help-27

[near trout]

r is like the multiplicative factor

so the series is like 0.857142 + 0.000000857142 + 0.000000000000857142

so the first term is a, 0.857142

and then r is how much you have to multiply each term to get to the next one

so that's 0.000001, or 10^-6

not just 1/10

try it again with those numbers and it should work

[glacial lava]

Oh...

Right, yeah, actually I had:

a = 857142/10^6

Which I guess is the right 'a'

But I didn't do the 'r' term right

[near trout]

yes

yep

ok so where you went wrong, easy mistake, is:

so you saw 1/10 in here and went 'r is 1/10'

but there's actually a 6 in the exponent there

[glacial lava]

Ohhh

So that makes it 1/10^6 as the 'r'?

[near trout]

yes exactly

[glacial lava]

I see

I definitely didn't think like that

Let me see if I get it right now

[near trout]

yeah easy mistake

[glacial lava]

Yup, it worked now

Thank you!

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[maiden zinc]

How can I show that?
All the variable €Z
I think that we need to use something about complex number (cause that the chapter I'm doing)

[near trout]

what are a, b, c, d, m and n

this is impossible without knowing those

[maiden zinc]

I said that all the variable €Z

[stone stump]

Consider these as the squares of absolute values of some complex numbers

[maiden zinc]

Hum

[near trout]

that doesn't help

it's still impossible, right? unless we know more about a, b, c, d, m and n

[maiden zinc]

I mean relative intzger

[stone stump]

I assume the question is "show that for all a,b,c,d in Z there exist m,n in Z with ..."

[maiden zinc]

Yes

[near trout]

hmm

ok now it seems possible

[stone stump]

Good. What is the square of the absolute value of the number x+iy?

[maiden zinc]

x^2 +2xiy - y^2 ?

[restive river]

square of absolute (!) value

|x+iy|²

[maiden zinc]

= | (x+iy)^2|?

[restive river]

whats the definition of |x+iy|?

[maiden zinc]

= |x| + |iy|

Ah ok

[stone stump]

No

[maiden zinc]

= sqrt(x^2 + y^2)

[restive river]

that

so |x+iy|² = x² + y²

[maiden zinc]

Yes

[restive river]

hope you can see what to do from there

[maiden zinc]

I can transform the expression with a b c and d to |z1|^2 × |z2|^2

[restive river]

yeah

and the square and abs then have some nice properties

[maiden zinc]

I'm blocked

[restive river]

now whats z1 and z2?

no need to even introduce these variables actually

[maiden zinc]

I don't see

[restive river]

just
(a²+b²)(c²+d²) = |a+bi|² * |c+di|²

and |z1|² * |z2|² = |z1*z2|²

[maiden zinc]

Ok i didn't know that

Sry but I'm still blocked x)

[restive river]

just expand it to get a number in the form of x+yi

[maiden zinc]

Is it that?

[restive river]

yeah

[maiden zinc]

Ok finally x)
Thx a lot

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[frosty dove]

This is what I need help with

[devout snow]

Please don't occupy multiple help channels.

[coral pecan]

@frosty dove ^

Close thism

[frosty dove]

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[runic pelican]

Hello

[runic pelican]

I need help

John is riding his motorcycle. His odometer shows that his motorcycle has 1345 miles on it. After riding 3.5 hours, Andrew’s odometer shows 1407 miles. What was Andrew’s average speed for his ride?

Do I use V=D/T?

[long kettle]

Yes

[runic pelican]

Could you help me figure it out?

[long kettle]

What's the distance he travelled?

[runic pelican]

Who?

[hallow prism]

John

[runic pelican]

1345

[hallow prism]

no, what is the distance he traveled during the 3.5 hour ride

[runic pelican]

Oh

4,707.5?

Divide that by Andrew?

[hallow prism]

wait

[runic pelican]

4,707.5/ 1407?

[hallow prism]

oh

it has no solution, since we dont know what was andrews odometer at the start

[runic pelican]

OH

[hallow prism]

and johns odometer info is useless, since it is unrelated to the quesiton

[runic pelican]

Dang should've known

[hallow prism]

are you sure there are those two names?

[runic pelican]

Yeah?

[hallow prism]

if all those names would be the same it would make sense

in that case your teacher messed up and changed name of the person while making this

everybody makes mistakes

[near trout]

what

lmao

that's hilarious

[hallow prism]

So while there is no real answer, i am not sure your teacher would be happy with that

[long kettle]

What?

Oh

[hybrid snow]

You assume that his odometer has 1345 on it, and his ride was the 3.5 hour excursion

[near trout]

no

the problem is we only know john's odometer to begin with

not andrew's

[hallow prism]

yeah xd

[hybrid snow]

But who is Andrew

[near trout]

exactly

[hybrid snow]

Lmao be a smart ass and say "who the fuck is Andrew?"

