help-38

[night zodiac]

Literally what I looked up lmao

[amber badger]

I used "why two's complement not used unsigned numbers"

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[lost silo]

can someone help me with this question

[long basin]

Are you gonna post it

[lost silo]

yea just wait

[long basin]

!status

[trim jolt]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[lost silo]

1

[long basin]

Start with understanding the domain of log(x) with any base, b

[lost silo]

the whole exercise seems easy enough, but its just this question that im stumped on

[long basin]

Therefore you can use the vertical asymptote to find h

[lost silo]

yep i know that

but how do i find k?

[long basin]

Plug in the two points for x and y

so you'll have a system of some sort

Luckily for 14a, they give you a pony whose y-value is 1

[lost silo]

so i use the asymptote to get h

[long basin]

And then (5,3)

Yes

[lost silo]

and then use one of the provided coordinates to get k

[long basin]

And then plug in (x,y) = given points

Yes

[lost silo]

how about the base

[long basin]

Find it simultaneously; there should only be 2 unknown variables

[lost silo]

is h 3 for the first one

because -3+3=0

so the equation is y=log_a(x+3)+k

[long basin]

h is -3

(x- (-3)) = (x+3)

[lost silo]

o

im at 3=log_a(5+3)+k

what do i do now?

@long basin

[long basin]

You need to do the time with the other point

[lost silo]

?

so 1=log_a(-1+3)+k

i still dont get how to solve for k or a tho

[long basin]

Here imma have to go for a bit but

https://www.tcc.fl.edu/media/divisions/learning-commons/resources-by-subject/math/trig-and-precalc/Finding-Logarithm-function-Using-Graphs.pdf

I found this

[lost silo]

let me see

wait its different to that in the book

<@&286206848099549185>

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[amber badger]

I'm going over my old quizzes for tomorrow's exam.

Why is #2 incorrect?

[digital bison]

how did you get that?

[amber badger]

plugged in the c

the method is on the right

but please if you know why it's incorrect, can you please just explain?

[digital bison]

ah.... well 4-10+6 = 0
and 4-6+2 = 0

[amber badger]

I don't have much time and really need to just re-do all the problems

[digital bison]

you need to factor the polynomials and reduce it
or use l'hopital's rule

[amber badger]

ah, i must have been extremely nervous to make such a simple mistake

i really hope this doesn't happen tomorrow

[amber badger]

#2 is -1, yes?

i just worked it out by hand

[digital bison]

sounds right

,w limit as x to 2 of (x^2-5x+6)/(x^2-3x+2)

[solid kiln]

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[amber badger]

ty

[amber badger]

can i cross cancel or lateral cancel here?

where the arrow is, i mean

[digital bison]

no

[amber badger]

neither?

[digital bison]

there's nothing to cancel except the thing you multiplied by

[amber badger]

these two?

[digital bison]

those aren't the same

[amber badger]

oh, right

[digital bison]

use the fact that the top is a difference of squares
(a-b)(a+b) = a^2 - b^2

[amber badger]

these two

[digital bison]

then some stuff should cancel, and you can plug in the limit again

the two numerators are multiplied together, not added. so those don't cancel

https://openstax.org/books/calculus-volume-1/pages/2-3-the-limit-laws
Check out example 2.18

[trim jolt]

@amber badger Has your question been resolved?

[amber badger]

did i multiply correctly here?

<@&286206848099549185>

,w limit as x to 1 of sqrt(2x-2)

[solid kiln]

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[digital bison]

looks right

[amber badger]

i think this is the last step and answer

[digital bison]

so what do you get?

[amber badger]

wait, no

i forgot to plug in c which is 2

here's the final answer:

correct?

[digital bison]

nope

[amber badger]

where did I err?

[digital bison]

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

[amber badger]

ah, I knew they would cancel out but I was nervous about switching out the signs

(-x+2) = (x-2)

yes?

[digital bison]

-(x-2)

[amber badger]

(-x+2) = -(x-2)

like this?

[digital bison]

yes

[amber badger]

correct, now?

[digital bison]

yes, now plug in your limit

[amber badger]

$-\frac{3}{4\sqrt{3}}$

[solid kiln]

Willow

[amber badger]

yes?

[digital bison]

almost

[amber badger]

ok, let me show my work

[digital bison]

x-1 became 2+1

and, just as a note, $2+2\sqrt3 \neq 4\sqrt3$

[solid kiln]

Zybikron

[amber badger]

oops, let me try again

[amber badger]

$-\frac{3}{2(1 + \sqrt{3})}$

[solid kiln]

Willow

[amber badger]

yes?

