help-38

[karmic acorn]

that makes a lot more sense

[uncut walrus]

I imagine the next part of the question asks for specific details about the continuity of the function?

[uncut walrus]

okay

I don't know how it wants it described?

If it doesn't give specifics on how it wants it described then I don't know because I forgot the lingo 😭

[karmic acorn]

yea idk it just says discribe the continuity or discontuinity

[uncut walrus]

Do you have another question?

[uncut walrus]

okay

[karmic acorn]

i guess this one might have more then one answer

[uncut walrus]

an odd function is any function where f(-x) = -f(x)

Ping me if that doesn't make sense or if you have an answer

[karmic acorn]

its weird

[uncut walrus]

?

[karmic acorn]

so like the function has to be negative and positive?

[uncut walrus]

It has to have positive and negative parts

think about the function f(x) = x which is an odd function

if you plug in, say... 3

you get f(3) = 3

if you plug in -3 you get f(-3) = -3

[karmic acorn]

how do you determine if one is odd tho

[uncut walrus]

I'm explaining it rn

[karmic acorn]

ohh okay thanks

[uncut walrus]

so for the function f(x) = x is -f(3) = f(-3)? (asking for your response)

[karmic acorn]

i think so i just dont know how the f got the negative sign behind it and the three doesnt and then they switch on the second part

[uncut walrus]

it's the definition of an odd function

A function is odd if...

at any point (x, y) (x is input, y is output)

using the negative of the input (-x) will result in the negative of the output (-y)

[karmic acorn]

so its even if they are switched

[uncut walrus]

I don't know what that means when you say it

[karmic acorn]

i mean like if y is the input and x is the output is it even

[uncut walrus]

..?

[karmic acorn]

lol nvm

[uncut walrus]

an even function is when f(x) = f(-x)

think of f(x) = x^2

[karmic acorn]

f(x) = x <- odd
f(x) = f(-x) <- even

[uncut walrus]

no

-f(x) = f(-x) <- odd
f(x) = f(-x) <- even

[uncut walrus]

yes

the reason is because...

f(x) = x is odd because in this case f(-x) = -x and -f(x) = -x

or if it suits you better...

f(x) = x and -f(-x) = x

does that make sense?

[karmic acorn]

a little bit

[uncut walrus]

Pick a function and I'll walk you through determining whether or not it's odd.

[karmic acorn]

thanks, i was just about to ask that

[uncut walrus]

okay

so f(x) = -(1/2)x^4 + 5

what is -f(-x)?

do you know how to find -f(-x)?

[karmic acorn]

no.....

[uncut walrus]

okay

let's start with something else first then

how about f(1)?

[karmic acorn]

f=1

?

[uncut walrus]

how did you come to that conclusion?

[karmic acorn]

because f(1) there is only two factors to look at here, f and 1

[uncut walrus]

have you learned function notation yet?

[karmic acorn]

i dont think so

[uncut walrus]

okay

so if I say f(x) = 2x then ask you to find f(2) that means to take the function f(x) which is equal to 2x and substitute 2 in place of x

so in this case f(2) = 2 * 2

does that make sense?

f(2) doesn't mean f times 2

it reads as f of 2

[karmic acorn]

so f(x)=3x is f(3)=3*3?

[uncut walrus]

YES!

now rq

to verify your understanding

if I said f(x) = x * x and asked you what f(2.5) was what would you say?

[karmic acorn]

f(2.5)=2.5*2.5?

[uncut walrus]

YES

perfect!

[karmic acorn]

awesome, i feel like we are making progress

[uncut walrus]

now back to this...

you know what f(1) is now right?

[karmic acorn]

f(1)=1*1?

[uncut walrus]

for this function

when f(x) = -(1/2)x^4 + 5

[karmic acorn]

well do i solve everything after the equal sign to find x?

[uncut walrus]

so f(x) = -(1/2)x^4 + 5
and I'm asking you what f(1) is

you replace x with 1 like you did before

[karmic acorn]

so i replace x with -(1/2)x^4 + 5?

[uncut walrus]

what would it look like if you did that?

[karmic acorn]

f((-1/2) x^4+5)?

[uncut walrus]

okay, that's not right

[uncut walrus]

think about this

you did this one correct

[karmic acorn]

i would multiply the function by itself?

[uncut walrus]

no

I'll give you the answer to this initial question so you can see what to do.

if f(x) = -(1/2)x^4 + 5
then f(1) = -(1/2) * 1^4 + 5

do you see what happened?

[karmic acorn]

so where did the 1 come from in the f(1)

[uncut walrus]

it's just a number I chose

[karmic acorn]

oh and we still come out with the right answer?

[uncut walrus]

function are interesting

x means any number

[karmic acorn]

you can say that again

lol

[uncut walrus]

there's a unique* answer for anything you replace x with

[karmic acorn]

oh, x can be anything really and we can still determine if the function is odd ?

[uncut walrus]

we're holding off on this problem for now

[karmic acorn]

okay but was this function odd?