[runic pelican]

LOL

[hybrid snow]

But let's say Andrew and John are the same person and the only reason they changed names was because they're running away from the police under a fake identity

[runic pelican]

Oh true

[hybrid snow]

Then you can find out John/andrew's average speed

[runic pelican]

But what did John do?

[hybrid snow]

Unspeakable things

[hallow prism]

Maybe he was speeding

[hybrid snow]

Recall that average speed is (distance)/(time)

The distance, in your case, is the change in mileage

[runic pelican]

How do I close the room?

[hallow prism]

.close

[runic pelican]

Oh thanks!

.close

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[loud beacon]

I am wondering if I did the negation right :

,tex \exists \epsilon > 0 s.t. \forall \delta > 0, \exists c \in R s.t. \neg P(x)

[woven radish]

Fluffy

\exists \epsilon > 0 s.t. \forall \delta > 0, \exists c \in R  s.t. \neg P(x)
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.57 \exists
             \epsilon > 0 s.t. \forall \delta > 0, \exists c \in R  s.t. \ne...
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

LaTeX Font Info:    Calculating math sizes for size <14> on input line 57.
LaTeX Font Info:    Trying to load font information for U+msa on input line 57.```

[devout snow]

@loud beacon Has your question been resolved?

[loud beacon]

<@&286206848099549185> i was wondering if i could get some help with this question?

[devout snow]

@loud beacon Has your question been resolved?

[loud beacon]

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[restive river]

Stuck on this linear algebra problem

[restive river]

a hint would be nice

thats all

[crystal rune]

can you show the question as is?

[restive river]

that is the question as is

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[restive river]

hello

[restive river]

can someone please help me understand how this is wrong?

is it because i missed absolute value? and if so why does it matter?

[crystal rune]

Could be the absolute value or maybe it's the c, keep trying

But absolute value is sometimes put to remind us that for ln(x), x > 0 since ln(x) is not defined for x<=0

[devout snow]

@restive river Has your question been resolved?

[restive river]

It says to include a capital plus c and I did

It’s weird lol

[devout snow]

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[balmy carbon]

how to i occupy this channel

[balmy carbon]

I guess that worked

how can I change this into Gaussian elimination

also how do i do Gaussian elimination

wait never mind I got it

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[short iron]

[sin^-1 (B) = \frac{2.02}{2}]

[woven radish]

dopediscorduser

[short iron]

For problem 17 finding angle B

What am I doing wrong here?

[csc(B)=\frac{2.02}{2}=1.01]
[csc(A)=\frac{2.02}{0.2835}=7.125]

[woven radish]

dopediscorduser

[short iron]

This is clearly wrong because

1.01 + 7.125 != 90

[tall knoll]

Why do they have to add to 90?

[devout snow]

@short iron Has your question been resolved?

[short iron]

Because it’s a right triangle

And angle C is already 90

[tall knoll]

Right the angles need to add up to 90, but their cosecants need not follow such a rule

[short iron]

Maybe I just don’t understand

I was taking the cosecant to solve for the angle?

[tall knoll]

You're taking the cosecant to solve for the cosecant of the angle

Once you get your ratios (1.01 and 7.125 in your case), you take the inverse cosecant to find the angles

[short iron]

Inverse cosecant would just be the sin?

,w sin(1.01)

[woven radish]

Results provided by WolframAlpha

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Requested by dopediscorduser#4906

[tall knoll]

No

Csc(x) = 1/sin(x)

They're reciprocals, not inverses

[short iron]

Okay

It’s very clear I need to review the basics here

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[eager crystal]

I said and for this one

But the answer sheet said or

-4 or 1/2

When can you say “and” and when do you have to say “or”?

[midnight dirge]

ye

its

or

it cant take both values at once

[eager crystal]

Why not

[thin gust]

x is only 1 variable

1 letter only

[eager crystal]

Im saying both answers are correct

[thin gust]

or and and have different like

[eager crystal]

So i put and

[thin gust]

meanings

[eager crystal]

Oh

[thin gust]

if u put and

it means

how do I exp this

[midnight dirge]

ok its like

[thin gust]

@midnight dirge

[midnight dirge]

x(x-1) = 0

so

from this

u can only conclude

x = 0

OR

x-1 = 0

u dont know if they both are 0

they can be both 0

but they might not be

so u cant conclude that

[midnight dirge]

im

bad at dis

but

[eager crystal]

Oh

[midnight dirge]

i feel i saw snow explain this b4

or smt similar haha

[thin gust]

snowy

[midnight dirge]

got it?

[eager crystal]

So since you don’t know the actual x

[midnight dirge]

so much snow

[thin gust]

explain to snowball

[eager crystal]

You’re kinda being safe by saying x?

[thin gust]

like

for eg

u got answer as 4

-1/2

[eager crystal]

Cuz it’s like one or the other

[thin gust]

plug in x=4

does it give u 0

[midnight dirge]

yep

[thin gust]

🥳💪

goddess is the new snow

❄️ Jr.