[digital bison]

no

[amber badger]

darn it! why do I keep making that mistake

[amber badger]

$-\frac{3}{4}$

[solid kiln]

Willow

[digital bison]

yes

[amber badger]

i'm going to do this problem again tonight and go to class early and do it again right before the big exam

[amber badger]

how about the problem below this problem?

i know i got it wrong because i froze up and could figure out what to do

square roots always intimidate me

[digital bison]

you can't just multiply by something.

You also don't need too

plug in 1

[amber badger]

it's indeterminate

[digital bison]

?

[amber badger]

so algebra is needed

$\sqrt{0} = indetermine$

[solid kiln]

Willow

[digital bison]

sqrt(0) = 0

[amber badger]

oh! so, indeterminate is only for fractions here?

but since sqrt(0) = 0, the limit is 0?

[digital bison]

there are non-fraction forms of indeterminate, but sqrt(0) isn't one of them

[amber badger]

so, I just should have written $\lim_{x \to 1} \sqrt{2x - 2} = 0$

[solid kiln]

Willow

[amber badger]

and that's it

yes?

[digital bison]

yep

[amber badger]

darn it

[digital bison]

maybe show you plugging in 1, but yep

[amber badger]

missed easy points

[amber badger]

oh, right. need to show work

[digital bison]

but I gotta go

good luck on the exam

[amber badger]

oh, ok

ty

[amber badger]

.close

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[weak vector]

I would like some guidance on this problem

[weak vector]

I understand that there are two parts to this, 8- and 8.

would the first part be found using

$lim_{b\to8}\left(cb+3\right)$

[solid kiln]

Forte

[weak vector]

Or idk, I might be completely lost

[burnt mulch]

$\lim_{b \to 8^{-}} f(b)=\lim_{b \to 8} (cb+3)$, yes.

[solid kiln]

Civil Service Pigeon

[burnt mulch]

Apply the same logic to the right hand limit

[weak vector]

I am used to solving limits where there is only one variable, how would I do so with both c and b?

[burnt mulch]

c is a constant, so it's also just a number 🤷‍♂️

[weak vector]

I understand that for b, I need to plug in values close to or equal to 8- but where do I start for c?

[burnt mulch]

just leave c alone ...

you'll form a linear equation to solve for c later on anyway

[weak vector]

For the sake of example, how would I go about this?

$lim_{b\to8-}\left(cb+3\right)=7.9b+3$

[solid kiln]

Forte

[weak vector]

I don't know if I'm on the right track

[burnt mulch]

Plugging in values rlly close is a good intro to limits

but after that

don't do it again

it's not rigorous and it's more like a demonstrative technique

just leave it as 8c+3, as I said earlier.

[weak vector]

Thats good to know

[weak vector]

I suppose the other one is c8^2-3

[burnt mulch]

You're mixing up $b$ and $c$ tbh

[solid kiln]

Civil Service Pigeon

[weak vector]

my bad

[burnt mulch]

wanna try again?

[weak vector]

^?

[burnt mulch]

Yes, so we can say $8c+3=8^2 c-3$. \ \ The algebra is trivial, I assume you can do that.

[solid kiln]

Civil Service Pigeon

[weak vector]

I see

c=6/56

[burnt mulch]

Now simplify and you're done

[weak vector]

thanks a lot for the help, I appreciate it

.close

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[burnt mulch]

👍

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[trim jolt]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[edgy sun]

1

I don't know where to begin

[burnt mulch]

Diagram?

[edgy sun]

Yeah I did a diagram

[burnt mulch]

Send it.

[edgy sun]

But I am not sure how is that going to help me

How can you find a smallest possible area from the information they tell you

I know how to do

cos or sine rule

but I am not sure how it helps me

[burnt mulch]

I'll take that as you're not going to send it so ig someone else can help 🤷‍♂️

[edgy sun]

.close

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[slender ocean]

Hey everyone, can someone please help me determine f’(x) using chain rule

[slender ocean]

I have done simpler questions but this one is more complex

,rotate

[vagrant marsh]

!status

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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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[slender ocean]

1

[vagrant marsh]

Ok, how many functions are composed?

