help-39

[sharp smelt]

x is prime.

trying to express this using logical quantifiers

[sharp smelt]

This essentially means $x= 1 \cdot x$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

and that's the only prime factorisation of the same

no?

Hello charbit

[grizzled dust]

maybe something along the lines of if d | x then d = x or d = 1

[sharp smelt]

hmm

yeah

makes sense

thanks

$\exists x( d|x \implies d=x \vee d=1)$$

[jolly parrot]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sharp smelt]

like this?

[merry carbon]

Remember you're making a claim about x (that it's prime), rather than [trying to/] say[/ing] "primes exist"

[sharp smelt]

mhm

[merry carbon]

Also "what's d?"

[sharp smelt]

right

a divisor

of x

hmm

so what exactly is wrong with my statement?

It's just a convolted way of saying primes exist?

[rustic gate]

it's not well formed

[sharp smelt]

mhm

maybe as $x = y \cdot z$ and then $y\neqz$

[jolly parrot]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sharp smelt]

$\exist x( x= y \cdot z \land y \neq z\land ( y \vee z =1))$

[jolly parrot]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sharp smelt]

ooh

yeah, I'm lost 😭

[rustic gate]

you need to introduce d here

[sharp smelt]

$\exists d( d|x \implies d=x \vee d=1)$

like so?

[rustic gate]

the statement that "x is prime" depends on x, so you don't need to introduce x

[sharp smelt]

I see, so just there exists a d such that d|x...

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[rustic gate]

is that how you tell a number is prime?

[sharp smelt]

no

a number is prime iff its prime factorisation is unique

sorr

one minute

[rustic gate]

9 is prime because there exists a number d=9 such that 9 | 9, and 9 = 9

that's not true is it

[sharp smelt]

no

a number is prime if the only d|x are 1 and x itself

is that right?

[quiet tendon]

yes

uniqueness of prime factorizations applies to all integers > 1, not just primes, as per the fundamental theorem of arithmetic

[sharp smelt]

$\exists d(d|x \land (d=x \vee d=1))$

is this it?

if not could I have another hint ?

[rustic gate]

a number x is prime if the only divisors are x and 1

[sharp smelt]

yes

[rustic gate]

here, you're only testing that fact for a single divisor d

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[rustic gate]

go back to this

this was on the right track

$\exists d( d|x \implies d=x \vee d=1)$

[rustic gate]

the problem is that you've only asked for a single d to satisfy this

[sharp smelt]

I don't follow, I just enclose the 2nd part in brackets and I'm done, right?

[rustic gate]

for x to be a prime, any divisor of x must satisfy the condition in the right

[sharp smelt]

yes

but here d is a free variable , right

[rustic gate]

and you've only asked for one d to exist

i can pick the d to be x

[sharp smelt]

so I introduce another variable similar to d

[rustic gate]

no

you quantify over d with a universal quantifier

every divisor of x must satisfy the condition

[sharp smelt]

$\exists d( d|x \implies \forall ( d=x \vee d=1))$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[rustic gate]

no thats not well formed

and what i mean is

$\forall d( d|x \implies d=x \vee d=1)$

[rustic gate]

every divisor of x must satisfy the condition

[sharp smelt]

yes, I get that bit, but I'm not sure of how to write it

[rustic gate]

this way

\forall d

[sharp smelt]

isn't that what I;ve written here?

[rustic gate]

no that expression is not well formed

the forall doesn't have anything it's attached to

[sharp smelt]

how would I do that

[rustic gate]

as i've done here

you want to universally quantify over d

[sharp smelt]

ah, I see

[rustic gate]

because you want to make a statement about all divisors of x

not just some particular one of your choosing

[sharp smelt]

got it

thanks!

.close

[pearl pond]

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[astral pebble]

im not sure how to do this

[astral pebble]

i do know that i have to calculate gpe and ke

but my topic doesnt have any formula for friction and im not sure how to minus off my friction so well kinda need some help with the friction

[pearl pond]

@astral pebble Has your question been resolved?

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[pearl pond]

@maiden swan Has your question been resolved?

[charred ocean]

!status

[pearl pond]

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

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[plucky nebula]

how would you determine if a subset is closed under addition and multiplication comprehensively

[sharp vigil]

for addition: define two arbitrary elements of the set, noting that they satisfy the condition of the subset. find out whether their sum necessarily satisfies the condition as well

the specifics depend on how the subset is defined

[plucky nebula]

ah okay

the lecturer just did an example of this lmao

[sharp vigil]

your lecturer combined the addition and multiplication closures into one expression

[plucky nebula]

yeah he said it is an alternative subspace test

[sharp vigil]

so to show it's closed under addition, we need f1 + f2 to belong to the set, and for it to be closed under multiplication we need λf1 to belong. so λf1 + f2 is in the space if and only if the previous two conditions are met

[plucky nebula]

yeah that makes sense

he only also proved the non-existence of the non-empty set by showing that the 0 vector of the vector space V is also part of the subspace S

[sharp vigil]

0 is a convenient vector to check since if it is not in the space then either:

1. the subset is empty
2. the subset is not closed under addition and scalar multiplication

[plucky nebula]

makes sense

[pearl pond]

@plucky nebula Has your question been resolved?