[midnight dirge]

another eg is

x^2 > 0

u can conclude that

[eager crystal]

So the mathematical term of and is different from the general and we use in english?

[midnight dirge]

wait nvm i shld use a,b

er

[thin gust]

same

[midnight dirge]

i think its quite similar to english

its only or

where if both is true

or is still true

[eager crystal]

When do you use and then,

?

Can you give me an example?

[midnight dirge]

um

i feel

[thin gust]

for eg

errr

[midnight dirge]

its kinda rarely used till later

well

except in questions maybe

they might have a

for all x,y such that x>0 and y>x

[thin gust]

-1+x<2x<5x+9

smth like rhat

[midnight dirge]

oh thattoo

hahaha

[thin gust]

i made that up

idk if it’s solvable

but ya

[midnight dirge]

u

shlda just used nice numbers then

like

[thin gust]

-1+x <2x 2x<5x+9

[midnight dirge]

1<x<3

[thin gust]

HAHAHA

[midnight dirge]

so we can conclude like

[thin gust]

YES

[midnight dirge]

x<1

and

x<3

[thin gust]

@eager crystal

1<x<3

[eager crystal]

So for example in x^2=25 you don’t know what x is so you’re saying it can be either 5 or -5

So thats where or comes from

[thin gust]

x >1 AND x < 3

[midnight dirge]

i guess that work

[midnight dirge]

ahhaa

[thin gust]

HAHAHA

[thin gust]

yesh

[eager crystal]

Cool thank you!

[midnight dirge]

ayy

[eager crystal]

I think ive been using and not knowing i was wrong lmao

[thin gust]

[eager crystal]

I got it now tho

[thin gust]

yayayaya

[midnight dirge]

i

did that too

[eager crystal]

Thanks again

.close

[midnight dirge]

hahhaa

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[midnight dirge]

❤️

[devout snow]

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[worn surge]

I dont know how to do this question. 5. Find the vertices of the region that contains the solutions to the system of linear inequalities

[grim agate]

!15m

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[worn surge]

<@&286206848099549185>

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[latent bay]

how do you write this in LaTeX but such that they're actually on top and bottom of the sigma instead of to the side?

[sand thorn]

display

Use $$

Whoops

[tall knoll]

$\sum_{i=0}^n$

[grim agate]

$\sum_{i}^n =0$

[woven radish]

Steakanator

[latent bay]

thank you!

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[unreal goblet]

I don't really have a visual for this question

So there is a square, and there is a circle that fits perfectly into the square. From one of the corners to the opposite side corner of the square is 4 units. what is the area of the circle?

[lunar harbor]

Note that the diameter is the same as the side length, which you can find ||using the diagonal's length||

[lunar harbor]

^

[near trout]

if you don't have a visual, then remember that you can literally always just draw a really crappy one

[lunar harbor]

touches but drawing exact is hard lol

[unreal goblet]

oh right.

[lunar harbor]

doesn't need to be perfect

just good enoguh to get the idea across

[near trout]

here we go

it's beautiful

[lunar harbor]

I love curvy squares

[unreal goblet]

crappy drawing

[unreal goblet]

halp...?

[lunar harbor]

is there some hidden meaning I'm unaware of lol

oop

[unreal goblet]

?

[lunar harbor]

🤷‍♂️

[unreal goblet]

is that question even solveable?

[lunar harbor]

yeah

[devout snow]

@unreal goblet Has your question been resolved?

[devout snow]

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[long snow]

can someone send an explanation for the question?

??

[devout snow]

@long snow Has your question been resolved?

[restive river]

The value on the 72 row is 25912 and the value of the 15th column is 11

i dont understand what the question wants us to do with those values as it was not specified.

[long snow]

should there only be one value?

at (72,15)

[long snow]

how did you get 2?

[restive river]

well think about it logically. It is a pattern that repeat after after the 5th row so the 70th row will be on the 5th row, 15 and the 71 will be 1241011, so the 72th row will be 25912

the same for the column. There are 5 columns repeating so the 15 column will be 11

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[devout snow]

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[restive river]

hi is it possible ask discrete amthematic question here?

[long kettle]

You can ask any math question here

[sonic smelt]

Yes, anything related to maths

[restive river]

ok i will make it

<@&286206848099549185>

anyone here

https://tenor.com/view/nervous-sign-of-the-cross-praying-hoping-nico-santos-gif-21421248

[devout snow]

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[echo turret]

seems im a bit stuck on this one

[prime egret]

What have you tried?

[echo turret]

various different methods but couldn't find a method specific to this sort of question?

[prime egret]

What various methods have you tried?

Have you worked out fg(x)?

[echo turret]

Nvm thanks I got it 🙂

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[jaunty gazelle]

Need help solving this

[jaunty gazelle]

I know I would do -log_4(2sqrt(2)), right?