[uncut walrus]

In the case of
if f(x) = -(1/2)x^4 + 5
then f(1) = -(1/2) * 1^4 + 5
f(x) = -(1/2)x^4 + 5 is the information provided, f(1) is the question, and -(1/2) * 1^4 + 5 is the answer

[uncut walrus]

we haven't got there yet, you need to understand function notation before you can answer this question.

[karmic acorn]

okay

[uncut walrus]

think about it like this

saying f(x) = 3x is the same thing as saying y = 3x

[karmic acorn]

thats right because f(x) is basically y ?

[uncut walrus]

yes

then f(1) means to substitute 1 in place of x in the equation y = 3x

[karmic acorn]

okay i get it

[uncut walrus]

so

1 more verification

[karmic acorn]

so where x is in the equation we sub 1

[uncut walrus]

back to this
f(x) = -(1/2)x^4 + 5

[uncut walrus]

which equation?

[karmic acorn]

f(1)= -(1/2) 1^4+5

[karmic acorn]

x becomes 1

[uncut walrus]

yes

so now

f(x) = -(1/2)x^4 + 5

what about f(2)?

[karmic acorn]

f(2)= -(1/2) 2^4 +5

[uncut walrus]

Yes!

That's it!

no what if I try something like this...

f(x) = -(1/2)x^4 + 5
what is f(a)?

[karmic acorn]

a is just another form of x no?

so

[uncut walrus]

yes, but you still need to substitute it!

[karmic acorn]

like

you could sub a with 1 or 2 the same way we did with x

[uncut walrus]

don't overthink it!

just do the same thing you did before

f(x) = -(1/2)x^4 + 5
what is f(a)?

[karmic acorn]

f(a)= -(1/2) a^4 + 5

[uncut walrus]

yes!

okay

[karmic acorn]

yea i didnt really quite understand what u were asking at first

[uncut walrus]

f(x) = -(1/2)x^4 + 5
what is f(-x)?

[karmic acorn]

but now i do

f(-x)= -(1/2)-x^4+5

[uncut walrus]

1 slight problem

[karmic acorn]

what is that?

[uncut walrus]

the correct answer is

f(-x)= -(1/2)(-x)^4+5

Do you see why?

[karmic acorn]

oh we would put it in perethesis so it doesnt look like a subtration sign right?

[uncut walrus]

and also

The negative sign needs to be affected by the ^4 as well

-x^4 reads as -(x^4)

which is not what we wanted

[karmic acorn]

i see

[uncut walrus]

we're so close to the odd function answer though!

[karmic acorn]

awesome

[uncut walrus]

f(x) = -(1/2)x^4 + 5
what is -f(x)?

[karmic acorn]

idk why but adding that sign screwed me up

[uncut walrus]

I'm not too shocked since I didn't explicity explain that

basically

if f(x) = 3x then -f(x) = -3x

[karmic acorn]

and if there was more to the equation where would that negative sign go

[uncut walrus]

around the whole thing

so

if f(x) = 3x + 5 then -f(x) = -(3x + 5)

[karmic acorn]

so if its in front of f then its in front of the rest of the equation ?

[uncut walrus]

yes

[karmic acorn]

can i make an example and you tell me if its right??

[uncut walrus]

yeah

[karmic acorn]

thanks

f(x)=7x + 1/3 + 5
-f(x)=-(7x+1/3) + 5

would it work the same with the added 5's?

[uncut walrus]

the 5 needs to be in the parentheses.

[karmic acorn]

okay so like

-f(x) = -(7x+1/3) + (5)

[uncut walrus]

no

like this

-f(x)=-(7x+1/3 + 5)

you just wrap the whole thing and make it negative.

[karmic acorn]

oh it would be added into the the first set of parentheses

[uncut walrus]

yes

[karmic acorn]

i see

[uncut walrus]

so now...

[karmic acorn]

lets try again

[uncut walrus]

f(x) = -(1/2)x^4 + 5
what is -f(x)?

[karmic acorn]

-f(x)= -((1/2)x^4 +5)?

[uncut walrus]

yeah

so now

we'll do 2 at the same time

f(x) = -(1/2)x^4 + 5
what is -f(x)?
and
what is f(-x)?

you should have 2 answers because there are 2 questions

[karmic acorn]

1
-f(x)= -((1/2)x^4 +5)
2
f(-x)=-((1/2)(-x^4)+5)?

[uncut walrus]

f(-x)=-((1/2)(-x^4)+5)? no this one's just f(-x)= -(1/2)(-x^4)+5?

wait a minute

I've been decieved by my eyes

[karmic acorn]

lol

[uncut walrus]

this one was actually subtly wrong

[karmic acorn]

dang they both wrong?

[uncut walrus]

you wrote
-f(x)= -((1/2)x^4 +5)
when it should be
-f(x)= -(-(1/2)x^4 +5)

[karmic acorn]

nooooooooo i forgot to add the subtraction sign in front

[uncut walrus]

XD

give it another go!

f(x) = -(1/2)x^4 + 5
what is -f(x)?
and
what is f(-x)?