[pearl pond]

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[halcyon plank]

The table shows three values of x and their corresponding values of y, where y = f(x) + 4 and f is a quadratic function.
What is the y-coordinate of the y-intercept of the graph = f(z) in the Y-plane?

Ans: f(x)=-4(x-21)(x-25)-12. but why?

[gloomy hare]

you asked this question more than 10 times

and you were answered each time

[dapper kraken]

actually bro

last time i saw you you straight up refused the method lmao

[dapper kraken]

13

[gloomy hare]

https://tenor.com/view/silly-cat-silly-car-car-stare-10-thousand-yard-stare-10-thousand-yard-gif-14200271775968563996

[dapper kraken]

meow i love cats

[halcyon plank]

I want to put it in the factored form.

[pearl pond]

@halcyon plank Has your question been resolved?

[pearl pond]

@halcyon plank Has your question been resolved?

[pearl pond]

@halcyon plank Has your question been resolved?

[pearl pond]

@halcyon plank Has your question been resolved?

[pearl pond]

@halcyon plank Has your question been resolved?

[last moth]

Can you explain what the previous methods explained to you were, and why they weren't what you were looking for?

[halcyon plank]

because i want to understand the method I am pleased with.

[last moth]

That didn't really answer my question. I'm trying to get enough information to help you, but I can't really help if I don't know what you're looking for and what's already been explained to you.

[halcyon plank]

Oh

I was taught the vertex form quadratic equation

and basically graphing points on a graphing calculator

i was also taught the factored form quadratic equation but surprisingly i couldn't understand it, so just like u can see lol. im trying to understand it

[last moth]

I think the answer you wrote has a typo. It should be -4(x-21)(x-25)-12.

Are you trying to figure out how to get from that answer into a factored form?

[halcyon plank]

yes

[last moth]

Okay

The first step would be to expand everything out

So, you want to multiply (x-21) by (x-25)

Do you know how to do that?

[halcyon plank]

Yes

[last moth]

Okay so then what do you get

[halcyon plank]

we dont need to expand tho

but ok

x^2-46x+525

[last moth]

We do if we want to figure out how to factor it

Okay good

Next we need to multiply by -4

So that gives us -4x^2 + 184x - 2100

And finally subtract 12 and we get

-4x^2 + 184x - 2112

[halcyon plank]

but why do we subtract by 12

[last moth]

do you mean why do we subtract by 12?

[halcyon plank]

yes]

[last moth]

It's because there's a - 12 in the original answer we got

[last moth]

See here ^^

[halcyon plank]

but i dont understand it

like why is there a -12

[last moth]

Are you asking why the answer is -4(x-21)(x-25) - 12?

[halcyon plank]

no

kinda

im only asking why is there -12

i understand from where do the other terms come

but not "-12"

here's where i'm confused

[last moth]

OK so let's try seeing what we get without the -12

[halcyon plank]

okay

[last moth]

So, f(x) = -4(x-21)(x-25)

Our equation is y = f(x) + 4

So that gives us y = -4(x-21)(x-25) + 4

And now let's try plugging in our values of x=21, x=23, x=25

[halcyon plank]

okay

[last moth]

If we plug in x=21, the first term becomes 0, so we get that y = 4

Same with if we plug in x=25, we get y = 4

Does that make sense

[halcyon plank]

okay

[halcyon plank]

no

[last moth]

What part doesn't make sense

[halcyon plank]

okay but why did you choose to put -12

why not any other number?

[last moth]

We haven't gotten to that part yet

Does what I said so far make sense

[halcyon plank]

yes

[last moth]

Okay so

Also if we plug in x=23, we get 20

So we can make a table of our results
x y
21 4
23 20
25 4

If we compare that to the table that we want

x y
21 -8
23 8
25 -8

You can see that we're off by 12 each time

4 - 12 = -8
20 - 12 = 8
4 - 12 = -8

So we need to subtract f(x) by 12 in order to get the table we want

[halcyon plank]

okay

[halcyon plank]

is there a shorter way to reach that step?

[last moth]

We're just plugging in numbers and then comparing them with the numbers in the table we want

[halcyon plank]

okay

[last moth]

I explained it in detail to help you, but the actual method is pretty short

Once you get used to it

[halcyon plank]

tysm

[last moth]

Np

[halcyon plank]

.close

[pearl pond]

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[pearl pond]

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[steel atlas]

hello

[steel atlas]

i need help with this

I don't know what I should do in this step.

[ivory swallow]

ok

[ivory swallow]

i can't read that

let me try to latex it

[steel atlas]

this is the answer

[ivory swallow]

$$\frac{5}{7} \cdot \frac{a^{11}}{a^7} \cdot \frac{1/b^8}{1/1} \cdot \sqrt[3]{2^3 \cdot 3^3 \cdot 3^2 \cdot a^2 \cdot b^{18}}$$

[jolly parrot]

Mr. Gamer 🇵🇸

[ivory swallow]

is that right

[steel atlas]

yes

[ivory swallow]

alright i'll let another helper take over because i'm gonna be hungover

[pearl pond]

@steel atlas Has your question been resolved?

