse tundra]

Should this be enough to proof it?

[smoky wing]

,rotate

[sharp thorn]

M and n are not consecutive so the n used would be different for both of them

The rest is correct

[smoky wing]

is that an x or n

[coarse tundra]

m=2x

[smoky wing]

they dont have to be the same x

m = 2x doesnt mean n = 2x+1 the next number necessarily

mayve you should say

\begin{proof}
if m is even then $m = 2x_1$ for some $x_1\in \mathbb{Z}$ \
if n is odd then $n = 2x_2 + 1$ for some $x_2 \in \mathbb{Z}$ \
Then consider [insert proof]
\end{proof}

am i dumb

[coarse tundra]

what?

[smoky wing]

idk how this thing works again one sec

damn

[coarse tundra]

take your time

[smoky wing]

like that

[glossy valve]

Pure

[coarse tundra]

oh

so i should use different unknown

[smoky wing]

yes]

[coarse tundra]

the answer should be
2(x1+x2) + 1 = m+n

thus odd

[smoky wing]

right

[coarse tundra]

ok nice thanks both of you

.close

[smoky wing]

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[undone bone]

How do I integrate $\int\frac{e^{2x}-e^{x}-2}{e^{x}}dx$ using the substitution $t = e^x$ and the equality
$\frac{t^{2}-t-2}{t^{2}}=-\frac{1}{t}-\frac{2}{t^{2}}+1$.

[glossy valve]

1311Discover

[undone bone]

I've done the following... but I got stuck at the moment

I'm not too good with integrations

[spice orchid]

Why did you start with e^2x on the bottom?

[undone bone]

typo

most likely

yeah

my bad

so it would be something like

(t^2-t-2/t)*e^t

??

[spice orchid]

Remember you have to replace the dx with something involving dt when you do a substitution

[undone bone]

like the updated expression?

[spice orchid]

Sure

[undone bone]

then what am I supposed to do with the equality?

[spice orchid]

If t = e^x then what is dt/dx?

[undone bone]

isnt it e^x

[spice orchid]

Yeah, so dt/dx = e^x, so dx = 1/e^x * dt

So taking that into account when you do the substitution you should get in the integral the expression on the right of the equality they gave you

[undone bone]

Am I going in the right direction?

sorry instead of

dt/du is just dt

could I do something like

t-2/t . t dt

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@undone bone Has your question been resolved?

[undone bone]

Not really

<@&286206848099549185>

[sharp thorn]

dt/dx is e^x not e^x dx

[undone bone]

Like this?

[sharp thorn]

No

[undone bone]

oh

I got it

dt / dx = e^x

[sharp thorn]

Yep

[full forum]

@undone bone Has your question been resolved?

[slate violet]

you can cancel out the ts here

[undone bone]

so it becomes -1/t -2/t^2 + 1

[slate violet]

oh wait you forgot the t^2 on the bottom

[undone bone]

multiplied

by 1/t

instead of t

[devout valley]

Still to be integrated, you haven’t done that yet

[undone bone]

doesnt it become -log(t) - 2/t + t

then replace t?

[devout valley]

+2/t but yep

[undone bone]

my bad

[devout valley]

And yea

[undone bone]

becoming -log(e^x) + 2 / e^x + e^x

my issue atm is just simplify it

but I'll try

[devout valley]

Yep yep

[devout valley]

Shouldn’t be too bad! Log(e^x) should be pretty simple to see what happens to it

[undone bone]

I have a voice in my head that is saying x

I dunno why

Maybe, but just maybe I know why...

exactly

confirmed by symbolab

So it all becomes -x + 2/e^x + e^x

or e^x - x + 2/e^x

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[wicked frost]

ahhhh I see

[devout valley]

Try again and let me know how it goes

[wicked frost]

Still messed up again somehow

Sry about the mess lol

[devout valley]

How did you find the "A" and "B"?

[wicked frost]

"cover up method"

saw it on youtube

basically lets say I wanna find A, the root of X+3 is -3 so I plug in -3 and cover up the (x+3) term I may have done it incorrectly tho lemme rewatch the vid

[devout valley]

that's a fair method, but not sure about its implementation, did similar and have different numbers!

...or wait

did you mean like this, but with 9x + 3 on the left hand side?

[rapid rain]

yes the remainder of the division by x^2-9 is wrong as you wrote it

[rapid rain]

the correct remainder is this

[wicked frost]

yeah I think I realized what I did wrong tho

I included the 6x+2 in my calculation I think im supposed to leave that out right

[devout valley]

Yeah, you're not supposed to include it

[wicked frost]

did you get 5/x-3 and 4/x+3 ?

[devout valley]

Yep that's what I have

[wicked frost]

perfect tysm

[devout valley]

all good!