[grim agate]

And it would be nice if you split the log into fractions

[jaunty gazelle]

Would I flip the base of 1/4 and turn it into 4/1?

[grim agate]

Uhh no

Thats what i mean

[jaunty gazelle]

I thought I’d have to do something with the base first..

[grim agate]

Well you can do that if you want

The thing you mentioned above

[jaunty gazelle]

So like this?

[grim agate]

Yes, but ur 2nd line it is wrong

[jaunty gazelle]

Why?

[grim agate]

Because the negative sign

[jaunty gazelle]

I thought i was suppose to me, the log negative?

[grim agate]

Oh no you did correct in following lines, just remove the second one

[jaunty gazelle]

Wait but why. That’s the line where I convert the log into negative

[grim agate]

Because what it says is log_(1/4)(..)= -log_(1/4) (…)

So its saying a = -a

Umm do you get it?

[jaunty gazelle]

No. Am I not suppose to make the entire expression negative?

[grim agate]

Uhh, no the property is $\log_{\frac 1b} a = - \log_b a$

[woven radish]

Deep. (kawaiiCat for emoji)

[grim agate]

And what you’ve written in 2nd line is $\log_{\frac 1b} a = - \log_{\frac 1b} a$

[woven radish]

Deep. (kawaiiCat for emoji)

[grim agate]

But 3rd line is fine

Umm do you get my point?

[jaunty gazelle]

So the fact that I treated b as a fraction instead of a whole number is why it’s wrong?

[grim agate]

No,

I know I would do -log_4(2sqrt(2)), right?
this is correct

But -log_(1/4) (2sqrt(2)) is not correct

That’s what i’m trynna say

[jaunty gazelle]

What’s the difference between the two? Is it because the 1/4 is it in parenthesis?

[grim agate]

Uhh how do i explain

Wait

If you just look at these two do you think equality sign holds true?

[jaunty gazelle]

Yeah. Does it not?

[grim agate]

So is $\log_{\frac 14} a = - \log_{\frac 14} a$

[woven radish]

Deep. (kawaiiCat for emoji)

[jaunty gazelle]

No they’re different

[grim agate]

…

Thats the point

The second line is wrong, because they are different

[jaunty gazelle]

So then the third line is also wrong…

[grim agate]

No

[jaunty gazelle]

Because it’s different from first

[grim agate]

No

They are same in fact

Well

Why don’t we do something

[grim agate]

Lets leave this thing aside

Okay?

[jaunty gazelle]

How is it the same when my second line is also my third

[grim agate]

Hmm, ur second line is not same as third line, think why

[jaunty gazelle]

I’ll just leave this aside. Now that I have my fraction..

[grim agate]

Well anyways I think its better to leave that

Yes

[jaunty gazelle]

I’m trying to find a rule to do the next step. I still have to finalize the answer right?

[grim agate]

So we have $\log_{\frac 14} 2\sqrt 2 = \frac{\log_c 2\sqrt 2}{\log_c \frac 14}}$

[jaunty gazelle]

Yes

[woven radish]

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

[jaunty gazelle]

Well I have log_c(4)

But same thing oh

Ig*

[grim agate]

Yes

And what do we know, 2 sqrt(2) = 2 ^(3/2)

Is that correct?

[jaunty gazelle]

Yes

Would I use quotient power rule here?

[grim agate]

And 1/4 is 2^…?

[jaunty gazelle]

Well I said 4 is 2^2

[grim agate]

Yes, so 1/4 is 2^(-2) you get that?

[jaunty gazelle]

So it wouldn’t be 2^-2

[grim agate]

Hm?

[jaunty gazelle]

So log_c(4) wouldn’t be transformed into log_c(2^2)?

[grim agate]

Yes

And now we use power rule like you mentioned

So now will you be able to do it?

[jaunty gazelle]

log_c(2^3/2)-log_c^(2^2)?

[grim agate]

/

Where the divide by?

[jaunty gazelle]

We'd be dividing instead of subtracting?

wait

would it be log_c(2^3/2)/log(2^2)?

[grim agate]

log_c(2^3/2)/(-log_c^(2^2)?)

[grim agate]

The - sign

[jaunty gazelle]

okay

so we wouldn't be subtracting

[grim agate]

Now you use power rule

[jaunty gazelle]

we'd still be dividing..?

[grim agate]

Yes

Its this

[jaunty gazelle]

But if we're dividing how can power property happen?

let me check

[grim agate]

We use property separately for numerator and denominator

[jaunty gazelle]

wouldn't it actually be (-log_c)(2^3/2)/log_c(2^2)

[grim agate]

$\frac{- \log_c 2^{\frac 32}}{\log_c 2^2}= \frac{- \frac 32 \log_c 2}{2 \log_c 2}$

[woven radish]

Deep. (kawaiiCat for emoji)

[grim agate]

Yes, both are same, aren’t they?