[pearl pond]

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[pearl pond]

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[stoic imp]

let $f(x) = \frac{24e^{x}}{3e^{x} + 1}$ find the range of f

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

,, \lim_{x \to +\infty} \frac{24e^{x}}{3e^{x} + 1} \= \frac{24}{3}\lim_{x \to +\infty} \frac{e^{x}}{e^{x} + \frac{1}{3}} \= \frac{24}{3}\lim_{x \to +\infty} \frac{1}{1 + \frac{1}{3e^{x}}} = 8

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[stoic imp]

,, \lim_{x \to -\infty} \frac{24e^x}{3e^x + 1}

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[tulip ore]

are you sure the question is to find the range of that function

right now all youre calculating are limits

you dont use limits to find the rnage

[stoic imp]

well

its the image, not range

[tulip ore]

from?

R?

[stoic imp]

idk the translation omg

can u google translate real quick=?

[meager trellis]

yeah that sounds like they're probably asking for the range

[tulip ore]

the english word image and range are interchangeable

[stoic imp]

okay

[tulip ore]

when you get to higher portions where the preimage or domain no longer mean the same thing, a distinction appears

[stoic imp]

okay

[tulip ore]

you can start at the range of e^x being (0, inf) and build up the function

that can be used if you only have one e^x in the function

[stoic imp]

0, inf?

how so?

can you elaborate in detail

[tulip ore]

the range of $y=e^x$ is $(0,\infty)$

[meager trellis]

not "0, inf", (0, inf)

[stoic imp]

why doe

[jolly parrot]

mtt

[stoic imp]

,w range e^x

[tulip ore]

the (0, inf) is interval notation

[jolly parrot]

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Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[meager trellis]

the open interval, "all numbers strictly greater than 0" ("up to infinity" meaning no upper bound)

[tulip ore]

it is short for "0 < something < infinity"

[stoic imp]

but why

is only positive

i dont get it

[tulip ore]

can you raise e to a power

and get 0?

remember youre only working with real numbers

[stoic imp]

never no

[tulip ore]

can you raise e to a power

[stoic imp]

u cannot

[tulip ore]

and get a negative number?

[stoic imp]

okay I feel u

but

[tulip ore]

that is what range means

remember youre only working with real numbers

ln(-1) is ambiguous and is also not real

please pay attention

are you paying attention right now?

I need a yes

[stoic imp]

yes

[tulip ore]

the range of e^x is believably (0, inf), right?

[stoic imp]

sure

[tulip ore]

can you elaborate on whats holding you back from completely believing it

[stoic imp]

nothing

[tulip ore]

using "sure" instead of "yes" is usually meant to mean "I see that you want me to believe that" and doesnt actually show whether you understand or not

[stoic imp]

yes

y

[tulip ore]

yea dw

[stoic imp]

so what?

[tulip ore]

you can start at the range of e^x being (0, inf) and build up the function
that can be used if you only have one e^x in the function

right now theres two e^xs so its not immediately useful

but theres an easy way to reduce the number of e^xs there are in the function to one

[stoic imp]

,, f(x) = \frac{24e^x}{3e^x + 1}

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[tulip ore]

can you find that easy way?

[stoic imp]

well

$\frac{24e^x}{3e^x + 1} = \frac{24e^x}{3e^x + 1} + \frac{24e^x}{3e^x + 1}$ is this legal

[tulip ore]

is $\frac32=\frac31+\frac31$?

[jolly parrot]

mtt

[stoic imp]

nope

[tulip ore]

then no its not legal

[stoic imp]

I messed up

[tulip ore]

if you want to split the fraction into something that cancels out nicely, you need to consider the entire denominator

see if you can get some version of a 3e^x + 1 in the numerator

right now it says 24e^x

[stoic imp]

omg

8(3e^x + 1) = 24e^x - 8

[tulip ore]

8(3e^x + 1) - 8 = 24e^x you mean

[stoic imp]

yeah that

sirry

[tulip ore]

its alr

[stoic imp]

now what

[tulip ore]

put that in the numeraotr

[stoic imp]

okay

[tulip ore]

then see if you can simplify f

[stoic imp]

,, \frac{8(3e^x + 1)-8}{3e^x + 1}

[tulip ore]

no not like that

youre forgetting the - 8 again

[stoic imp]

mb

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[tulip ore]

yea

try simplifying f

[stoic imp]

okay

,, \frac{8(3e^x + 1)-8}{3e^x + 1}=8-\frac{8}{3e^x + 1}

[jolly parrot]

938c2cc0dcc05f2b68c4287040cfcf71

[tulip ore]

thats correct

[stoic imp]

😎

[tulip ore]

now with one e^x, you can start from "the range of e^x is (0, inf)" and apply each calculation that f does to the range

for example 1 + e^x's range is (1, inf)

[stoic imp]

sure

,w range 8 - 8/(3e^x + 1)

[jolly parrot]

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[tulip ore]

bruh

[stoic imp]

(0,8)

but for y to be 0

3e^x + 1 has to be 1

but e^x is never negative

[tulip ore]

did you figure it out yourself or did you just use wolfram alpha to give me that answer

[stoic imp]

nono, I still dont understand it

I am gathering info

[tulip ore]

I told you the general method

you are not using it

do you want a different one

[stoic imp]

wdym different one? I usually do this shit using limits ngl

[tulip ore]

you gotta ask when you dont understand something

forget this

we're doing something completely different

you arent even looking at what I suggested to you

[stoic imp]

I mean

you said

e^x range is 0,inf

• 1

1, inf

then multiplied by 8

8, inf

then?