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[torn jolt]

I'm somehow forgetting basic math

[torn jolt]

I'm not getting these numbers

[ocean lark]

,calc 8/pi*4.57+10

[glossy valve]

Result:

21.637409438879

[ocean lark]

,calc 6/pi*4.57+10

[glossy valve]

Result:

18.72805707916

[ocean lark]

,calc 6/pi*1.71+10

[glossy valve]

Result:

13.265859432246

[ocean lark]

Well the numbers are right

what did you do?

[torn jolt]

first I expanded the bracket by multiplying pi/6 by t and -10

then shifted the -10(pi/6) to the right

making it +10(pi/6)+1.71

=(pi/6)t

[ocean lark]

nvm

[torn jolt]

oh, was going to say maybe I should've solved

[ocean lark]

yeah go on i misread

[torn jolt]

but then what i did from there is multiply both sides by 6

and divide by pi to isolate for t

and got t=3.64

[ocean lark]

Uh that sounds correct to me, so you made some mistake putting it into your calculator

,calc 10(pi/6)+1.71

[glossy valve]

Result:

6.945987755983

[ocean lark]

,calc6.945987755983/6

[glossy valve]

Result:

1.1576646259972

[torn jolt]

maybe, but main difference I'm seeing in both of the equations is 2pi/12 was what I had and pi/6 was what they had

theoretically should be the same

but I'll try doing that again

[ocean lark]

oh

it should be multiply by 6 divide by pi

in your last two steps

not the other way around

thatll be your issue

[torn jolt]

ohh right

lol thanks for catching that

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[full forum]

Please don't occupy multiple help channels.

[chrome lantern]

-close

.close

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[smoky jackal]

Hi sorry for so many helps but I need help again. I don’t get this question at all. I made these questions but doesn’t make sense
17 cups = flour
4.5 = sugar

3.5f + 0.75s = 1 loaf bread
2.5f + 0.75s = batch of muffins

[short niche]

the part on top where you write 17 cups = flour is pretty misleading

[smoky jackal]

Oh I meant like Helena has 17 cups in total to use

For baking

[short niche]

what doesn't make sense about the question?

[smoky jackal]

Idk how to solve it

And if my equations are correct

And what I should do with them

[short niche]

check if the you have enough ingredients to make the given bread and muffins for each option

[empty sapphire]

So I will say no they're not correct.

How you should break them down into two equations: one for total flour and one for total sugar.

We know that Helena needs 3.5 cups of flour for bread (we can say we need x amount of loaves) and 2.5 cups for muffins (we can say we need y amount for muffins): 17 = 3.5x + 2.5y

Similarly, she needs 0.75 cups of sugar for bread and 0.75 cups of sugar for muffins. She has 4.5 cups: 4.5 = 0.75x + 0.75y

Now you have a system of equations and can solve for your solution 🙂

[smoky jackal]

Omg this makes so much more sense

You’re awesome thank you for your help 🙏

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[deft summit]

may i know how to solve this?

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[rocky scaffold]

could someone help me with this please? im not sure where to even start

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[wary spindle]

There are 10 balls in an urn, numbered from 1 to 10. We draw 5 balls. How many cases are where exactly on 3 drawn balls stands a number less than 5?

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@wary spindle Has your question been resolved?

[trail hawk]

3 ?

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[slate violet]

are the balls drawn with or without replacement?

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[glad solar]

Can someone help me

[brave crater]

3n*(n-2)

[glad solar]

ohhhh

Weird wording

[serene hazel]

,rccw

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[limber flicker]

hey, just wondering for this matrix, when finding a basis for the nullspace, I got it into RREF, I couldnt decide which variables to make free, it wasn't immediately obvious to me, other than knowing I needed 2,

[limber flicker]

for rref i got

and x_4 being free

I think I missed something earlier in the notes or am just being stupid, but I cant decide with a good reason as to which other variable should be free?

[limpid moat]

the image has clearly dimension equal to 2

[limber flicker]

yep

[limpid moat]

so dim (null space)=4-2=2

[limber flicker]

yep

[limpid moat]

x4 free, thet x3+2x4=0 then x3=-2x4

finally x1-2x2-x4=0

[limber flicker]

yeah, I then let x_2 be free which got me the correct basis but idrk why I chose it, I kinda just did

maybe I dont really understand wtf a free variable is

like if I choose x_4 = s and x_2 = t, I get one basis, would the result still be correct if I let x_4 = s and x_1 = t?

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[limber flicker]

.close

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[oak pebble]

can someone check my project?

[oak pebble]

<@&286206848099549185>

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[compact briar]

What project?

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[full forum]

@graceful aurora Has your question been resolved?