[slender ocean]

Wdym

Dy/dx=du/dx * dy/du

[vagrant marsh]

A composition is a when a function is in another function

[slender ocean]

Ok

So for cos

I can do -sin^3

[vagrant marsh]

No

Youre getting ahead

[slender ocean]

Then how do I start this?

😬

I don’t have much time

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[plush axle]

Y2-y1/x2-x1

Find out the change of y between 0 and 4

Then divide it by 4

Find out the y value of 4 then find out the y value of 0. Then subtract it

Yeah

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[errant crag]

hi guys any idea how to do this one without using trig sub

[errant crag]

i tried u subbing many ways but i cant find one that works

[haughty oasis]

how about u = sqrt(x)?

[mortal breach]

looks like u=sqrtx+1 would be the "easiest" one

[errant crag]

i cant theres no du

[mortal breach]

wdym theres no du

[mortal breach]

yeah prob would work better

[errant crag]

1/2sqrt(x)

i dont see it

[haughty oasis]

dx = 2sqrt(x) du = 2u du

[errant crag]

wait let me try

oh ojk

thanks

.close

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[errant crag]

,rotate

[errant crag]

where did i go wrong ._.

[vagrant marsh]

2sqrtx != u

[errant crag]

oh

im stuipd thanks

.clopse

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[mortal granite]

Can anyone help me find hypotenuse?

[delicate bobcat]

what's this

[upbeat tundra]

y+4

[delicate bobcat]

aa

[upbeat tundra]

or + 9

[mortal granite]

Y

Is it possible to solve it?

[split chasm]

so which one is it...

[mortal granite]

It os y

[split chasm]

we can sorta see that the first char is y

what's the thing after the + sign

[mortal granite]

Next one is 4

[split chasm]

a 4, 9 or something else

try considering similar triangles

[mortal granite]

Wdym

Are they similar?

[split chasm]

justifiable with AA

[mortal granite]

I don’t get it

[trim jolt]

@mortal granite Has your question been resolved?

[granite valve]

still here ?

@mortal granite

[mortal granite]

Yes

[granite valve]

ok to start up let's divide our triangle into two

and let's call that line R

apply Pythagorean theorem in the whole triangle

[mortal granite]

I called it h

[granite valve]

wrong

$a^2 +b^2 = r^2$

so when we apply it in the whole triangle we'll get :

[solid kiln]

mino65

[granite valve]

a here is 2x

and b is 3x

and r is y + 4 + y which is 2y + 4

understandable till now ?

so we'll be left with :

[mortal granite]

How is r y + 4 + y

you meant c?

a square + b square = c square

[granite valve]

yeb

[mortal granite]

Ok

[granite valve]

you decided to call it h didn'tcha ?

[mortal granite]

.

I called the height H

[granite valve]

yeb

[mortal granite]

The line which goes in the middle

[granite valve]

yeye

any ways apply Pythagorean theorem on the bigger triangle

[granite valve]

$9x^2 + 4x^2 = \left( 2y + 4 \right)^ 2$

[solid kiln]

mino65

[mortal granite]

Ye ye

[granite valve]

go and expand the $\left( 2y + 4 \right)^ 2$

[solid kiln]

mino65

[mortal granite]

What can i do now?

[granite valve]

sorry for any delay my internet sucks

[mortal granite]

Its ok

[granite valve]

8y❌
16y⭕

it's 2 * 2 times 4

[mortal granite]

Oy yeah my bad

[granite valve]

so it's $4y^2 + 16y + 16$

[solid kiln]

mino65

[granite valve]

now make the equation equal to 0 by taking every thing into one side and changing the signs respectively

[granite valve]

we'll be left with $13x^2 -4y^2 -16y -16 = 0$

[solid kiln]

mino65

[granite valve]

very well

now call this equation(1) and keep it for later

let's head into the two small triangles

[mortal granite]

Ok

[granite valve]

apply Pythagorean theorem on the first one

[mortal granite]

To find the middle line?

[granite valve]

yeb

hello ?