[jaunty gazelle]

I didn't think so; but if they are okay. Let me try the next step and show you

[grim agate]

Sure

[jaunty gazelle]

So would it then become this?

[grim agate]

yes

[jaunty gazelle]

Really?

[grim agate]

and now log_c (2) will cancel out

do you see

$\frac{- \frac 32 \cancel{\log_c 2}}{2 \cancel{\log_c 2}}$

[woven radish]

Deep. (kawaiiCat for emoji)

[grim agate]

so you are left with -(3/2)/2

and that’s -3/4

that’s it

[jaunty gazelle]

online it says the answer would be 2/3 though

wait no

wrong one

you are correct

wow

[grim agate]

hmm 🤔

[jaunty gazelle]

you made it seem easy

but it isn't..

[grim agate]

i am glad you understood

[jaunty gazelle]

thank you senpai. Is there an online directory where I can find examples on how to solve logairthm questions?

[grim agate]

hmm, honestly i’m not quite sure, you could maybe ask in #book-recommendations or maths discussion channel

[jaunty gazelle]

I tried doing this out by myself. Does it seem right?

[devout snow]

@jaunty gazelle Has your question been resolved?

[devout snow]

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[restive river]

hello

[restive river]

can I get help with this question: An atom has a half life of 3 days. you start with 10g how many grams will remain after 4 days?

[devout snow]

@restive river Has your question been resolved?

[devout snow]

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[scarlet peak]

I have some differential equations, my answer doesn't match the books solutions, they didn't multiply tan x by sec x, I'm not sure if this is intentional or not, I missed this class so I'm not too confident on it to know if its a mistake

[north flame]

Why?

The derivative of sec(x)*y is sec(x)*dy/dx + (sec(x)tan(x))*y

[scarlet peak]

🤦‍♂️

Yeh

Ofc

Thanks for pointing that out

[north flame]

Np

[scarlet peak]

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[vague ibex]

can someone explain why sqrt(28) had to be factorised into sqrt(4(7)) in step 2-3? thanks

[vague ibex]

can you solve without doing that?

[main gull]

It's best practice to simplify radicals down

sqrt(28) can be simplified and should be simplified to that

[arctic field]

it's also asking for x

[vague ibex]

yeah, need to use t now, thank you all

[arctic field]

in terms of p +- q sqrt(r)

[vague ibex]

wait

if i skipped step 3

and did 6+sqrt(28) all /2

i get 3+sqrt(7)

[arctic field]

so at the moment

since you have t

and you're solving for x

what you do doesn't really matter

[vague ibex]

ohhh

thank you !

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[restive river]

Prove that $\sqrt{\frac{a^2 + b^2}{2}} \le max(a; b)$, where $max(a; b)$ is the greater number of the two.

[woven radish]

trololol !

[restive river]

What I tried is:

[stone stump]

That's false. Unless at least one of a or b is positive

[restive river]

oh right, forgot to mention that

yes, a and b are positive

both

Let $m = max(a; b)$

We now have:
$\sqrt{\frac{m^2 + \left(m -x\right)^2}{2}}$

[woven radish]

trololol !

[restive river]

I tried squaring both sides of the inequality, and also got rid of the fraction, and I got

$2m^2 -2mx + x^2 \le 2m^2$

[woven radish]

trololol !

[restive river]

i substracted $2m^2$, and got

[woven radish]

trololol !

[restive river]

$x^2 - 2mx \le 0$

[woven radish]

trololol !

[restive river]

and idk what to do from this point

[inland seal]

Notice that x != 0

Since x is positive.

So you can divide by x.

[stone stump]

You are overcomplicating it

[inland seal]

How?

[stone stump]

Just use this

[inland seal]

You will get ||x <= 2m|| which is always true

[restive river]

i dont understand

so

[stone stump]

a^2+b^2<=2m^2

Plug this into the sqrt

[restive river]

oh

🤦‍♂️

idk how i havent thought of this

thanks!

[stone stump]

Happens. Sometimes we miss the obvious

[restive river]

yeah

thanks, once again!

.close

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[short iron]

[-\frac{24}{25}^2+cos(\theta)^2=1]

[cos=\sqrt{1}-\frac{24}{25}]

[cos=\frac{1}{25}]

[woven radish]

dopediscorduser

[short iron]

What am I doing wrong here?

[prime egret]

You can’t square root a sum like that

$\sqrt{a+b} \neq \sqrt a + \sqrt b$

[woven radish]

Pure

[short iron]

Oh

[cos=\sqrt{1-(-\frac{24}{25})^2}]
[cos=\sqrt{\frac{625}{625}-\frac{576}{625}}]
[cos=\sqrt{\frac{49}{625}}]
[cos=0.28]

[woven radish]

dopediscorduser

[short iron]

Does this look right?

Im gonna just close this

.close

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[paper girder]

How do I solve for x

[restive river]

why cant you use u = cosx?