[gentle oracle]

I wouldn't be dead sure

[torn jolt]

I feel like you can better ask this in MODS server

Wait I found the solution

You can prove that angle AKP=angle YXP

YXCP is a cyclic quadrilateral, so angle YXP=YCP=ACP=AKP

And then you are done

CXP and CYP are both 90 degrees

YXCP, YAZP, ZBXP are all cyclic quadrilaterals

You can prove the Simson line with them so they are quite standard

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[lapis flower]

Uhh it's 1/3 right

9 possible combinations I think

3 of them with teleport

Idk

[cursive dock]

P(teleport at least once) = P(teleports in the morning) * P(anything in the evening) + P(doesn't teleport in the morning)*P(teleports in the evening)

[lapis flower]

whaf

ahT

What

[cursive dock]

P(...) = probability of

[lapis flower]

Uhh

Is my answer right

[undone vector]

no

[lapis flower]

😔

[undone vector]

If he teleports in the morning, then all 3 options of the evening means he still teleported

but he could also not teleport in the morning, but teleport in the evening. So you know its more than 3 combinations with teleport

[lapis flower]

5/9

[undone vector]

ye

[lapis flower]

Hurray

BRO HALF THESE QUESTIONS ARE ESSAYS😭 😭 😭

I left out the three paragraphs of unneeded lore

Uhhh

1/2

Right

Idk I just counted

@undone vector

[undone vector]

ye

[lapis flower]

Easy

Light work

[lapis flower]

7/16 right

[undone vector]

I do not think so

[lapis flower]

😔

Oh wait

9/16

Right

Right

[undone vector]

yeah

[lapis flower]

ES

EZ

Ty

I'm so bad at statistics 😭 😭 😭

[lapis flower]

okokokok

It's like form 2-12 right

from

4 of them meat the constraint

Meet

idk how to write that in probability

[undone vector]

You can say P(sum >= 9) = P(sum = 9) + P(sum = 10) + P(sum = 11) + P(sum = 12) ig

[lapis flower]

I can only write numbers

[undone vector]

For questions like these you can think of a chart with dice from 1-6 as the axis and the entries as the sum. Then it should be easy to count

[lapis flower]

But that's like cheating frfr

[undone vector]

Idk how else you would do it w/o using more advanced ideas (random variables), but logically you can see that there is only 1 way to get 2 , 2 ways to get 3, 3 ways to get 4, ... 6 ways to get 7, 5 ways to get 8, ... 1 way to get 12

[lapis flower]

10/36

[undone vector]

yah

[lapis flower]

2/18

wait

[undone vector]

lol

[lapis flower]

no

I divided wrong

5/18

[undone vector]

ye

[lapis flower]

1/720

@undone vector

[undone vector]

yeah

[lapis flower]

Easy

Light work

No sweat

Uhh so initially for the first pick it's 1/3

Thennn

IDK WHAT TO DO AFTER

5/17

After

Soo

5/51

Chance

I think

@undone vector I need your mastermind thoughts

To look over my reasoning

🙏

[undone vector]

Ye

[lapis flower]

EZ

[lapis flower]

IM GETTING BETTER RIGHT

[undone vector]

Already perfect

[lapis flower]

This looks easy

.18

Right

So the total of all them equal to hundred percent (one)

If my mental arithmetic is right

@undone vector

[undone vector]

Yeah

[lapis flower]

easy

[lapis flower]

Also ur so cool tysm

[undone vector]

np

[lapis flower]

This is just B right

Obv not A or C

3/7 • 280 =120

[undone vector]

Yeah

[lapis flower]

8/25 riggt

[undone vector]

Yeah

[wild sleet]

B is still a bit too strong, if 100−140 is "close", it's only 1/2 to be close to 120

[lapis flower]

Are my answers okkkk

Slightly unsure about the last one but I just did what I thought I was supposed to do

@undone vector

[undone vector]

Yeah

[lapis flower]

LIGHT WORK

RAAAAHHHH

[lapis flower]

Bro idk why it does it like that

But Phil, rory, rickie

[undone vector]

Yeah

[lapis flower]

OK ONE LAST ONE

Brb lemme try it first

[lapis flower]

Uhhh 1/15

I did 1/3 • 1/5

Cuz like 2/6 aren't robots

And like

Yeah

[undone vector]

Yeah

㊗️

[lapis flower]

RAHHHHHH

Ty

.close

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[main elm]

ive worked out the frequency density of them all ik thats my first step i just dont know the rest

27 is 1.5
13 is 7
21 is 3.5
33 is 4.4
20 is 2.2

they are the frequency densities of them i just dont know what to do next

can sm1 help

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[main elm]

.close

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[glad whale]

how that one wrong plz

terminal has to be the opposite?

or something

[shut thorn]

you have to give it as x,y i guess

the first box fro x coordinate and the second for y coordinate

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[twin wolf]

To determine the nature of a stationary point what do you first have to do? It’s like

f’(x) = 0 to find the stationary point and then do something with f’’(x)? Do you check the concavity or solve an inequality in a little confused

[cosmic canopy]

What are the types of stationary point

List them

[twin wolf]

A horizontal inflection point, maxima, minima

[cosmic canopy]

yes

What can you say about f”(x) for each of those

[twin wolf]

if f’’(x) < 0 then a maxima occurs and if f’’(x) > 0 a minima occurs

[cosmic canopy]

And for the horizontal point of inflection?