[granite valve]

hmmm ... you got the way

but the answer is wrong

here step by step

for the first one

$h^2 + \left( y + 4 \right)^2 = 9x^2$

[solid kiln]

mino65

[granite valve]

===>$9x^2 -\left( y + 4 \right)^2 = h^2$

[solid kiln]

mino65

[granite valve]

tfq?

any ways

it'll be :

$9x^2 - y^2 - 16 -8y = h^2$

[solid kiln]

mino65

[granite valve]

for the second triangle

it's $y^2 + h^2 = 4x^2$

[solid kiln]

mino65

[granite valve]

so

[mortal granite]

I was following this one

[granite valve]

delete this junk

learn it by hand first and when you become a pro then go use these apps

[mortal granite]

Ok

[granite valve]

if you use them too much now you'll fall into stupid mistakes in the future

as inverting signs and so

any ways

$h^2 = 4x^2 - y^2$

[solid kiln]

mino65

[granite valve]

so the first equals to h^2

and the second one as well

so we equal them to each other

we'll get :

$9x^2 -y^2 -8y -16 = 4x^2 -y^2$

[solid kiln]

mino65

[granite valve]

and in the last

$5x^2-8y-16 = 0$

[solid kiln]

mino65

[mortal granite]

What should i do after?

[granite valve]

sorry

trash internet

now

it's the most important step

extract the x squared

$x^2 = \frac{8y+16}{5}$

[solid kiln]

mino65

[granite valve]

ayo

[mortal granite]

Hi

[granite valve]

now go and plug that x squared in our previous equation

$13x^2 -4y^2 -16y -16 = 0$

[solid kiln]

mino65

[granite valve]

you know quadratic right ?

[mortal granite]

Yes

[granite valve]

good

go ahead

[mortal granite]

Ok i will try

Thanks a lot brother

[granite valve]

you found it ?

show me your answer

you may missed something

[mortal granite]

Its 16

But i am very tired i need to sleep

[granite valve]

NO SLEEP !!! ONLY MATH FOR LIFE !!!

ayo just what is the value of y

your equation should look like this :

and this is your answer

well at least i hope we didn't make any mistake all the way long cuz that'll make me look ridiculous

any ways

it was a pleasure

[mortal granite]

👍🏻

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[glass basin]

https://cdn.discordapp.com/attachments/908075848752590918/1068345642826608742/image.png

[glass basin]

help

[quartz yoke]

!status

[trim jolt]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[glass basin]

@quartz yoke 1.

idk where to start

and this is due in an hour

[quartz yoke]

lets take care of a

[glass basin]

ok

[quartz yoke]

do you know what x to 6- means

(I'll leave the rest to you)

[glass basin]

no i dont

[vagrant marsh]

It means you are approaching from the negative side (aka the left)

[glass basin]

okay

[trim jolt]

@glass basin Has your question been resolved?

[glass basin]

no i got a 100% anyways

[errant crag]

W

[trim jolt]

@glass basin Has your question been resolved?

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[agile prism]

how do i solve this quickly?

[agile prism]

]

[frozen plover]

$\sin 2\theta = 2\sin \theta \cos \theta$

[solid kiln]

NEONPerseus

[frozen plover]

Use this fact

[agile prism]

ok

what nxt

[trim jolt]

@agile prism Has your question been resolved?

[frozen plover]

Write 4x in terms of 2x

[trim jolt]

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[calm saffron]

I know how to find volume through shells, slice, washer, and disk method but this word problems seems so impossible to me. Any help will be amazing

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[gritty burrow]

help

[gritty burrow]

help pls

[worldly monolith]

hint: ||its the area of triangle PDQ||

[gritty burrow]

bruh

i know that 😭

[worldly monolith]

so you know the solution since.. well, you have all the numbers you need for the formula

[gritty burrow]

i dont 😭

5 and 8 are not the side lengths

[worldly monolith]

$A=b\cdot \frac{1}{2}h$, where b is the base and h is the height of the triangle from the base

[gritty burrow]

and i cant use herons formula because i dont have the semiperimeter

[solid kiln]

Jigglyproff

[vagrant marsh]

Jiggly are you ok

[gritty burrow]

5 and 8 are not the side lengths

what are you talking about jiggly

[tepid rock]

Lol.

What time is it, jiggly?

[gritty burrow]

bro som1 help

[tepid rock]

,ti jiggly

[gritty burrow]

someone 😭

[solid kiln]

User cancelled member selection.

Jan 28, 2023, 8:59 PM

[tepid rock]

pumpkin

let AB = x

Let's find the side anyways.

Of the square.

Alright?

[vagrant marsh]

I have some pythagorean spam but im looking for a better way

[tepid rock]

Similar triangles

[vagrant marsh]

I was thinking about it

[tepid rock]

handy.

[vagrant marsh]

Ah got it

Take it away my friend

[gritty burrow]

how

i know that pab and pdq are similar, but putting it into x and solving like that gives me a quadratic

[tepid rock]

ABP is similar to DPQ

if it gives you a quadratic, then solve it?