[tulip ore]

first off

this is your function

you are not finding the range of 8 (e^x + 1)

second off

not a single word abotu this until after I look like Im dropping the idea entirely

[stoic imp]

sorry

[tulip ore]

surely its not an exclusively english idea to you to say when you do something

you believe this?

[stoic imp]

I am not sure what ur asking tbh, my english is bad

I believe what ?

[tulip ore]

this is your function

what is the range of 3e^x

[stoic imp]

0,uif

[tulip ore]

what is the range of 3e^x + 1

[stoic imp]

1,inf

[tulip ore]

what is the range of 8 / (3e^x + 1)

[stoic imp]

ohh

3e^x + 1 > 0

[tulip ore]

no

already youre doing something bizarre

why not > 1?

[stoic imp]

mb

why > 1 doe

[tulip ore]

you just said the range of 3e^x + 1 is (1, inf)

that means 3e^x + 1 > 1

[stoic imp]

💯

[tulip ore]

@stoic imp are you stuck

[stoic imp]

lowkey yeah

[tulip ore]

you didnt say anything

care to explain?

[stoic imp]

okay

how to get the range of rhe fraction

[tulip ore]

youre doing a good job explaining so far

please keep this up ok?

[stoic imp]

okay

[tulip ore]

say you have a number between 2 and 4

and you take its reciprocal

that number then is between 1/4 and 1/2, right?

[stoic imp]

okay but what does reciprocal even mean

you flip it

[tulip ore]

reciprocal is an english term for doing 1 / (something) or (something)^-1

flipping is ambiguous and should not be used

[stoic imp]

okay

0,25

0,5

[tulip ore]

thats not ideal

you dont need to convert 1/4 and 1/2 to decimals to understand what Im going to say

[stoic imp]

okay

[tulip ore]

now say you have a number that is at least 127

[stoic imp]

is just the ordering

that scared me a little bit

[tulip ore]

1/4 and 1/2 are numbers you know

are you familiar with them

[stoic imp]

1/2 and 1/4

yes

[tulip ore]

wait you are?

what about 1/127?

[stoic imp]

0.5 and 0.25

,calc 1/127

[tulip ore]

no not like that

[jolly parrot]

Result:

0.0078740157480315

[tulip ore]

no

Im asking if youre familiar with the fractions 1/2 and 1/4

instead of reading them as a weird way to write a decimal

[stoic imp]

okay

[tulip ore]

this is taking too long, you can ignore that question

say you have a number thats 127 or higher

and you take its reciprocal

the reciprocal is between which two numbers?

[stoic imp]

can i respond with unsure

[tulip ore]

say you have a number thats between 127 and 131

and you take its reciprocal

the reciprocal is between which two numbers?

[stoic imp]

127^-1 and 131^-1

[tulip ore]

now if numbers are bigger

reciprocals are smaller

right?

[stoic imp]

y

[tulip ore]

I gotta break it to you but "y" is ambiguously "yes" or "why"

[stoic imp]

yess what then?

so what

[tulip ore]

this number here doesnt have an upper bound

think about its reciprocal

its positive, rightg

[stoic imp]

yeah

✅

[tulip ore]

its also below 1/127, right

[stoic imp]

+inf, 1/127

[tulip ore]

now you gotta be careful with what youre saying there

can you repeat the two numbers the reciprocal is between?

[somber mason]

how did you guys get to this point

[tulip ore]

this was supposed to be the easy way out

but you gotta build up the basics first

[stoic imp]

127^-1 and 131^-1

[tulip ore]

127 or higher here

that means it can go lower than 131^-1

[stoic imp]

okay

[tulip ore]

it can go lower than any positive number

but it cant go lower than 0

right?

[stoic imp]

right

[tulip ore]

it cant be 0 either, right?

[stoic imp]

right

[tulip ore]

so knowing 127 < number,

what would that mean for the reciprocal?

[stoic imp]

(0, 1/127]

[tulip ore]

0 < reciprocal < 1/127

[stoic imp]

now ?

[tulip ore]

writing $f(x)$ as $8-8\cdot\frac1{3e^x+1}$,

[jolly parrot]

mtt

[tulip ore]

and knowing the range of 3e^x + 1 is (1, inf),

what is the range of 1 / (3e^x + 1)?

[stoic imp]

(0, 1]

[tulip ore]

thats correct

whats the range of 8 * 1 / (3e^x + 1)

[stoic imp]

(0,8]

[tulip ore]

wait

[tulip ore]

slight difference here

[tulip ore]

here I included 127 so the reciprocal could go up to 1/127

[tulip ore]

but here 1 cant be included

so 1 is left out

whoops

so the range of 1 / (3e^x + 1) is (0, 1)

[stoic imp]

(0,8)

[tulip ore]

yea

then whats the range of 8 - 8 * 1 / (3e^x + 1)

[stoic imp]

(0,1)

[tulip ore]

...whats the range of 8 * 1 / (3e^x + 1)

[stoic imp]

(0,8)

[tulip ore]

so whats the range of 8 - 8 * 1 / (3e^x + 1)

[stoic imp]

unsure

[tulip ore]

if you have a number from 0 to 8

and you subtract it from 8

what numbers would it be between

[stoic imp]

-8,0

[tulip ore]

I didnt say to subtract 8 from it

subtract it FROM 8

[stoic imp]

mb

[tulip ore]

I said **8 - **

so if you have 8 - a number,

what numbers would it be between

[stoic imp]

8 to 0

[tulip ore]

yes

so (0, 8)

right?