[twin wolf]

Is it f’’(x) = 0 will be a horizontal POI under the condition that f’(x) also equals 0

[cosmic canopy]

Yes

That’s correct

[twin wolf]

Okay

But do I have to check every single one of the things I just stated to determine the nature?

[cosmic canopy]

it’s only two things

[twin wolf]

Like I have to solve for f’’(x)<0 and f’’(x) > 0 and f’’(x) = 0

[cosmic canopy]

why would you solve for that when you can sub in the point

[twin wolf]

Wdym

[cosmic canopy]

say you know f’(a)=0

Then just check the sign of f”(a)

No need to solve anything

[twin wolf]

How do I do that?

[cosmic canopy]

what

By checking what f”(a) is equal to??!

[twin wolf]

Like for example I have f(x) = x^3

f’(x) = 3x^2

f”(x) = 6x

[cosmic canopy]

yea

so x=0 is a stationary point

[twin wolf]

Well I need f’(x) = 0 for that

[cosmic canopy]

f”(0)=0 so horizontal point of inflection

[twin wolf]

So 3x^2 = 0

x= 0 is a stationary

[cosmic canopy]

yeah

[twin wolf]

And I plug in my x value for the stationary point into the second derivative?

So f”(0)??

[cosmic canopy]

yes

what else would you do

[twin wolf]

So it’s a horizontal inflection point since the first and second derivative are 0

[cosmic canopy]

Yes

[twin wolf]

ok

And let’s say my second derivative was like 4

4>0 so it’s a minima?

[cosmic canopy]

Ye

[twin wolf]

ok 👍

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[rocky bay]

Yes

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[azure burrow]

having trouble with 18

[azure burrow]

by side do they mean height

and base do they mean length and width?

I know our constraint formula is cost function and the objective formula is volume

but i’m not sure about the variables

a cube volume is lwh

[tiny flax]

A cube has volume a^3

[narrow ermine]

I think they mean side = walls no?

[tiny flax]

$a^3

[azure burrow]

yeah sorry rectangular prism has lwh right

[tiny flax]

,help

[glossy valve]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[azure burrow]

yeah i think the same

and so for base i assume it’s just a square

and by base they mean length x width

i think

so that means we can do V= hx^2

i think

solve for h in terms of x

[narrow ermine]

Yeah well I would set, say b = "sidelength of the square base" and h = "height of the walls"

So yeah use the V = hx^2 one to solve for say h and then use the price equation and make it single variable by subbing int.

Or use the price equation to single out h and sub into the volume equation, they're interchangeable

[azure burrow]

oh ok thanks

.close

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[glad whale]

hello

[glad whale]

so for this one

How would i know what quadrant is this one in

like i know the refrence number will be pi/4

But the quadrant confuse me with this one -3pi/4

ah i guess i am blind

I saw it in quadrant 2

but does the negative make it with quadrant 3?

.close thnx

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[tough pawn]

Can I have some help to know what should I do after this?

[torn jolt]

partial fractions decomposition sounds good here

[tough pawn]

But we have quadratic on the top, how can I doing partial fraction with that quadratic on it?

[sharp thorn]

The quadratic can be written as x^2+4+7(x+2)-16

[torn jolt]

^

you will get an equivalent expression with two fractions one of which has a linear numerator and the other a constant

[tough pawn]

Ok thank you very much!

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[unique garnet]

for a trigonometric function can we simplfy sin(x squared) sin(ax) and sin(ax+b) for all real a,b

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@unique garnet Has your question been resolved?

[fair oracle]

what exactly are you looking for?

is there a question you are looking at?

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[hardy snow]

Hi

[hardy snow]

Please help

[shy isle]

Whats the question

[hardy snow]

This

It's not my hw don't worry

[hardy snow]

Alr sorry

[shy isle]

Oh thats chill

8+ what = 12?

And can that number be divided into 8?

And you’ll get ur answer

[hardy snow]

4

[shy isle]

Exactly and if the above recipe makes 8, what do you divide it by to make 4?