Or can you not solve one?

[gritty burrow]

i can

but its not the correct answer?

[vagrant marsh]

What is DQ in terms of x

[tepid rock]

Your quadratic might be flawed. Show work.

[gritty burrow]

kk

so basically

x is = the shorter side leg

and the longer is x+3

right?

[tepid rock]

Label it.

[gritty burrow]

ok

[tepid rock]

No shorter longer play.

Well?

[gritty burrow]

yeah

[tepid rock]

Which side is x?

[gritty burrow]

the shorter leg of pdq

[tepid rock]

DQ?

[gritty burrow]

yew

[tepid rock]

Then say that!

[gritty burrow]

kk

[tepid rock]

Alright, that's your x. Cool.

Go ahead.

[gritty burrow]

x:5

x+3:x+8

so now would i have the equation of

[tepid rock]

x+3 is what?

[gritty burrow]

(x/x+5)=(x+3/2x+11)

x+3 is pd

is my equation correct?

[tepid rock]

Okay I want you to do something before you write anything with X's

First write what you're trying to write in terms of A B C D P Q

Alright?

And then, write the equation.

[gritty burrow]

ok

[tepid rock]

That'll make it easier.

[gritty burrow]

dq:ap=x:5

pd:ab=x+3:x+8

[tepid rock]

Seems good.

[gritty burrow]

ok

now i multiple by x+5 on both sides?

[tepid rock]

No?

[gritty burrow]

oh

x/5=x+3/x+8?

is my equation

[tepid rock]

yeah.

[gritty burrow]

so i multiply by 5?

[tepid rock]

Yes.

And (x+8) as well.

[gritty burrow]

x=5x+15/x+8

ok

x^2+8x=5x+15

x^2+3x-15=0

then i get x = -3 + or minus sqrt 69

divided by 2

[tepid rock]

Don't.

[gritty burrow]

oh

[tepid rock]

Solve the quadratic.

[gritty burrow]

dont solve the quadratic?

[tepid rock]

Yeah.

[gritty burrow]

oh

[tepid rock]

What does the question ask for?

[gritty burrow]

then what do i do

the area of pdq

[tepid rock]

Exactly.

It doesn't ask you about your x.

The area is pdq would be

x(x+3)/2

Yes?

[gritty burrow]

yes

[tepid rock]

(x^2 + 3x)/2

Yes?

[gritty burrow]

yes

[tepid rock]

x^2 + 3x is 15

!

[gritty burrow]

oh

OHHHH

THANK YOU

😄

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[smoky viper]

<@&286206848099549185> im doing the derivatives of an equation , and i think my answer is correct but the choice given as the answers are not the same as my answer

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[smoky viper]

am i doing something wrong ?

[ashen condor]

Nothing wrong.. Just try to factorise by 2x(2x...)

[smoky viper]

i dont get where the 32 comes from

[delicate bobcat]

2x^3 + 30x^3 = 32x^3

[smoky viper]

and where does the exponents go ?

why the 10 and the 9 disappeared?

[delicate bobcat]

he forgot about ^9 I guess

[smoky viper]

so the correct answer is B right

i just need it to factorise

[delicate bobcat]

ye

[smoky viper]

awesome, thank you soo much 🙂

.close

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[flat wing]

Can someone explain question 3A pls

[delicate bobcat]

what part don't you understand?

notice that g(k) = 5^(2k) - 6k + 8

they used it

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[kind fractal]

I'm having some issues with a Trigonometry problem

The problem requests the height of the tree
With the angle CAD being 18°
CBD being 35°
AB being 15m
ACD being 90°
And A B and C being on the same line.

I made a triangle with the 18° and the last side starting on the tip of the 35° angle indicated, used the tangent of 18° to calculate that side's length, but past that I'm quite lost. Can anyone help?

[kind fractal]

I'm sorry that the text is in Portuguese. I translated all the important stuff, rest isn't relevant.

Oh and lastly, the result needs to be rounded up to the unit (pretty sure that's how it's said in English), in other words no decimals.
I checked the result and it's supposed to be 9, though I'm not quite sure how.

<@&286206848099549185> anyone?

[trim jolt]

@kind fractal Has your question been resolved?

[kind fractal]

<@&286206848099549185>

[inland orbit]

I think you can make a system of equations

and solve for BC and CD simultaneously

[kind fractal]

What would the system be?