[stoic imp]

sure

[tulip ore]

the range of 8 * 1 / (3e^x + 1) is (0, 8)

so whats the range of **8 - **8 * 1 / (3e^x + 1)

[stoic imp]

8 to 0

[tulip ore]

yes

so (0, 8)

right?

[stoic imp]

mhmmkay

i usually do it with limits doe

[tulip ore]

how would that work

[stoic imp]

or finding the domain of the inverse

[tulip ore]

neither this or the other one reliably works

[stoic imp]

reliably

no but

usually its better to know where the asymptotes are to find the range quicker no?

[tulip ore]

would it work for 1 / (1 + x^2)?

[stoic imp]

by doing a sketch of the function

[tulip ore]

you dont define functions by their asymptotes

[stoic imp]

I was referring to finding the range

[tulip ore]

what?

how does that go against what I said? youre implying I didnt know you were doing that?

[stoic imp]

no

its just

nothing

[tulip ore]

so?

[stoic imp]

well

[tulip ore]

its not necesary to know about the asymptotes to find range

[tulip ore]

this gives you some slight confirmation on what you found but can be overridden

so in terms of that its not very good confirmation

the asymptote can at least guarantee that the range should include a (asymptote or asymptote) in at least one interval

[stoic imp]

asymptote or asymptote

okay

[tulip ore]

you should repeat the ()s

[stoic imp]

but

then asymptotes don’t work for finding image?

wdym

[tulip ore]

you found that as x -> inf, f(x) -> 8

this tells you that the image should have a 8) or a (8 somewhere

[stoic imp]

yeah

[tulip ore]

the image of f(x) is (0, 8) which we found without ever caring about the asymptotes

[stoic imp]

okay

[tulip ore]

you can see there the 8) as suggested by the asymptotes

[stoic imp]

but that is only vertical asymptote we still need to find horizontal and oblique asymptotes before we find range

[tulip ore]

x -> inf is not a vertical asymptote

[stoic imp]

that is horizontal mb

[tulip ore]

now you do have something neat if you look for vertical asymptotes

if you have a limit that -> inf or -> -inf,

x -> something, f(x) -> inf would confirm a inf)

[stoic imp]

sure

[tulip ore]

x -> something, f(x) -> -inf would confirm a (-inf

[stoic imp]

okay

sure

what about oblique asymptotes

let me bring the formula

[tulip ore]

that sounds like a lot of work for asymptotes that we ultimately dont need

if you do find an oblique asymptote, it still fits what I said about vertical asymptotes

alr I gtg

[pearl pond]

@stoic imp Has your question been resolved?

[pearl pond]

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[sharp smelt]

what am I doing wrong here

[main oxide]

not sure what you're doing but think of (n+1)! as a single chunk of which you have 1 and n+1 of

[verbal whale]

How did you get this? The rest is correct

[sharp smelt]

ooh

got it

yeah, silly mistake on my end

thanks!

,clos

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[edgy otter]

How do I simplify this/ what do I do next?

[ivory swallow]

first of all you can simplify f(t)

(6t+18)/(t+7) = [6 (t+3) ] / (t+7) = (6(t+7) - 42 + 18) / (t+7) = (6(t+7) - 24)/(t+7) = 6 - 24/(t+7)

[edgy otter]

How do I get that?

Does this work?

[stoic pecan]

do you know quotient rule?

i assume yes since theres lims

[edgy otter]

I think I have it in my notes

[stoic pecan]

wait

i have an apropiate image

[edgy otter]

LMAO

[stoic pecan]

oh wait yeah you can simplify it

thats probadly easier than this

but this should work just fine

[solid pier]

lol

[edgy otter]

so what do i do 😭

[kind rampart]

Quite Peculiar

[stoic pecan]

either works

[edgy otter]

so what I did works?

[stoic pecan]

oh idk what you did

[stoic pecan]

i meant quotient rule or differntiating that guys simplified form

[edgy otter]

I should understand this but I'm also bad at minecraft

I like farming

[stoic pecan]

thats fine

[stoic pecan]

what did you do here?