[hardy snow]

2

[shy isle]

Exactly so divide the ingredients by 2 and add them to the pre existing ones

Ones

[hardy snow]

Breh I don't get it

Are we doing number a or b

Hello?

[cursive dock]

so for a): you know how to make 8 (the recipe given). you know how to make 4, as w_rry explained. Therefore to make 12 you need to make 8 + 4.

[hardy snow]

Ohhhh

Alright my Brian

Thanks that really helped

[hardy snow]

How about B

[cursive dock]

so I'll start with an extreme example: let's say I have 20000g sugar, 25000g butter, 35000g oats, 5 tablespoons of syrup
how many can I make?

and how did you work that out?

[hardy snow]

Ummmm

It's kinda confusing

Let me see

[cursive dock]

I have tons of sugar and butter and oats but only enough syrup for 8

so I can only make....?

[hardy snow]

Still 8?

[cursive dock]

yeah exactly

[hardy snow]

Oh

Because you only have that amount of syrup

Is it 32 for B???

[cursive dock]

so with mary's ingredients, you can go "okay, if she has enough of everything else - how many flapjacks can all her sugar make? how about all her butter? how about all her oats? how about all her syrup?"

and the smallest one will be the amount 🙂

in my case I had enough sugar to make 800 flapjacks and enough butter for 800 and enough oats for 800 but only enough syrup for 8, so I could only make 8

[hardy snow]

Yea kinda

[cursive dock]

600g sugar can make __ flapjacks
1kg butter can make ___ flapjacks
700g oats can make __ flapjacks
20 spoons syrup can make __ flapjacks
the smallest of these is __ flapjacks

[hardy snow]

Wit

Wait

It could make
3
4
2
4

[hardy snow]

It could make
3
4
2
4

[cursive dock]

perfect, so which is the limiting factor?

and don't forget that those numbers are in terms of how many copies of the original recipe you can make - and the original recipe makes 8 flapjacks 🙂

[hardy snow]

Yeah ik

[hardy snow]

Oats?

[cursive dock]

yep, so with 700g of oats she can make __ flapjacks

[hardy snow]

16

Breh

[cursive dock]

nailed it

[hardy snow]

Thank you so much

But I'm still kinda confused

[cursive dock]

sure

[hardy snow]

I'ma read through it again

Will it still be here tomorrow?

Because I'm about to sleep

[cursive dock]

the server will still be here tomorrow yes

[hardy snow]

Alright thank you so much

Very patient teacher

Bye bye

Could I please add your friend so you could explain how to do the math questions?

Because I think you're the best teacher I've ever met on this server

[cursive dock]

I'm usually hanging around here

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[hardy snow]

Yessir

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[hardy snow]

Alr

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[worldly wind]

is my answer right

[worldly wind]

.rotate

shit idk how to do it

[worldly wind]

model i drew is not to scale but that doesn’t really matter to anyone is my math right?

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[brave crater]

what's the issue

!status

[full forum]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

@ruby stag Has your question been resolved?

[ruby stag]

1

<@&286206848099549185>

[brave crater]

let your number of cups of Teh Halia be x

if that is so, then what are your number of cups of coffee?

[ruby stag]

I don't know

[brave crater]

read the question for yourself

and try answering that

[ruby stag]

3x?

[brave crater]

sweet

now, let your cost of 1 cup of coffee be y

then what's the cost of the cup of Teh Halia?

[ruby stag]

y+90¢

[ruby stag]

.close

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[glacial glacier]

can anyone help me with (c)

[glacial glacier]

i’m not sure how to work out the area

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@glacial glacier Has your question been resolved?

[odd terrace]

.status

[glacial glacier]

what does .status do

[devout valley]

!status was probs what was intended

[full forum]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[frozen maple]

Bu sunucu ID'sine ait elimde hiçbir bilgi yok.

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[glacial glacier]

I don’t know where to begin I’ve managed the first 2 parts but not sure about part c

.close

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[loud hull]

[\text{Find interval of convergence}]
[\frac{(-1)^{n+1} (x-3)^n}{n}]
[\frac{x-3}{1} < 1]
[x<4]
[\text{c = 3, roc = 4}]

[glossy valve]

dopediscorduser

[loud hull]

Does this look right?

[gritty rose]

You lost absolute value signs from the very beginning

[loud hull]

So not right

[\text{Find interval of convergence}]
[\frac{(-1)^{n+1} (x-3)^n}{n}]
[|\frac{x-3}{1}| < 1]

[glossy valve]

dopediscorduser

[loud hull]

?