[inland orbit]

well could you use a trig function to relate them?

[kind fractal]

sin35xBC=DC
could be one of the equations

Hmm

[inland orbit]

tan, not sin

but yes that's what I'm thinking

[kind fractal]

Ah my mistake, get them confused sometimes

[inland orbit]

all good

You can do something similar with the 18 degree angle as well

[kind fractal]

Ah! I see it!

tan18°= DC/ (BC+15)

[inland orbit]

yep

[kind fractal]

Alright, I'm gonna try and resolve this now

[inland orbit]

okay, great 👍

[kind fractal]

sin35xBC=DC
tan18°= DC/ (BC+15)
<=> tan18°=(sin35°xBC) /(BC+15)
<=>tan18° - (sin35°/15) = BC/BC
Feel like I'm doing something wrong here..

[inland orbit]

yeah you can't break up the ratio like that

$\frac{ab}{c+d} \neq \frac{a}{c}+\frac{b}{d}$

[solid kiln]

tatpoj

[inland orbit]

multiply by (BC+15) instead

(BC+15)tan18° = BCtan35°

also make sure you have tan35, not sin35

[kind fractal]

Forgot to correct it

[kind fractal]

So I put the incognits in different parts of the equation?

[inland orbit]

incognits?

[kind fractal]

Unknown numbers

Usually X or y, in this situation BC and DC.

[inland orbit]

I just took your equation here:

this one

tan18°=(sin35°xBC) /(BC+15)

that's correct

multiply both sides by (BC+15)

and you get this

(BC+15)tan18° = BCtan35°

[kind fractal]

Ooh, I see

So it'll end up

tan18°BC+ 4.87365=BCtan35°

(tan18°-tan35°)BC= -4.87365

[inland orbit]

looks good

[kind fractal]

0.32491-1= -0.67509

[inland orbit]

wait, what is this?

[kind fractal]

Those're the tangents

[inland orbit]

oh, tan35 is not 1

[kind fractal]

.. put in 45

Mb

[inland orbit]

tan45 is 1

yeah lol no problem

[kind fractal]

Late night maths what can I say

Right gimme a second to fix that now

-0.37529BC= -15
<=> BC = 39.96909

[inland orbit]

-15?

[inland orbit]

you had this before

[kind fractal]

I forgot

[inland orbit]

maybe you meant -15tan18°

which is what that -4.87... is

[kind fractal]

Yeah that's what I meant, not writing any of this down right now

12.98... that's better

[inland orbit]

Yeah that looks good

[kind fractal]

tan35°xBC=DC
<=> tan35°x 12.98=DC
<=> DC=9.088

🎉

[inland orbit]

ayyyye 👍

[kind fractal]

Thank you very much for this, I'm quite good with this material but one of the problems wasn't computing with my sleep deprived brain haha
Appreciated 🙏

[inland orbit]

No problem 👍

Good work

[kind fractal]

thanks, take care

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[dense zealot]

[quartz yoke]

one second but why do you have the "number theory" role

anyway can you please rewrite it neater

although what I can read is

want to prove I_n = floor( (2+sqrt(3))^n ) is odd

[dense zealot]

[quartz yoke]

!status

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[dense zealot]

i think its quite intuitive its 1

[quartz yoke]

try binomial expansion

also it's over n so you want induction

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[mortal burrow]

Anyone know if this is true or false? Classmates say it is false but chatgpt says its corect

[thick frost]

that is correct

arcsinx is inverse sinx, so it just undoes it

[mortal burrow]

I thought so too

ok thanks

.close

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@wraith hinge Has your question been resolved?

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@wraith hinge Has your question been resolved?

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[viscid flower]

i had some thoughts

[viscid flower]

you can identify issues

like you know if an ln comes out

not rational

i think you really just want uhh

well idk

did you solve it

maybe i shouldnt have pung

what you really want is for the top and the bottom to be off by a degree, i believe

or maybe thats silly

off by a degree, or evenly divide

idk nvm

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[cursive forge]

How many six digit odd numbers less than 200000 can be formed using the digits 1 1 2 2 3 and 5
Not sure how to approach this one. Maybe something like 2*5!/3! = 40?