[edgy otter]

gave them a common demoninator?

i think

[stoic pecan]

am i misunderstanding the question or are you not supposed to just differntiate it

and thats it

[edgy otter]

i don't get how to do that 😭

I'm so confused

[stoic pecan]

uh

like

f(t) = 5x^2

f'(t)=10x

like that?

whats this btw

is a just a value

or something else

[edgy otter]

derivative?

i think

im dying here

[stoic pecan]

yes

okay so you jsut derive it

lets do this guys method first

[edgy otter]

im doomed

[stoic pecan]

6-24/(t+7)

do you know how to derivate 6

lets start there

[edgy otter]

all i know is dy/dx

[stoic pecan]

yeah

do you not know like derivation basics

if not this is prob impossible

[edgy otter]

😭 im so screwed for my exam on saturday

[stoic pecan]

you got time

go watch a video from organic chem or khan academy

its not that complicated

[edgy otter]

i'll go do that and be back in the morning

[stoic pecan]

alright

good luck

[edgy otter]

Could also be cause its like 3am for me

tyty

🫡

[glacial sequoia]

Writing it as
[ 6 - \f{24}{t + 7} ]
might make it easier for you

[stoic pecan]

nah she doesnt know like derivation of 6

its alr we all start without knowing it

[jolly parrot]

neon

[glacial sequoia]

I see

[edgy otter]

I'll go watch some khan academy b4 i come back

ty guys for your help!!

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[pearl pond]

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[midnight haven]

The reflection of the plane P1:2x−3y+4z−3=0, in the plane P2:x−y+z−3=0, is the plane ?

[midnight haven]

Status 1

A.4x−3y+2z−15=0
B)4x−2y+z−15=0
C)3x−2y+2z−15=0
D)None of these

[pearl pond]

@midnight haven Has your question been resolved?

[midnight haven]

<@&286206848099549185>

[midnight haven]

Considered a random point on the plane 2x-3y+4z-3

(-2,-1,1)

found the image of the point wrt to the plane x-y+z-3=0

${-2+2sqrt3,-1-2sqrt3 ,1+2sqrt(3)}$

[jolly parrot]

Cnidarian

[midnight haven]

this point did not lie on neither A nor B nor C

so I choose D

i was wrong

This is what i did

2 min i will check

:p

i used the wrong formula

I always forget that the sqrt doesn't exist

thank you!

got it

.close

[pearl pond]

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[pulsar stump]

is $k^m$ called "algerbric field"? and what does it equate to?

[jolly parrot]

Ayanokoji

[rustic gate]

no you'd call it a vector space

in your picture the elements are being denoted with column vectors

[sweet snow]

Why?

[rustic gate]

its just what people call it

[sweet snow]

But a vector space is not just a set of vectors

It also needs a set of scalars and a binary operation

[pulsar stump]

it doesnt matter if it's a column vector or a row vector, or does it?

[rustic gate]

K^m can be equipped with those operations in a natural way

[rustic gate]

thats just a matter of taste

[pulsar stump]

great, ty mate

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[pulsar stump]

also it's a space because i can divide the vector space into a summation of different vectors right? v1(1,0,0,..)+...+v_n(0,...,1)

[rustic gate]

no, its a vector space because it satisfies the vector space axioms

(technically, you need to equip K^m with the vector space operations and then say that its a vector space but thats being pedantic)

[pulsar stump]

alright, i think then this will be formally defined in a future lecture, ty mate

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[full tree]

This seams to easy to be true <@&286206848099549185>

[muted shale]

!15m

[pearl pond]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[full tree]

ok

.close

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[muted shale]

you can ask your question

just dont ping until after 15 minutes

[full tree]

ok

thx

I got it tho

[pearl pond]

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[raven spade]

How do I proceed with this? All I got was simplifying the numerator to (x^3 + 1)(x^3 - 1)

[brisk steeple]

what is R?

[rugged jolt]

real numbers

[brisk steeple]

oky

[light helm]

consider: $x^6 - 1 = x^6 + 1 - 2$

[jolly parrot]

ℝαμΩℕωⅤ

[raven spade]

ah gotcha

hold on

[raven spade]

idk what to do after this though

oh wait

i can try long division

[glacial sequoia]

You could also trig sub and repeatedly use 1 + tan^2 x = sec^2 x

[rugged jolt]

i had the sudden idea of doing [\frac{x^6}{1+x^2}-\frac{1}{1+x^2}] to get a -arctan term, but probably would lead to bad integration by parts.

[jolly parrot]

fish

[glacial sequoia]

It'll only take like 2 iterations

[raven spade]

how do i go about doing that

[glacial sequoia]

Did you do the trig sub?

x = tan u

[raven spade]

oh wait

so would that be [tan^6(x) + 1 - 2]/tan^(x) + 1

[raven spade]

denom being this

[light helm]

with the route i suggested, you don't really need trig sub

[glacial sequoia]

You can try both

What Ramanov said is faster but less obvious if you're newer

[light helm]

if you want to do long division, just do it directly
the way i expressed it, a sum of two cubes identity can be applied to (x^6 + 1) which makes the division very simple

and an integral identity can be applied to the other fraction if you're allowed to use it
(or use x = tan(u) for that part)

[raven spade]

ooh okay

i'll try that out

thanks

[raven spade]

wait there isn't any (x^6 + 1) though

it's x^6 - 1 in the numerator of the question

[light helm]

you can rig it so that x^6 + 1 appears

as mentioned earlier

[raven spade]

i'm sorry if i'm sounding daft i'm half sleepy rn 😭

[raven spade]

ah right

[pearl pond]

@raven spade Has your question been resolved?

[pearl pond]

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[sharp smelt]

could I have a hint?