[gritty rose]

No. You were also supposed to take a limit

The n can't just magically disappear

[loud hull]

Alright one sec

[loud hull]

[\text{Find interval of convergence}]
[\frac{(-1)^{n+1} (x-3)^n}{n}]
[\lim_{n \to \infty}|\frac{\frac{(-1)^{n+1} (x-3)^{n+1}}{n+1}}{\frac{n}{(-1)^{n} (x-3)^n}}|]
[\lim_{n \to \infty}|\frac{x-3}{1}|]
[|x|<4]
[\text{c = 3, roc = 4}]

This was my thought process

[gritty rose]

Still no

You lost absolute value signs here
|x - 3| < 1

[loud hull]

Alright

[glossy valve]

dopediscorduser

[loud hull]

Am I misunderstanding? Or could you just add 3 to the left side to get -3 out of the abs val?

[gritty rose]

Yes you're misunderstanding absolute value signs

https://tutorial.math.lamar.edu/classes/alg/solveabsvalueineq.aspx

[loud hull]

Hmm I see

[gritty rose]

?

[loud hull]

I was doing it incorrectly

[|x-3|<1]
[-1<x-3<1]
[2<x<4]

I have to admit I don't see how this helps solve the problem?

[gritty rose]

Solve for x

[glossy valve]

dopediscorduser

[loud hull]

Okay so thats our convergence interval?

[\frac{(-1)^{n+1} (2-3)^n}{n} = \frac{(-1)^{n+1} (-1)^n}{n}]
[\frac{(-1)^{n+1} (4-3)^n}{n} = \frac{(-1)^{n+1} (1)^n}{n}]

[glossy valve]

dopediscorduser

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[torn jolt]

what does the modulus X < 2 mean

[torn jolt]

I got to this

But im not sure what the modulus x<2 part means

[devout valley]

General binomial expansion will only be valid for values between -2 and 2 exclusive

You can't put e.g. x=3 into that expansion and get something valid

[torn jolt]

okay so what is the correct answer 😭

it is saying me to " give your answer in its simplest form"

[devout valley]

Assuming you've applied the binomial expansion right, your answer seems fine

That would be the simplest form (basically it doesn't want your final answer like that)

Basically remember how the geenral expansion says something like this (and also note the restriction on x, that |x| < 1, replacing x with -x/2 gets you the |x| < 2 restriction)

[torn jolt]

right

so that thing is just there for no reason rlly and i am fine just putting what i put

or do i need to add a for x <2 on the end of it

[torn jolt]

like they did on here

but with 1

?

[devout valley]

You can just give the answer you wrote as the expansion, you don't really need to restate it as they told you, but if they asked for the range of values for which your answer is valid you'd have to be able to come up with |x| < 2

[torn jolt]

oh okay

thank you

[devout valley]

all good

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[abstract bay]

I need to calculate the angle of the 2 lines

[mighty ridge]

you're looking for theta where a and b are the respective directions of the lines

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[devout turtle]

This is the question, I have got to a stage where I am not entirely sure what to do as the powers are all negative.

[devout turtle]

I assume I have to divide by something but not sure what.

[twilit leaf]

Multiply byx^2

[devout turtle]

Would that not keep the negative power?

[twilit leaf]

Oh right

x^4

[devout turtle]

Doesn’t really make sense to me

[twilit leaf]

If you multiply by x^4 you get a quadratic

[devout turtle]

What would let’s say 3x times 3x^-4 equal?

[twilit leaf]

3x × 3x^-4 = 3x × 3/x^4 = 9/x^3 = 9x^-3

[devout turtle]

I think I mean to say x^4 times by 3x^-4

Meant

The powers would need to be postitive for a quadratic right

[light mortar]

Can you multiply both side by x^4

The expression will be 3-2x-x^2=0

Now solve the equation

[devout turtle]

I’m actually stupid

So x^4 x 3x^-4 goes to 3 right

[devout turtle]

Thank you, I had a mind blank on multiplying powers

[light mortar]

No worry, it's fine.

[twilit leaf]

All good, it happens

[devout turtle]

The nature would mean if they are minimum/maximum points, correct?

So I’d do the second derivative

And subsititute in my x values

From the quadratic

[light mortar]

Correct

[light mortar]

Right

Check if it is coming +ve or -ve

[devout turtle]

I got the answer thanks

One was a max and one min

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[torn jolt]

answer to c is 4-x^2

[obsidian horizon]

What's f?

[torn jolt]

ah sorry

[obsidian horizon]

<@&286206848099549185>

Sorry I'm kinda busy atm

[spice prism]

js ping helpers like he did

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[foggy pollen]

These 3 confuse me very much

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@foggy pollen Has your question been resolved?

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@foggy pollen Has your question been resolved?

[foggy pollen]

Worked for the first one only

I need the answer for the third

Hello?

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[torn jolt]

How can I better understand that representing this problem via a quadratic equation x(x + 1) = 132 that there is more than one solution?