[stuck sable]

since the numbers are less than 200 000

so the first digit must be a 1

[cursive forge]

It has to be odd as well, and the 2s are interchangable

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[cursive forge]

<@&286206848099549185>

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[spark minnow]

Did I simplify correctly

[past pendant]

Hey mate

hope you are doing well

your answer is correct

but your working isn't

$\log_2(5) + \log_2(x) = \log_2(5x)$

[solid kiln]

StopTheSkap

[past pendant]

not

$\log_2(5) + \log_2(x) = \log_2() + 5x$

[solid kiln]

StopTheSkap

[past pendant]

does that make sense?

[spark minnow]

Yes it does I didn’t realize I did that

Thanks

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[daring karma]

is (p implies q) and (q implies p) equivalent?

[stoic iron]

draw a truth table

[daring karma]

they have the same amount of true values and same amount of false values, so that's what made me curious

i did

the values just appear in a different order

[stoic iron]

what do you think "equivalent" means?

[daring karma]

equal in value

so since they have the same value of true values, are they equal?

[tidal igloo]

are the third and fourth columns same?

[daring karma]

well idk if order matters, that's the question im asking

[tidal igloo]

no theyre not equal

[daring karma]

oh okay

[tidal igloo]

is a-b and b-a the same

[daring karma]

like in multiplication, order doesn't matter

[daring karma]

no, but in some scenarios like multiplication, order does not matter

unless its matrix multiplication

[tidal igloo]

a b a*b b*a
1 2  2  2
7 3 21  21

now in this table do the 3rd and fourth column look equal to you?