[karmic acorn]

-f(x)=-(-(1/2)x^4 +5
and
f(-x)=-(1/2)(-x^4)+5

??

[uncut walrus]

double check your first one

you may have forgotten a character

You're so close right now, there's a slight issue in both answers though

[karmic acorn]

the )

[uncut walrus]

yes

and the second one

you wrote (-x^4) when it should be (-x)^4

does it make sense why it should be that way?

[karmic acorn]

not really

[uncut walrus]

I didn't see the mistake the first time... whoops

[karmic acorn]

lol

[uncut walrus]

so my correction in that case was also wrong

[karmic acorn]

its fine

[uncut walrus]

so\

[karmic acorn]

so this is the way to do it (-x)^4

[uncut walrus]

yes

[karmic acorn]

okay i got it, i think

[uncut walrus]

because (-x^4) looks like (-(x^4))

so after those corrections your answers would be
-f(x) = -(-(1/2)x^4 + 5)
and
f(-x) = -(1/2)(-x)^4 + 5

the question is... are they equal?

[karmic acorn]

wdym by that

[uncut walrus]

is -f(x) the same as f(-x)?

[karmic acorn]

i would assume they would be since we just switched some things around

[uncut walrus]

(you need to simplify the expressions to find out)

do you know how to simplify these expressions?

[karmic acorn]

no.....

[uncut walrus]

okay, no problem

lets start with -f(x) = -(-(1/2)x^4 + 5)

the part we're simplifying is -(-(1/2)x^4 + 5)

do you have any clue where to start?

[karmic acorn]

uh

with

the parentheses

[uncut walrus]

okay, so what do you do with the parentheses?

[karmic acorn]

uh

not really

idk

[uncut walrus]

okay

let's try something similar but easier

-(-2 + 5)

do you know how to simplify this?

[karmic acorn]

do 2-5 then add the negative?

[uncut walrus]

that's a way to do it

now what if I make it -(-2x + 5)?

[karmic acorn]

do 2x+5 then -(7x)?

[uncut walrus]

you can't add -2x and 5 to get -7x

[karmic acorn]

😦

[uncut walrus]

-2x + 5 can't be simplified further

instead

distribute the negative

-(-2x + 5)

[karmic acorn]

?

[uncut walrus]

for example

[karmic acorn]

?-2x and -5

[uncut walrus]

there is another way to simplify -(-2 + 5)

you can distribute the negative and get --2 - 5

but a double negative is a positive

so it becomes 2 - 5

you can apply the same logic to -(-2x + 5)

[karmic acorn]

so like this but the end result would be just 2-5

[uncut walrus]

not quite

you forgot x

[karmic acorn]

so -(-2x+5)
(-2x+-5)
-(2x-5)<----?

[uncut walrus]

no

[karmic acorn]

bruh lol

[uncut walrus]

start with
-(-2x + 5)
distribute the negative to get
--2x - 5
double negative makes a positive
2x - 5

[karmic acorn]

so is there 3 negatives?

to start with?

[uncut walrus]

explain what you mean?

[karmic acorn]

oh wait, im thinking the sub sign is a negative

[uncut walrus]

it technically is

[karmic acorn]

so at one point in the process we have this right -2x-5?

[uncut walrus]

no

almost though

[karmic acorn]

my head bruh, lol, do you mind if we continue this tommorow

[uncut walrus]

we go from -(-2x+5) to --2x-5

[karmic acorn]

so the plus sign turns minus

[uncut walrus]

yeah

[karmic acorn]

and we put the two negatives together

and those become a positve

[uncut walrus]

yes

[karmic acorn]

therefore no more negative

[uncut walrus]

so we get...?

[karmic acorn]

which brings us to 2x-5?

[uncut walrus]

yes!

[karmic acorn]

finally lol

[uncut walrus]

now try it on
-(-(1/2)x^4 + 5)

[karmic acorn]

-(-(1/2)x^4 +5
uh im sorry this just has more numbers and it has a fraction which i suck at and an exponent unlike what we just did which was just 2x - 5

[uncut walrus]

okay

the exponent and fraction don't need to change at all!

only the + and - signs!

[karmic acorn]

okay

[uncut walrus]

Here I'll write it in LaTeX if it makes it easier!

[karmic acorn]

sorry my brains frying

[uncut walrus]

$-\left(-\frac{1}{2}x^{4}+5\right)$

[solid kiln]

RYZEN 9 3950X

[karmic acorn]

is that the entire functoin unsimplified?

[uncut walrus]

yes

[karmic acorn]

okay so still needs to be simplified

(1/2x^4-5)?

[uncut walrus]

YES!

$\frac{1}{2}x^{4}-5$ is correct!