[paper girder]

WHY

[restive river]

lol

[paper girder]

Sorry

[restive river]

can you maybe delete ur stuff

ok

.close

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[tight estuary]

Is my work correct?

[tight estuary]

.close

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[jaunty gazelle]

Tryna find the domain and range of this inverse function. The function I’m being asked to find the inverse and domain and range of inverse is f(x)=x-2/x+3

[jaunty gazelle]

The inverse I got was y=-2-3x/x-1, but what would the domain and range of that inverse be?

Wouldn’t the domain be the literal domain of the inverse? And what would the range be?

[devout snow]

@jaunty gazelle Has your question been resolved?

[bright burrow]

The range of the inverse would be the range of the inverse

[jaunty gazelle]

And the domain of the inverse would be the domain of the inverse right?

[bright burrow]

Yes

Also the domain of the inverse is the range of the original

And range of inverse is domain of original

[jaunty gazelle]

Wait what

[jaunty gazelle]

The domain of the inverse…is the range of the original?

I thought the domain of the inverse would be the literal domain of the inverse?

[bright burrow]

Yes

[bright burrow]

It is

[jaunty gazelle]

Oh

[bright burrow]

Which is also equal to the range of the original

[jaunty gazelle]

So for the example here the domain would be negative inf to 1, Union 1 positive inf? And the range would be the domain of the original function which is negative inf to -3 Union -3 to positive inf?

[bright burrow]

Which domain are you talking abt in the first sentence

[devout snow]

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[unreal saffron]

$-1 \leq 1 - \frac{2}{x} \leq 1 \iff \frac{2x - 2}{x} \geq 0 \wedge \frac{-2}{x} \leq 0$

[woven radish]

Gigabyte

[unreal saffron]

can somebody like

explain this

I thought they did something with the '1' at the left side

but that would result in

$\frac{x - 2}{x}$

[woven radish]

Gigabyte

[long kettle]

Split the inequalities. You have -1 ≤ 1 - 2/x and 1 - 2/x ≤ 1

Deal with each inequality seperately

[unreal saffron]

okay wait let me write that down

on the left side i get

$\frac{2 - 2x}{x} \leq 0$

[woven radish]

Gigabyte

[unreal saffron]

instead of

[long kettle]

Explain

Nvm

It's the same

[unreal saffron]

$\frac{2x - 2}{x} \geq 0$

[long kettle]

Multiply both sides by -1 and it's the same thing

[woven radish]

Gigabyte

[unreal saffron]

ohhh okay any reason they did this though

[unreal saffron]

instead of this

[long kettle]

To put the x in front, I guess

Positive coefficient and all

[unreal saffron]

aha okay thank you!!

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[unreal saffron]

really appreciate it

[devout snow]

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[restive river]

Prove that, for any positive real numbers $a$, $b$ and $c$, the following is true:

$\frac{b + c}{a} + \frac{a + c}{b} + \frac{a + b}{c} \ge 6$

[woven radish]

trololol !

[restive river]

I have no idea what to do.

[royal current]

Apply am≥gm for p/q cyclic

[restive river]

what

[royal current]

Um

U know AM≥GM?

arithmetic mean and geometric mean

[restive river]

what is am, what is gm, what is "p/q" cyclic

[restive river]

oh, yeah

[royal current]

And p/q cyclic is

p=a, b, c

q=a, b, c

Like a/b+a/c+b/a+b/c+c/a+c/b

[restive river]

i don't understand

[royal current]

Apply am≥gm in these terms

[restive river]

i don't see how i can get an arithmetic mean and geometric mean out of these terms

so that i can apply the formula

[royal current]

um wait

[restive river]

right, in order for $S \ge 6$ to be true, the arithmetic mean of these six fractions needs to be greater, or at least equal to 1

[woven radish]

trololol !

[restive river]

but what do i do from that point on

[royal current]

The arithmetic mean is what you needed

In the question

[restive river]

no

[restive river]

.

this is the question

[royal current]

It is the 6(arithmetic mean)

[restive river]

yes

but if we do that, i'd have to prove that the arithmetic mean is greater than, or equal to 1

and idk how to do that

[royal current]

Am≥Gm is a proof for that

For all positive real numbers this holds

[restive river]

where do we have a geometric mean in this inequality

i don't see any

[royal current]

When u multiply all the terms they just get cancelled

And it equals 1

Hence the (1)^(1/6) in the rhs

[restive river]

oh

OH

thanks!

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[restive river]

is something like tan^-2(x) considered arctan

[restive river]

or is it when its only -1

[topaz beacon]

I think blackpenredpen made a video about this

[inland seal]

I think you would consider it as 1/tan^2(x)

[restive river]

so like for instance

-2tan^-3(x) sec^2(x)

would be the derivative

[inland seal]

tan^-3(x) will be cot^3(x)

[restive river]

wait why

[inland seal]

cot(x) is 1/tan(x)

$\tan^{-3}(x) = \frac{1}{\tan^{3}(x)} = \cot^{3}(x)$

[woven radish]

RedstonePlayz09

[inland seal]

If you don't know what cot(x) is, just keep it as 1/tan^3(x)

[restive river]

so if it has -3 i should make the exponent positive and then take the derivative?