[sharp smelt]

I think the last bit is wrong

maybe use something similar to the sandwhich theorm ?

or a trig substitution?

just a hint please

[autumn trellis]

write $(1+a)^{n+1}=(1+a)(1+a)^{n}$

[jolly parrot]

quickdoom

[sharp smelt]

mhm

[autumn fossil]

What does that even mean?

[sharp smelt]

nothing, was just jotting down my thoughts

[autumn fossil]

ohk

[sharp smelt]

could I use a sub for $a\in [-1,0)$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[sharp smelt]

like $a =cos(2u)$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[autumn trellis]

There is no need of a sub, if you continue with this

[sharp smelt]

I don't see how writing it like that helps

sorry

[autumn fossil]

You already know something about (1+a)^n...

[sharp smelt]

it's more than $na+1$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[autumn fossil]

and so you can conclude something about (1+a)*(1+a)^n

correct

so what can you conclude about (1+a) * (1+a)^n?

(which is equivalent to (1+a)^(n+1))

[sharp smelt]

it's more than $(1+a)(1+na)$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

quickdoom

[autumn trellis]

Usually with inductipn proofs you xan just follow the algebra

[autumn fossil]

Wait are you sure about that?

oh there is just one mistake

[sharp smelt]

right

[autumn fossil]

okay so now you have this

expand and see what you get

[sharp smelt]

$(1+a)(1+(n+1)a)$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[autumn fossil]

hmm

that doesnt look like (1+a)(1+na)

[autumn trellis]

What js it?

[autumn fossil]

this was correct

[autumn fossil]

after the equal

a(1+n)

[autumn trellis]

That is correct though

[jolly parrot]

quickdoom

[rustic gate]

texsp fail

[autumn trellis]

That term comes from just doing the algebra

[autumn fossil]

uh

what kind of algebra?

[autumn trellis]

Read the spoiler

[autumn fossil]

(1+a)^3 = (1+a)^2 + a(1+a)?

[autumn trellis]

Yeah?

[autumn fossil]

we can clearly see that its not true, just by comparing the degrees. Or am I missing something?

[autumn trellis]

Forgive me

Im quite stupid

[autumn fossil]

(1+a)^3 = (1+a) * (1+a)^2

thats the right one

[rustic gate]

i think you applied the induction hypothesis halfway there

[autumn fossil]

Okay anyway, this was correct

now you can just expand that

[sharp smelt]

ok,, so that gives me $1+na+a+na^2$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[autumn fossil]

okay, good

now you need that to be greater than (or equal to) 1 + (n+1)a

so think about how would you verify that

[autumn trellis]

The line after the equality was a typo, here. I forgot to put a ^n on the second term

[autumn fossil]

@sharp smelt Any progress? You are not far

[sharp smelt]

not really

one minute

I feel that subbing a=cos(2u) may be easier here

[autumn fossil]

no, dont overthink it

all you need to prove is

1 + na + a + na^2 >= 1 + (n+1)a

[sharp smelt]

right

[autumn fossil]

treat it like those inequations where you are solving for x

just manipulate it smh

[sharp smelt]

I can cancel outt terms

[autumn fossil]

indeed

[sharp smelt]

so this gives me $na^2 \geq0$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[autumn fossil]

correct

when is product of 2 things greater than 0?

(or equal)

[sharp smelt]

which is when both are positive or negative

[autumn fossil]

yeah

or one of them is 0

that would work too

can you determine their signs?

[sharp smelt]

n>0

[autumn fossil]

correct

[sharp smelt]

as we're working on N

as is a^2

[autumn fossil]

a^2, not a

[sharp smelt]

so this is always true

[autumn fossil]

yeah

squares are non-negative

[sharp smelt]

got it

tysm!

[autumn fossil]

np

but wait a moment

it would be good if you wrote it in one line

[sharp smelt]

ok, let me try

$(1+a)(1+a)^n \geq (1+a)(1+an) \implies 1+na+a+na^2\geq na+a \implies na^2>0$

[jolly parrot]

ƒ(Why am. I here)=I don't Know

[autumn trellis]

Write = instead of =>

And, that series of implications doesnt lead us to what we want

[sharp smelt]

it does, doesn't it?

[autumn trellis]

We want to show that (1+a)^{n+1} \geq 1+(n+1)a

[sharp smelt]

yes

[autumn fossil]

the thought process behind that was correct, but now you just need to transform it to a proof

which should ideally look like this:

(1+a)^(n+1) >= ... >= ... >= ... >= 1 + (n+1)a

[sharp smelt]

wait, what

[autumn fossil]

the part where I told you to "solve" the inequation 1 + na + a + na^2 >= 1 + (n+1)a was mainly supposed to guide you towards the right manipulation

[sharp smelt]

yeah

but isn't that enough?

[autumn fossil]

It is, but its not that elegant

the proof wouldnt be that readable

You already had (1+a)^(n+1) = (1+a) * (1+a)^n >= (1+a) * (1 + na) = 1 + a + na + na^2

now you just need to continue this and reach >= 1 + (n+1)a

Which term was the only one that didnt cancel when you did the manipulation?

[sharp smelt]

uh

one minute

I'm lost sorry, I'll try this again tomorrow

is that fine?