[solid kiln]

RYZEN 9 3950X

[karmic acorn]

[uncut walrus]

now

this one
f(-x) = -(1/2)(-x)^4 + 5

we need to simplify $-\frac{1}{2}\left(-x\right)^{4}+5$

[solid kiln]

RYZEN 9 3950X

[karmic acorn]

does the -1/2 need to stay negative?

[uncut walrus]

yes, the only part that needs simplified is actually the $\left(-x\right)^{4}$

[karmic acorn]

(1/2 x^4-5)?

[solid kiln]

RYZEN 9 3950X

[karmic acorn]

and we cant remove x from the equation?

[uncut walrus]

explain what you mean by remove x

[karmic acorn]

(-1/2^4 - 5)

[uncut walrus]

no we can't

[karmic acorn]

okay

(1/2(x^4)-5)?

[uncut walrus]

no

[karmic acorn]

am i close?

[uncut walrus]

we only need to change the $\left(-x\right)^{4}$ part

[solid kiln]

RYZEN 9 3950X

[karmic acorn]

(1/2 x^4-5)?

[uncut walrus]

start with $-\frac{1}{2}\left(-x\right)^{4}+5$ and simplify the $\left(-x\right)^{4}$ part

[solid kiln]

RYZEN 9 3950X

[karmic acorn]

no

[uncut walrus]

that's all good

lets try some numbers and see if we find a pattern

what is $\left(-1\right)^{4}$

[solid kiln]

RYZEN 9 3950X

[karmic acorn]

1

[uncut walrus]

okay

[solid kiln]

RYZEN 9 3950X

[karmic acorn]

16

[solid kiln]

RYZEN 9 3950X

[karmic acorn]

16

[solid kiln]

RYZEN 9 3950X

[karmic acorn]

16

[uncut walrus]

but they're the same value

[karmic acorn]

yes

2 and 4

[uncut walrus]

do you know why they end up the same?

[karmic acorn]

i dont know why (2)^4 comes out to positive 16

[solid kiln]

RYZEN 9 3950X

[uncut walrus]

and 2 * 2 * 2 * 2 = 16

$\left(-2\right)^{4}$ means $\left(-2\right)\cdot\left(-2\right)\cdot\left(-2\right)\cdot\left(-2\right)$

[solid kiln]

RYZEN 9 3950X

[karmic acorn]

ohhhhh i see

[uncut walrus]

so

what does that mean about $\left(-x\right)^{4}$?

[solid kiln]

RYZEN 9 3950X

[karmic acorn]

x^4

[uncut walrus]

YES

[karmic acorn]

so simple

[uncut walrus]

so simplifying $\frac{1}{2}\left(-x\right)^{4}+5$ makes what?

[solid kiln]

RYZEN 9 3950X

[karmic acorn]

1/2 x^4 +5

[uncut walrus]

YES

so

we simplified
-f(x) = -(-(1/2)x^4 + 5)
and
f(-x) = -(1/2)(-x)^4 + 5

[karmic acorn]

yes

[uncut walrus]

the answers were $\frac{1}{2}x^{4}-5$ and $\frac{1}{2}x^{4}+5$

[solid kiln]

RYZEN 9 3950X

[uncut walrus]

since those are not equal

[karmic acorn]

they are odd?

[uncut walrus]

no

[solid kiln]

RYZEN 9 3950X

RYZEN 9 3950X

[karmic acorn]

they are not odd so they are even?

[uncut walrus]

not always true

some functions are not odd or even

[karmic acorn]

so the one we just did is neither?

[uncut walrus]

we didn't check that

so we don't know

if you'd like I can quickly do the check for you and show my work so you can see how

[karmic acorn]

sure

[uncut walrus]

$f\left(x\right)\ =\ -\frac{1}{2}x^{4}+5\$

find $f\left(-x\right)\$...
$f\left(-x\right)\ =\ -\frac{1}{2}\left(-x\right)^{4}+5\$
simplify
$\f\left(-x\right)\ =\ -\frac{1}{2}x^{4}+5\$

$-\frac{1}{2}x^{4}+5$ is equal to $-\frac{1}{2}x^{4}+5\$
so $f\left(x\right)\ =\ -\frac{1}{2}x^{4}+5$ is an even function

[karmic acorn]

so its even?

[solid kiln]

RYZEN 9 3950X

[uncut walrus]

it is even!

[karmic acorn]

awesome

[uncut walrus]

I'll give a few examples so you can see one of each kind (odd, even, neither)

$f\left(x\right)\ =\ x^{2}$ is even
$\f\left(x\right)\ =\ x^{3}$ is odd
$\f\left(x\right)\ =\ x^{3}+x^{2}$ is neither

[solid kiln]

RYZEN 9 3950X

[karmic acorn]

i see

[uncut walrus]

you can also have a function that is both

[karmic acorn]

so would x^4 be evven?

[uncut walrus]

yes

[karmic acorn]

and x^5 would be odd

[uncut walrus]

yes

[karmic acorn]

i understnd now, all this is more simple then it lookes then when i first started

[uncut walrus]

Yeah

There are just a decent number of prerequesites

and finally, I have an example of a function that is both even and odd...