[inland seal]

Derivative?

No

$a^{-b} = \frac{1}{a^b}$

[woven radish]

RedstonePlayz09

[inland seal]

Exponent rules

[restive river]

yes

but then after you do that, then youd take the derivative?

[inland seal]

Are you taking the derivative?

[restive river]

yeah i was confused about taking the derivative of tan^-1(x) vs tan^-2(x)

arc tan is just -1 and has its own formulas?

[inland seal]

arctan(x) is often written as tan^-1(x), which is NOT 1/tan(x)

[devout snow]

@restive river Has your question been resolved?

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[devout snow]

Please don't occupy multiple help channels.

[rare roost]

I understand how to solve this problem, but I'm having a bit of a time actually writing it into a proof

The language is proving hard to work with

Gimme a sec and I'll post what I have...

I keep bumping up against this problem when I try to merge the two inequalities

The rightmost element in the inequality needs to be 1

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[wanton rapids]

26. Is z<0? x and y are unknown.
    (1) xz > yz
    (2) xy > yz

Sufficient information for the solution is obtained:
A in (1) but not in (2)
B in (2) but not in (1)
C in (1) together with (2)
D ¡ (1) and (2) separately
E not in either statement

[devout snow]

@wanton rapids Has your question been resolved?

[umbral cobalt]

Just check each case seperately

[wanton rapids]

What would each case be here

[umbral cobalt]

The answer choices

Check if each case is sufficient

If you don't know how to do that, just tell me :)

[wanton rapids]

I don't

[devout snow]

@wanton rapids Has your question been resolved?

[devout snow]

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[west girder]

Can anyone help me understand the steps to solve these problems?

[hybrid snow]

Erase low order terms

Legit just grab the eraser and erase low-order terms from the numerator and denominator

[midnight dirge]

hahaha eraser

just understand that

with landau u can get through this

[hybrid snow]

@west girder

@honest aurora yo you good?

You're typing a whole ass paragraph

[west girder]

ok so bottom would be x^3 -3x + 1

[honest aurora]

but if u need to show your work and understand why this works i would recommend dividing all parts by the largest power of x. so in example 32 divide all parts of both parts of the fraction by x^3, once u set the limit to infinity then u can see that the lower order terms go to 0

[hybrid snow]

Don't factor anything out

Literally just erase low order terms

Save yourself the time

[honest aurora]

telling people to just erase stuff and trust you wont exactly teach them very much im afraid

[west girder]

sorry i dont understand

[hybrid snow]

I'll show you as an example

[west girder]

erase low order terms

ok

[midnight dirge]

hahhaha

[hybrid snow]

The first one as an example

$$\lim_{x\to \infty}\frac{1-3x^3}{2x^3-6x+2} \equiv \lim_{x\to\infty} \frac{-3x^3}{2x^3}$$

[woven radish]

Umbraleviathan

[hybrid snow]

You're essentially just gonna keep the leading terms

For the most part, they're the highest-power term

[west girder]

wait why does that work?

[hybrid snow]

Imagine adding soemthing like 5 to "infinity"

It's still gonna be infinity

[west girder]

oh ok

[hybrid snow]

So we look at whichever term is gonna have the "biggest infinity" (which is an incorrect statement, but for conceptual purposes, I will still say it)

Because any pebble against a large ass infinity ain't gonna do anything

So we just keep whatever term is bigger

[honest aurora]

if u show the work in between its easy to see why this must be the case as x goes to infinity

notice how the limit was applied in the 2nd last equality

[west girder]

so all i do is divide everything by the biggest term

[honest aurora]

yes, which is why they are saying just erase all lower order terms

but u need to know why you can do that

erasing all does the same as what i just did but this way you can understand the logic behind it

[west girder]

ok

so -3/2

is that the answer?

[honest aurora]

yes

[west girder]

so its not infinity?

[honest aurora]

no it is not

if however the largest term in the top would be x^4 for example

then you would get infinity as once u apply the limit u essentially get 0 in the denominator after dividing by x^4 and applying the limit

[west girder]

cuz it would all be 0

cuz the 2 would be 0

[honest aurora]

exactly

[west girder]

ok

so for the nextg one

it would be x^2 over -x^3

[honest aurora]

doesnt matter whether u do negative or positive

or, actually u will get the opposite limit

so dont do negative, always positive

actually i think thats wrong lol. doesnt matter as long as u do it on both the top and the bottom. either way there is really no reason to do over -x^3 as doing it over x^3 does the same thing

[west girder]

ok

is the answer infinity?

cuz what do u do when its x^2 over x^3

[honest aurora]

reposting so i dont gotta scroll

so you decided that dividing by x^3 is the correct move right?