@autumn fossil

[autumn fossil]

I'll just show you im sure youll understand

if you dont mind

its really quick

[sharp smelt]

ok

[autumn fossil]

(1+a)^(n+1) = (1+a) * (1+a)^n >= (1+a) * (1 + na) = 1 + a + na + na^2 >= 1 + a + na = 1 + a(n+1)

and done

the key observaton was that most of the terms cancel, except for na^2

so we needed to remove that one

[sharp smelt]

yeah, makes sense

[autumn fossil]

and then its equal

[rustic gate]

,align[2\textwidth] (1 + a)^{n + 1} & = (1 + a)\2b{(1 + a)^n} \ & \ge (1 + a)\2b{(1 + na)} \ & = 1+ a + na + \2o{na^2} \ & \2o{\ge} 1 + a + na \ & = 1 + (n + 1)a

[sharp smelt]

hmm, yeah

makes sense

[sharp smelt]

uh, I really have to go now, sorry. Thanks for all the help

can I close this?

[rustic gate]

yeah you can go

[autumn fossil]

sure

[rustic gate]

the non tex'd math was just making my eyes bleed

[sharp smelt]

sorry, and thanks a lot for the help

.close

[pearl pond]

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[autumn fossil]

I need to start using tex again

ill start practising

[pearl pond]

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[inland ivy]

$$\text{Prove that} \frac{\int_0^{\frac{\pi}{2}}\log(\sin x)dx}{\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\log(\sin x+\cos x)dx}=2$$

[inland ivy]

ugh that looks horrible

[jolly parrot]

kheerii

[inland ivy]

sure I guess that kinda works

how I did this was write the bottom integral as $I_2=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\log\left(\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\right)dx=\frac{\pi}{4}\log 2+\int_0^{\frac{\pi}{2}}\log(\sin x)dx =\frac{\pi}{4}\log 2+I_1$

[jolly parrot]

kheerii

[inland ivy]

From here, I solved for $I_1$ to get $I_1=-\frac{\pi}{2}\log 2$ which gives us the result combined with the other equation we got, $I_2=\frac{\pi}{4}\log 2+I_1$

[jolly parrot]

kheerii

[inland ivy]

can anyone think of a way to solve this without explicitly solving for I_1?

[glacial sequoia]

yes

but from the beginning

[inland ivy]

what's that

[glacial sequoia]

for the bottom integral sub x --> -x and add it back to itself

[inland ivy]

ohh cos2x

[glacial sequoia]

you get cos 2x inside the log, then sub 2x

[inland ivy]

gotcha

[glacial sequoia]

yes

[inland ivy]

cool

that was way easier\

than I thought

bruh

thanks neon

.close

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[pearl pond]

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[oak falcon]

How would you slolve this?

[brisk steeple]

use the binomial theorem to expand (1+2/3)^-4

[oak falcon]

whats that?

.close

[brisk steeple]

nvm.

[pearl pond]

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[brisk steeple]

first convert it into a improper fraction

[oak falcon]

.reopen

[pearl pond]

✅

[oak falcon]

so 5/3

[brisk steeple]

yeah. now solve

[oak falcon]

oh 3/5

[worthy lance]

That’d be if it were -1 not -4

[brisk steeple]

yeah

[oak falcon]

3/5 ^4?

[brisk steeple]

yup

[oak falcon]

ty

[brisk steeple]

yw

[oak falcon]

.close

[pearl pond]

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[brisk steeple]

,w 0.8^4

[jolly parrot]

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[worthy lance]

.6

[brisk steeple]

yes.

,w 0.6^4

[jolly parrot]

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[pearl pond]

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[split lark]

pls help is this solution correct?

[warm current]

Looks good

[split lark]

ok thx

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[jolly nova]

How does one solve
$$4x^3+30x^2+66x+40 = 0$$
I got this as one of the last steps in the assignment, and I don't know how to continue from here. The answers in my book didn't really help me as they fastforwarded straight to the answer.

[jolly parrot]

ReaMart

[jolly nova]

as far as I know this is a cubic/third degree equation. This is what the answer says:

[quiet tendon]

there technically is an explicit cubic formula but that's kind of overkill

if you can find a single solution you can reduce it to a quadratic

[jolly nova]

hmm yes

[quiet tendon]

rational root theorem helps you check for any rational roots here

which is always a good way to quickly find a root to reduce this

[jolly nova]

btw the function that is equal to 0 is called f(x)

and my book says
$$f(x) = 0 \text{ gives } x=-1 \vee x=-2\frac{1}{2} \vee x=-4$$

[jolly parrot]

ReaMart

[quiet tendon]

so is the problem coming up with those solutions?

[jolly nova]

yeah

[quiet tendon]

do you know what rational root theorem is?

[jolly nova]

let me translate it real quick

[quiet tendon]

okay

[jolly nova]

no I don't know it

[quiet tendon]

if you have a polynomial with integer coefficients

then there's a finite possible set of rational roots

of that polynomail

and you can find out all possible candidates for the rational roots using this theorem

take all possible factors of the constant term and divide them by all possible factors of the coefficient of the term of highest degree

so for example, you can firstly rewrite your equation as 2x^3 + 15x^2 + 33x + 20 = 0

[jolly nova]

yeah

[quiet tendon]

if you divide by 2 again you no longer ahve integer coefficents