[karmic acorn]

lol

lol what

[uncut walrus]

I was wrong, I chose the wrong function

$f\left(x\right)\ =0$ is even and odd

[solid kiln]

RYZEN 9 3950X

[uncut walrus]

in fact, it's the only function that is both even and odd

[karmic acorn]

that makes sense

youve been a great help man thanks for all this.

i better head to bed its 1

130 am

[uncut walrus]

aight

be sure to .close

[karmic acorn]

thank you man your awesome

[karmic acorn]

yes sirrrr

.close

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[fossil grail]

!help

[fossil grail]

How do I solve Q4 with working?

?

Is it just 540g

?

?

He

?

@everyone

Help

22 minutes

Been waiting

@maiden fjord

.close

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Send your question here to claim the channel.

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• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[quasi apex]

i'm so confused for finding limits of piecewise functions

i thought the lim t->2 p(t) is undefined

because the right sided limit of p(t) is undefined, so therefore the limit from both sides is undefined

can someone help me with this?

please ping me when you have a response

[hybrid path]

So you mean to say that right sided limit for t tends to 2 is undefined?

[quasi apex]

mhm

i thought it's undefined

[hybrid path]

And why do you think so?

[quasi apex]

because there's a jump from the right handed section of the graph to the other side

[hybrid path]

That is from 3 to 4

Not from 2

[quasi apex]

i don't understand like

what's the general rule of thumb for these limits of piecewise functions

[hybrid path]

There is no jump for t=2

[quasi apex]

hmmm

so if the limit was from t->3 there is a jump

so in that instance

[hybrid path]

Yes

[quasi apex]

would it be undefined?

from the right hand side btw

[hybrid path]

Like in question (b)
They specified left hand and right hand limits differently

[hybrid path]

No

None is undefined in this graph atleast

For 3 The left hand limit would be somewhere between 6 and 10 according to the graph
And the right hand limit would be 0

Since the function from 3 to 4 is y=0 , that is, p(t)=0

[quasi apex]

.close

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[chrome vigil]

I don’t understand how the domain isn’t 0 ≤ x ≤ 20. The problem is:

Jen and Giovanni are using 2 way radios to communicate while studying at a reserve. The average range for a 2 way radio is 15 miles. But, the actual range of the two-way radio can vary from the average from up to 5 miles.

Jen is in the camp and Giovanni is in the preserve. If x is the distance Giovanni is from the camp and y is the variation in range from the average, the equation y = |x -15| models the variation in possible range.

a) determine the domain and range that make sense for this context.

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[chrome vigil]

<@&286206848099549185>

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<@&286206848099549185>

[uncut walrus]

@chrome vigil to find out why it's not 0 ≤ x ≤ 20 we should start by asking why you think it is 0 ≤ x ≤ 20

[chrome vigil]

I think it’s 0 ≤ x ≤ 20 because x is the distance he is from the camp. Can’t he be anywhere from 0-20 miles away, since that’s the range the radio can work from?

0 is right next to the camp, and 20 is the maximum he can go if the range is 5 miles above the average of 15.

[uncut walrus]

The question isn't asking how far away he can be from the camp, it's asking what the possible value for range on the radio is.

[chrome vigil]

oh, so 10 ≤ x ≤ 20?

[uncut walrus]

yeah

[chrome vigil]

ok, thank you.

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[simple slate]

Use an algebraic proof to prove ∑𝑛―1,𝑘=0, 𝑃(𝑚+ 𝑘,𝑚)= 𝑃(𝑚+ 𝑛,𝑚+ 1)/(𝑚+ 1) .

[simple slate]

I'm not sure how to prove this.

[zinc ginkgo]

can you just write this out. formatting is impossible to read

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[simple slate]

yes

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[digital bolt]

what have you tried

[wraith hinge]

just filling on 3 for x

[stark bison]

And how did it go?

[split chasm]

and where did that lead you

[split chasm]

continue

also keep that 2 more centred

so it doesn't look like its going in the wedge

[wraith hinge]

okok

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[wraith hinge]

How can I find the length of i?

[solid kiln]

3317

3317

3317

[stark bison]

You mean xtan(theta) = 1?

Also the line AB can be represented as y = xtan(theta)*

[wraith hinge]

True, true. Missed that part.

[stark bison]

(as long as theta is the angle that OA makes with x-axis)

[solid kiln]

3317

3317

[stark bison]

Yeah, so the coordinates of B are 1/tan(theta) and 1

[wraith hinge]

That's what I needed, thanks a lot.