[red swallow]

Get this into the form ax^2 + bx + c = 0, and then by using the quadratic formula, you’ll obtain two solutions

[torn jolt]

ok

interesting

math works wonders

[red swallow]

Or did you mean you want some intuition on it?

[torn jolt]

intution

[red swallow]

If you want a more graphical understand, you could graph this

Graph x(x + 1)

Graph y = x(x+1)

And see where y = 132

You’ll be able to see that, after the first solution, the quadratic eventually shifts directions and starts heading towards 132 again

[torn jolt]

yeah i see that hmm

it just intrigues me

how it all comes together

theories and all

amazing.

[red swallow]

Yeah it is fascinating

[torn jolt]

thank you

[red swallow]

I hope I was able to help somewhat 😅

For further reading you could look at the fundamental theorem of algebra and/or derivation of the quadratic formula

Like completing the square and stuff

[torn jolt]

yea im studying factoring right now

[red swallow]

Nice, best of luck 😎

[torn jolt]

ty

bye

.close

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[tribal gull]

Suppose {M_i} is a finite set of invertible and diagonalizable matrices that pairwise commute. I know this means that every pair of matrices is simultaneously diagonalizable, but how does this imply that the whole set is simultaneously diagonalizable?

[tribal gull]

like if Q diagonalizes M_i and M_j and R diagonalizes M_j and M_k, how do I know that Q=R?

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@tribal gull Has your question been resolved?

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@tribal gull Has your question been resolved?

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@tribal gull Has your question been resolved?

[tribal gull]

.close

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[tranquil rivet]

!status

[full forum]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[tranquil rivet]

1

[compact briar]

Induction maybe?

[brave crater]

2^n = 2^(n-1) + 2^(n-1)

2^(n-1) = 2^(n-2) + 2^(n-2)

2^(n-2) = 2^(n-3) + 2^(n-3)

follow the pattern

you'll get 2^n - 1 = 2^(n-1) + 2^(n-2) + 2^(n-3)... 2^(n-n)

[tranquil rivet]

it was supposed to be \prod not \sum

[brave crater]

oh I read it wrong then

is it 2^2n or 2^2^n?

[tranquil rivet]

2^2^n

my latex is being so weird

ik induction should be used here

but im new to induction so im not sure what to do

<@&286206848099549185>

[compact briar]

Lets do induction: We start by proving the equation for n=1:

$R_1 = 1 + 2^2 = 5$ by definition

[glossy valve]

Gapi

[compact briar]

$R_1 = 2 + R_0 = 2 + 1 + 2^1 = 2+1+2=5$

[glossy valve]

Gapi

[compact briar]

So we have proven that the statement is true for n = 1.

[tranquil rivet]

its meant to be prod not sum

[compact briar]

Its a product of only one thing, which is $R_0$

[glossy valve]

Gapi

[compact briar]

And then you calculate $R_0$ by the formula

[glossy valve]

Gapi

[compact briar]

You can try to prove it for $n \geq 1$ on your own.

[glossy valve]

Gapi

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@tranquil rivet Has your question been resolved?

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[fresh granite]

I am not sure how they got from step 1 to step 2

[fresh granite]

context

[tired birch]

They just took 4 common?

ok wait

You're not allowed to use an already occupied text channel

@fresh granite I'm here to explain

[fresh granite]

What do you mean by that?

[tired birch]

Are you only confused in the pic that you sent?

$\frac{4^n}{3^{n-1}}=4\frac{4^{n-1}}{3^{n-1}}$

[glossy valve]

diaas_(yt)

[fresh granite]

Since its a geometric series, were trying to find a common ration r (being 4) and something to multiply by.

[fresh granite]

yeah, I dont know the algebra behind this

[tired birch]

Bro

$4^n=4 \times 4^{n-1}$

[glossy valve]

diaas_(yt)

[tired birch]

ik that

@fresh granite got it?

[fresh granite]

.close

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[torn jolt]

Find the domain of $\log_7(21-4x-x^2)$

[glossy valve]

Matt

[torn jolt]

How do I find the domain of this?

[fair oracle]

when is this undefined?

[torn jolt]

I think you solve for $21-4x-x^2>0$ but im not sure how to solve that

[glossy valve]

Matt

[torn jolt]

when its < 0

[fair oracle]

what about zero?

[torn jolt]

wdym? i just know you cant plug in 0 into a logarithm

or anything less than for that matter

[fair oracle]

you said, it's not defined when <0 right?

[torn jolt]

yes

[fair oracle]

so, what if it's exactly zero ?

the input

[torn jolt]

also undefined

[fair oracle]

exactly, so solve <=0

[torn jolt]

im not sure how to solve that tho

$21-4x-x^2>=0$

[glossy valve]

Matt

[fair oracle]

21-4x-x^2<=0 right?