i would say so. therefore, multiplication is commutative

a b a-b b-a
1 2  -1  1
7 3   4 -4
```now in this table do the 3rd and 4th column look `equal` to you?

the magnitudes are same, yes. but are the columns? no. i would say for subtraction, order matters
your table makes it clear that order also matters for your particular operation.

[daring karma]

got it, thank you

.close

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[sinful haven]

I need help with this question

[sinful haven]

Everything is correct but D. I don’t know how to make it into interval notation

@stoic saddle can u help me

Hello guys

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[meager night]

How did you find this value?

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[proper plume]

how do you graph this on a number line x^2-1>0

[reef plaza]

Add one to both sides

[proper plume]

yea

then what

@reef plaza

[reef plaza]

Square root

Than should have x > +- 1

So x > 1 and x < -1

[proper plume]

mhm

oh alr

thx

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[proper plume]

lmao

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[alpine kiln]

whats the definition of a domain and range

And the x-values are the ones that go into a function and the y values are your values that come out

So to that extent, what are all the values going in

[main olive]

domain is [-1,4)

it’s not defined for values less than -1

or greater than or equal to 4

[alpine kiln]

you have it backwards lol

kind of*

Nope but close

The range is how far the y values span

so you basically look at how far up and down the values go and thats your range

The highest y isnt 4

Also not that the lowest y is touching -4 so it would be [

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[alpine kiln]

It touching -4 since the circle is filled in

That means its touching the point

If the circle isnt filled in then it approaches the point but never touches it

There are two intervals for the range since there is a gap

Also I think you are looking at your x-value for your upper bound. You have to look at the y-value

Close there is a gap in the interval

Yes

np

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[midnight cape]

Help algebra

[midnight cape]

Ex 1

Part 1

Deduce that x^(2) + 1 divides f(x)

[odd ridge]

Hi, I see the exam is from January 2023, are you currently in the exam? If so we are not allowed to assist

[midnight cape]

No

Of course no

This is my friend's exam

He do it before one week

[odd ridge]

Alright

[midnight cape]

I'm trying to solve it

For the exam tomorrow

[odd ridge]

So it would be sufficient to show that $x + i$ also divides $f$, because then $(x+i)(x-i) = x^2 + 1$ divides $f$, agreed?

[solid kiln]

Yuese

[odd ridge]

Yes that's great!

[midnight cape]

Wait wait

I will try

[odd ridge]

So you can check that $-i$ is indeed also a root, or there is another trick you may have learnt in class: because $f$ has only \textit{real} coefficients, if $a + bi$ is a root, then $a - bi$ is also a root. This directly gives you that $-i$ is a root.

[solid kiln]

Yuese

[midnight cape]

Maybe I understand

[odd ridge]

Ah, I see what you are stuck on

[midnight cape]

Oh

Maybe I'm stupid

😫

[odd ridge]

Try to understand that
$$x^2 + 1 = x^2 - i^2 = (x+i)(x-i)$$

[solid kiln]

Yuese

[midnight cape]

it's easy man

[odd ridge]

No you're not!

[midnight cape]

Sorry

[odd ridge]

Can you see why this is true?

[midnight cape]

Wait

@odd ridge true?

[odd ridge]

Yes that's perfect

[midnight cape]

Sorry, it's easy

[odd ridge]

Haha no problem, sometimes it's hard to come up with those kinds of things

[midnight cape]

Since I'm studying from 12 hours

Thank you ⭐

[odd ridge]

No problem! Good luck with your exam!

[midnight cape]

Thx

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[wraith hinge]

Can every ODE system be written as a single higher order differential equation?

[wraith hinge]

just a random thought I had

[green sand]

yeah

[wraith hinge]

I mean, It is well-known that, given a differential equation of order n, it can be written as an ODE system of first-order equations using the change of variables

just was wondering if the converse is also applicable

[green sand]

I'm actually not completely sure - best to wait for someone else to answer tbh

[wraith hinge]

okay i figured it out thanks

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[wraith hinge]

consider the sequence space of
[
h^{2,1} = \begin{cases} c: \bN \ni j \mapsto c_j \ni \bC; \sum_j (1+j^2) |c_j|^2 < \infty \end{cases}
]
how do i show that
[
h^{2,1} \times h^{2,1} \ni (c,d) \mapsto \langle c,d \rangle = \sum_j (1+j^2) c_j \bar{d_j}
]

[solid kiln]

Lixera

[wraith hinge]

okay my latex is kind of fucked but i hope it is clear

nvm

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[glad bolt]

I have never done multivariable calculus, but I have a question
How do you calculate the volume of the region ${(x,y,z) \mid |x|+2\sqrt{y^2+z^2}<t}$, where $t$ is a fixed given (positive) constant? I got $\pi\cdot t^4/12$, but I think that's wrong

[solid kiln]

Croqueta

[glad bolt]

This is what I did

$|x|+2\sqrt{y^2+z^2}<t\iff y^2+z^2<(t-|x|)^2/4$, but thats the volume of the sphere with radius $(t-|x|)/2$, and $-t\leq x\leq t$ therefore, we have to compute the integral
[
\int_{-t}^t \frac 43\pi \left(\frac{t-|x|}{2}\right)^3,dx
]

[solid kiln]

Croqueta

[glad bolt]

and you can ask wolframalpha to compute taht, which gives pi*t^4/12
(that's doable by hand ofc)

@warm kernel

<@&286206848099549185>

bruh

that is not a sphere wtf

yeah ok fixed it

(actually, there's a reason why I used the volume of a sphere, it is because when I did the computations with the area of a circle I thought I got the wrong value, but it was correct)

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[flat wing]

What did I do wrong?

[wraith hinge]

firstly, try not to use 'x' as multiplication, especially if you have x as a variable

you multiplied the left by 4, but divided the right by 4

[flat wing]

Thanks

[flat wing]

Thanks for the tip

I probs use the weird looking x instead

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[wraith hinge]

Hmmm

[wraith hinge]

I wonder how to find the t ?

[random citrus]

Log

[trim lichen]

that's a t?

that looked like a misplaced plus sign

do not write your t's this way

[main glade]

Take the log of the base on both sides

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@wraith hinge Has your question been resolved?

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[wraith hinge]

.reopen

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✅

[wraith hinge]

Im not sure how to do it, do i need a calculator ?

[main drift]

Yea

[main glade]

Yep

You need a good one tho

Normal ones only shows log10

[wraith hinge]

This is how far i got

Is it correct?

[wraith hinge]

But this is not log10?

[main drift]

Most calculators should have natural log

Denoted by ln like your wrote

But if they don’t, you can do the same

[wraith hinge]

ln and log is sonething different?

[main drift]

Yes

[wraith hinge]

[main drift]

If you just see log on a calculator, they mean log base 10

[wraith hinge]

Ln = log10 ?

[main drift]

If you see ln, it is log base e

[wraith hinge]

My calculator has these

[main drift]

You can use either then

[wraith hinge]

But when using log i have to enter a 10 below it?

Its different results hm

[thick frost]

the question you're working with is ln

so base e

[wraith hinge]

I was just wondering if log10=ln

[main drift]

No

[thick frost]

ln is base e

[wraith hinge]

So this is the t im looking for right ?:D

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[hollow elk]

Bro

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[oblique meadow]

why is y=3 an asymptote

[oblique meadow]

Please may you explain?

I was reiterating my question