[west girder]

i think

[honest aurora]

ok, so what happens after you simplify

[west girder]

its just x^2 over x^3

[honest aurora]

aka remove as many x's as you can

and what is x^2 divided by x^3?

[west girder]

0?

[honest aurora]

dont apply the limit yet

just normal division

[west girder]

what do u mean by applying the limit?

its just root x right?

[honest aurora]

$\frac{x^2}{x^3}=?$

[woven radish]

Duh Hello

[honest aurora]

its very simple

not a trick question

[west girder]

root x?

[honest aurora]

not quite

$\frac{x^2}{x^3}=\frac{1}{x}$

[woven radish]

Duh Hello

[west girder]

am i stupid

i still dont get it

why is x^2 / x^3 = 1/x

[honest aurora]

because $x^3=x\cdot x\cdot x$ and $x^2=x\cdot x$ so $$\frac{x^2}{x^3}=\frac{x\cdot x}{x\cdot x\cdot x}=\frac{1}{x}$$

[woven radish]

Duh Hello

[honest aurora]

you just cancel out 2 x's

[west girder]

oh ok

[honest aurora]

i would recommend working on your algebra if this is something that confuses you

[west girder]

so 1/x is the end?

[honest aurora]

so posting again

the first term at the top

what is that after you have divided by x^3?

on 34

[west girder]

its 1/x

[honest aurora]

good, now what is the next one? following the same logic

[west girder]

its 1/x^2

[honest aurora]

correct, now next one we obviously cant cancel any x's

[west girder]

so its 0?

[honest aurora]

no, just keep it as -5/x^3

i will come to what i mean by applying the limit

in a bit

but first we go through simplifying all

so first on the bottom

the first one we also cannot cancel anything correct? so we just have 1/x^3

[west girder]

yes

[honest aurora]

what about the next one?

[west girder]

-2/x^2

[honest aurora]

good, and then the last one

[west girder]

1

or -1

[honest aurora]

good

so now since we are checking what happens when x goes to infinity we use this lim symbol to symbolize that

now what happens to each term when x does go towards infinity?

[west girder]

it becomes nothing

[honest aurora]

does all of them become nothing?

[west girder]

except -1

[honest aurora]

ok, so what do we have left?

[west girder]

uh

0

1/x

and root x

[honest aurora]

not sure why root of x keeps coming up for you

but going 1 at a time. we have $\frac{1}{\infty}=?$

[woven radish]

Duh Hello

[west girder]

ohhh

[honest aurora]

thats what applying the limit means

u essentially plug it in (in loose terms), technically we are just approaching it as infinity, hence why we call it a limit

[west girder]

ok

let me try

sorry im stuck again

idk where im at

[honest aurora]

this is the next "step". mathematicians will be angry at me for writing it like this, but this is essentially what u are doing

but what happens when u divide a finite number by an infinite number?

[west girder]

becomes 0

[honest aurora]

yes

so what does this become

[west girder]

0 / -1

[honest aurora]

yes

and what is that?

[west girder]

undefined?

[honest aurora]

why is that not defined?

type it into a calculator

[west girder]

just 0

[honest aurora]

yes, u divide 0 by something that isnt 0 and u get 0

and that is the limit as it approaches infinity

do you understand what happened here?

[west girder]

so we divided by x^3

i thnik so

[honest aurora]

u think u can complete the next question?

[west girder]

i think so

is it 0/1

[honest aurora]

how did u get to that?

[west girder]

i did everything divided by x

1/x - 2x^3/x

over

x/x + 1/x

then put the limit in

and got 1/infinity - 2infinity^2

over

1 + 1/infinity

[honest aurora]

u are talking about 36 right?

[west girder]

yes

[honest aurora]

so u divided by what?

[west girder]

i divided by x

should i have divided by x^3

[honest aurora]

x^3 is a larger power than x

[west girder]

ok so thats where i went wrong

i divided by a lower power

[honest aurora]

try again

[west girder]

-2 over 0

-2/0

[honest aurora]

good, what is that then?

[west girder]

undefined

[honest aurora]

yes, but for now we are just gonna call it infinity

now know that we have actually only done part of the problems

they ask for both infinity and negative infinity

and in this last one, since it tends towards infinity, we need to know whether its negative or positive infinity

but i think i might have to throw in my towel here. i believe you need to revisit some algebra here so that it can be accomplished easier. i believe it would take me quite some time to be able to properly explain to you exactly what it is the question is asking for

[west girder]

alright well i appreciate the help!

thanks alot!

[honest aurora]

no worries. good luck

[west girder]

.close

[devout snow]

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[silk ridge]

how do I factor this?

[silk ridge]

this is the answer

[near trout]

remainder theorem

2(3)^3 - 3(3)^2 - 27 = 0, so (x-3) is a factor

[silk ridge]

How could you tell right away that p(3) = 0?

[near trout]

i couldn't