[torn jolt]

yes

[fair oracle]

$21-4x-x^2 \leq 0$

[glossy valve]

.doc

[fair oracle]

can you factor ?

the left side?

[torn jolt]

im not sure, maybe?

would this be possible to do with the quadratic formula?

[fair oracle]

maybe multiply by a negative and make it look good

$x^2+4x-21 \geq 0$

this looks good?

[torn jolt]

yes but why did you do that?

[glossy valve]

.doc

[fair oracle]

maybe its easy to see the factors now

look for two numbers, whose sum is 4 product -21

[torn jolt]

7 and -3

i think

right?

[fair oracle]

yes, so (x+7)(x-3) works?

[torn jolt]

wait no

[fair oracle]

try mutliplying?

[torn jolt]

oh yeah it does

[fair oracle]

yep

[torn jolt]

usuallly you have to

expand the middle term

and factor by grouping whenn doing that

right?

oh wait nvm the coefficient is 1 its all good haha

[fair oracle]

but, i would recommend you to master this sum and product trick

[fair oracle]

it's sort the first step we try for solving a quadratic

[torn jolt]

ok sounds good

ok so i know the factors are (x+7)(x-3), whats the nenxt step

[fair oracle]

$(x+7)(x-3)\geq 0$ right?

[glossy valve]

.doc

[torn jolt]

yes

[fair oracle]

now it should proceed

[torn jolt]

would i just now solve the inequalities individually for each factor?

$(x-3)\geq 0$ and

[glossy valve]

Matt

[torn jolt]

$(x+7)\geq 0$

[glossy valve]

Matt

[torn jolt]

@fair oracle

[fair oracle]

sorry

i was away

[fair oracle]

try to make sense about this equation

it's saying product of two things are positive

right?

[torn jolt]

yes

that makes sense

[fair oracle]

what does that tell us?

[torn jolt]

it tells us that it cannot be negative?

[fair oracle]

either (x+7)>=0 AND x-3>=0 OR (x+7)<0 AND (x-3)<0

it says, either both of them are positive

or both of them are negative

[torn jolt]

ohh ok

[fair oracle]

does that make sense?

[torn jolt]

yes

[fair oracle]

can you solve further

?

[torn jolt]

yes i think i got it from here, thank you!

.close

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[cosmic scaffold]

its not math but help

[cosmic scaffold]

well sorta

but yeah help

[muted flare]

Well do you know what the various quantum numbers represent?

[cosmic scaffold]

no please explain it to me

[muted flare]

That would take a while

You should look over your notes

[cosmic scaffold]

oh wait yea I do

i just dont know how to get the answer

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[oblique skiff]

where did they get 3 and -5

im assuming theyre just plugging in random values

i did that here as well

with -2 and 4

f(a) =/= to f(b) for me

[torn jolt]

Shit I posted in the wrong one

So sorry

[red swallow]

They chose 3 and -5 as inputs because, when you plug them into the equation, you get the same answer

[oblique skiff]

well..thats not faitr

fair*

i dont have time to go through every combination

[red swallow]

Which tells you that it's not one-to-one, since two inputs correspond to the same output

[red swallow]

you could solve the equation with y = 0 and that would yield two values of x that will give 0, right?

you could solve teh equation with y = anything, showing that there are two answers for x in each case

[lilac pagoda]

2(x1 + 1)^2 - 4 = 2(x2 + 1)^2 - 4

(x1 + 1)^2 = (x2+1)^2

(-5 + 1)^2 = (3 + 1)^2

[oblique skiff]

whats this

[lilac pagoda]

like i just plugged in x1 and x2

[oblique skiff]

wdym x1

[red swallow]

or you could see that if you plug in x = 2, that would give the same answer as, say, x = -6, since inside the brakcet you'd get 4 and -4 respectively, both of which are 16 when squared

[lilac pagoda]

two arbitary numbers

[oblique skiff]

ah

[lilac pagoda]

you need to find some x1, x2 s.t. (x1 + 1)^2 = (x2+1)^2 and x1 != x2 for it to be injective

and you can just pick any x1 lets say 3

then you want to find (3+1)^2 = (x2+1)^2

or 16 = (x2+1)^2 and x2 != 3

and x2=-5 is a solution to that so its not injective

[oblique skiff]

oh i get it

thanks

[lilac pagoda]

u just need to show any example x1 and x2 to prove its not injective

np

[oblique skiff]

.close

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[lime ether]

The slope of the tangent to a certain curve at any point x,y is -(4/9)(x/y) and the curve passes through the point (3,2). Show that its equation is (x^2)/18 + (y^2)/8 =1