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[wet olive]

in a) is that a good enough solution for this linear system of equations and in b) how do I prove that it has no solution

[wet olive]

<@&286206848099549185>

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[acoustic trout]

note that there is a better way here by substituting u=cosx

or you could also substitute u=sin^2(x) too if you want

[tough pond]

that’s what I was thinking at first but I’m limited to integration by parts

can’t use u sub

[acoustic trout]

ah, right then

[tough pond]

yea have to integrate by parts twice I’m pretty sure just not sure if the problem itself is ugly or I made it that way

[acoustic trout]

then you might want to actually group the terms differently

the idea is to not introduce anymore x terms into the integrand

and keep it as only between trig functions

[tough pond]

hmm let me think about it

[acoustic trout]

for example, here, you could try to integrate sin^3(x)cos(x) and differentiate cos(x)

[tough pond]

so I’d let u = cosx and find du (-sinx)

and sin^3xcosx would be dv (integral of dv = v)

[acoustic trout]

and v is?

[tough pond]

Integral of dv (sin3xcosx)

[acoustic trout]

i know, but what is the antiderivative of that?

[tough pond]

uhh 1/4(sin^4x)

• c

[acoustic trout]

yes

no need for the plus c

[tough pond]

oh right

[acoustic trout]

now just continue with what you have

[tough pond]

and then I can just plug

👍

@acoustic trout had one more question about priority list - lets say they both are the type of function do you use the power to determine which one gets priority

or can you use any

[acoustic trout]

no, you just see if i integrate this thing, will it look simpler, or will it give me another term that i can integrate together with the du, similarly, if i choose this to be u, will du look simple enough so that my integral vdu does not look nasty

[tough pond]

oh okay got it

thanks

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[jolly basin]

i got 2x=x^2-1/x+3

and then got that is x=-3+-2√2

is that right so far?

then do i have to plug it back in?

i got 17+-12√2

what do i do after that?

im so confused

.close

[burnt mulch]

that's the point on the line

then you have that combined with (1,-3)

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[quasi frost]

Help with finding out he one that could not represent the line

[jolly basin]

what have u tried so far?

[quasi frost]

I have eliminated i so far as yknow

I’m kind of lost trying to figure out 2 and 3

[jolly basin]

try to think about it

if i and ii

were equal to each after simplifying either of them

same thing with iii

[quasi frost]

I just have a question i’m curious on how you would get -4 over 3

[jolly basin]

point slope formula

well there are 2 steps

first is finding the slope using 2 points

then use point slope formula

and then

[quasi frost]

I do not understand how you would get a negative tho

[jolly basin]

well let's try it

[quasi frost]

As it’s (0,8) and (6,0)

[jolly basin]

ok let's make (0,8) our first point and (6,0) our 2nd

let me make this latex

$\frac{8-0}{0-6}=\frac{8}{-6}=\frac{-4}{3}$

[solid kiln]

Scarecrow

[jolly basin]

the negative sign can be infront of the fraction, position doesnt matter but it's considered better to write it either infront of the fraction of in the numerator

[quasi frost]

Why did you subtract 8-0 and 0-6?

[jolly basin]

(0,8) is our first point

and (6,0) is our 2nd point

so we subtract the y vals

8-0

in the numerator

and 0-6 because 0 is the x value of our first point

[quasi frost]

Ohhhhhh

Ok I get it

[quasi frost]

for ii -4/3 times 1 is -1.3 repeat right

Or am I doing it wrong?

[jolly basin]

it's -4/3 time -3

[quasi frost]

Do you not multiply x too?

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[fringe leaf]

Is there a shortcut to simplifying this bottom part

[boreal dome]

,rotate

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[wind bloom]

how do i solve these type of questions?

[woven trail]

can a number be divided by 0

basically u need to find the values for x that make ur denom 0

so (2x-1)(x+8)=0

it’s like solving a quad now

the values of x u get are what it can’t be

[wind bloom]

so -8 and 1/2 are correct answers?

[woven trail]

yeshh

🥳

[wind bloom]

tysm ❤️

[woven trail]

welcc!

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[cobalt jungle]

i need help with this question, it seems like to easy lmao is the answer 40? or is there something im missin g

[vernal briar]

The sum of angles in a triangle is 180 degrees.

50x + 40x + 90 = 180

Solve this equation.

[cobalt jungle]

1

[vernal briar]

Yeah, x = 1

[cobalt jungle]

so then angle a would be 40?

[vernal briar]

Yeah

Thought you were saying x = 40

[cobalt jungle]

okay thought so, thanks man.

nah lol

thanks bro

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[rancid kayak]

anyone knows how to solve this?

[wraith hinge]

"show"?

is there any more context to the question?

[rancid kayak]

That's it.

[wraith hinge]

the y^-1 cancel, so there's no way for equality to always happen

[feral surge]

Just basically solve it and your answer has to be the same thing as the one on teh right hand side

[feral surge]

what if you remove the ^-1 by inverting everything

I don't know man, I need to go

[rancid kayak]

Is this question even right?

[fair bison]

no lol

[feral surge]

would inversion work?

[fair bison]

I don't see any way to interpret this question that makes sense

it's just wrong

[feral surge]

Oh ok I guess

[rancid kayak]

Alright thanks

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[sour herald]

Can somebody help figure out AB and BU

[fading kayak]

look at the smaller triangle

[feral surge]

use proportions

[sour herald]

Oh

[feral surge]

AB:CD = AU:DU

[sour herald]

Top is 256?

Or nah

Meters

[feral surge]

Use proportions to find AB

the use pythogaras rule

[sour herald]

But is BU 256?

[feral surge]

$(32 + x)^2 = 225^2 + \qty(AB)^2$

[solid kiln]

DuxDave

[feral surge]

Idk man I may be wrong

[feral surge]

use this when you find AB using proportions

[feral surge]

....

this is the correct thing

I made a mistake at first

sorry

[sour herald]

i think i got it

BU is 256 and AB is 160

Hopefully its corect

Is it?

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[lament jewel]

how does this work when the poles and zeros are of higher order than 0? like if weve an order 3 pole at z=0, wouldnt f'/f have a simple pole at z=0?

[lament jewel]

btw for this, g'/g and h'/h are holomorphic at those zeros/poles

[feral surge]

Yeah maybe

[feral surge]

I'm glad I helped 😄

[lament jewel]

NO OPEN ANOTHER CHANNEL

[feral surge]

Sorry

[lament jewel]

ok thats all phew

[sour herald]

I did

[lament jewel]

my savior is here

is allg

[amber python]

the residues of the simple poles of f'/f are exactly the orders of the poles of f

that was in one of the previous slides

also zeroes ig

so by adding up the residues of the simple poles

you calculate the sum of the orders of the zeroes and poles of f

it's called the argument principle

[lament jewel]

idk

isit this

[amber python]

yes

ν is the order

taking logs just brings down the exponent

[lament jewel]

w8w8

[amber python]

so your simple poles inherit the exponent as their residue

[lament jewel]

im lagging

[amber python]

brain lag

smh

[lament jewel]

yes

[amber python]

well im gonna dip

[lament jewel]

:O

thanks sweetie

❤️

[amber python]

❄️

.close

jk lol

[lament jewel]

LOL

wait

how did that not close

[amber python]

i said jk didn't i

the bot understood me

[lament jewel]

WOW

[amber python]

psychic

[lament jewel]

no i dont get

how they

product ruled

:c

can u

[amber python]

:c

[lament jewel]

help me with product rule

[amber python]

just do it

[amber python]

just write it out

product rule the top equation

[lament jewel]

[amber python]

ν(z0) is a constant

it's not a function

doesn't depend on z

[lament jewel]

ok got it

but

why cant they show

working

thanks

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[lament jewel]

❤️

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[strong root]

In this proof why do they use different values of delta for (3) and (4)

[strong root]

This is the rest of the proof for the If (1) then (2) proof

[strong root]

It seems to me that if they are using two different values of delta then they should be using different epsilon values as well but that is not the case

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[strong root]

After a little more Im thinking it may be because in the definition of the limit they say for all epsilon > 0 so we can pick really any arbitrary epsilon but the delta is unique to the limit that is in question

<@&286206848099549185>

[waxen narwhal]

I believe that's right 🤔

[strong root]

Great thank you

do you understand the steps they take in the second picture because that's where my head starts to hurt

[waxen narwhal]

I did pretty badly in complex analysis 😆 but let me take a look

[strong root]

😂 appreciate it

[waxen narwhal]

How is |i| typically handled? I think the triangle inequality makes sense in the first line but unsure about that one small detail

[strong root]

basically norm of complex number is just the sqrt of the sum of the squres of the real part and the img part so |i| = 1 i believe

[waxen narwhal]

Gotcha thank you

Seems to me like a basic definition of limits for complex functions, with a lot of symbols thrown in because you have double the variables to think about

I think you also should bear in mind that the epsilon/2 is also arbitrary -- they could have said epsilon instead in both (3) and (4) and written 2epsilon at the end of the proof and it still would have been valid

[strong root]

They just do that cause it looks nice there right

[waxen narwhal]

yeup

[strong root]

ok this gives me more to think about really appreciate the help

[waxen narwhal]

[strong root]

.close

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[hallow spruce]

HELP

[hallow spruce]

PLEASE

help

SOMEONE

My teacher didn't explain us

We have a test

Please

Someone

[trim jolt]

@hallow spruce Has your question been resolved?

[hallow spruce]

HELP

[unreal loom]

ping helpers

[pine coyote]

you haven't even given your question

[gloomy pendant]

lol

[pine coyote]

https://dontasktoask.com/

[hallow spruce]

<@&286206848099549185>

Fuck wait

Lemme go take a picture

<@&286206848099549185>

HELP

HELP

[main cargo]

Where? What? What have you tried? Where did you struggle?

[hallow spruce]

I don't know how to solve it

The question

I don't understand

What the answer is supposed to be

The teacher didn't teach us nor give us notes

I don't know what to do

In this question

[hallow spruce]

This one

Everything on question one

[main cargo]

a is -2

[hallow spruce]

How?

[main cargo]

It says the slope is -2

[hallow spruce]

So what is 11?

[main cargo]

11

[hallow spruce]

So I write answer is -2?

[main cargo]

the solution should just be y =-2x + 11

[hallow spruce]

Ok

What about 2?

1. B)

[main cargo]

You know 2 points

[hallow spruce]

(2,-5) and +-3,13)

[split socket]

Chalk, if you're gonna try to help, explain what you're doing, don't just give solutions

Help them be able to come up with the solution themselves

[hallow spruce]

Bro the teacher hasn't even give us the answers and never will, i need to know the answers to know if I'm doing my math right

[main cargo]

slope = a = (y2-y1)/(x2-x1)

[hallow spruce]

So I'm not confident

If I'm right

[hallow spruce]

Ok

I did that

And got -.36

-3.6

[main cargo]

Possible, I didn’t do the math

[split socket]

If your process is correct, your answer will be correct. So just explain how you get your answers and if there's anything wrong, we'll point it out

[main cargo]

So now you have the slope from y=ax+b

[main cargo]

Put values from one point in there and solve for b

[hallow spruce]

Teacher never gave us anything

That's why I'm using online help

Because if teacher ain't gonna do her job

Then I need someone to tell me if it's right

From online

[split socket]

So you learned how to do these online?

[hallow spruce]

Yes

Because my new teacher doesn't teach us it

She gives us the sheet

And tells us to do it

[main cargo]

Here do this

[hallow spruce]

And doesn't show us how to do itt after

[split socket]

Alright, well, can you explain what you learned? The steps for solving?

[hallow spruce]

Nor the answers

[hallow spruce]

Learned nothing

I'm learning from people online

[split socket]

I mean online

What did you learn online

[hallow spruce]

Those are for the other questions and speaking of this, please hurry up because I don't have enough time, i got two test on Wednesday, history test and a math test and i can only study up to 2 hours max per day since I got classes for my sports

So after this i gotta go

In like 1 hour

It's my only studying time

The rest gotta go for history

And i got a science test on Thursday

Please just hurry up and give me the answers so I can look into it

But again

I'm confused how to solve this

[split socket]

Sorry, this server isn't for handing out solutions

[split socket]

If you want to understand how to do it, someone will gladly help

[hallow spruce]

How do you do 2.d)

You got the x intercept

And the two coordinates

I wasn't taught that

How do you do that?

[split socket]

Do you know the one above it?

[hallow spruce]

Yes

You find the slope with the formula

And then you pick a random coordinate

And apply it

To find your B initial value

[split socket]

This is the same idea, you have a second point, if you'll notice

[hallow spruce]

But they only gave me x intercept

With one coordinate

I don't understand

[split socket]

x-intercept is -3, that means the line goes through (-3,0)

And there's your second point

[hallow spruce]

Ok okok

Now how do you do 1.f and 1.g?

They don't give me a number

[split socket]

x-intercept means the x coordinate of the point at which the line intercepts the x axis, so at this point, y will be zero

[hallow spruce]

Ok

What about f and G

[split socket]

A horizontal line has a slope of zero, always

[hallow spruce]

HOLY SHIT

Help

[split socket]

???

[hallow spruce]

How do you solve this

[split socket]

Dawg

What

[hallow spruce]

Wait

I'll do it

Tell me if it's correct

[split socket]

Okay

[hallow spruce]

So

Ok?

Then I subtract 8 with 0

Or is it divide?

[split socket]

Yeah, that's correct
Then what's the final answer?

No, subtract

[hallow spruce]

Isn't it not divide?

[split socket]

You want the b to be alone

[hallow spruce]

So what is dividing for

[split socket]

When you have a = bx, for example.
If you want x to be alone, divide by b

To get a/b = x

[hallow spruce]

Huhhh

[split socket]

Here's one way to think about it

You want to the b to be alone, right?

[hallow spruce]

I don't know I'm confused

[split socket]

Okay, look

[hallow spruce]

Is this formula

To find y intercept?

The dividing

[split socket]

I'll explain from the start so you're following

[gloomy pendant]

,rccw

[solid kiln]

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

Sep 26, 2022, 8:54 PM

[gloomy pendant]

rotate!

[hallow spruce]

Ok

So

Let's finish this first

[split socket]

What this question wants you to do is find an equation of the form

y = ax + b

To find this, you need to know what a and b are

[hallow spruce]

Ok

So you find a by y2-y1 divide x2-x2

And then

You pick two coordinates to find B

[split socket]

So for example, if you find that a is 6, and b is 11, your final answer will be

y = 6x + 11

[hallow spruce]

Apply it

So

Put Y in Y

[split socket]

Yes, that's how you find a and b

[hallow spruce]

And X in X

[split socket]

Correct

Yes

[hallow spruce]

Times 6 with X

And then

Bring it over

To the other side

Subtract

?

[split socket]

No, no, once you find a and b, that's it

Just write the equation with them

That's the final answer

Don't move anything

[hallow spruce]

When do I move?

[split socket]

When you're trying to find b

[hallow spruce]

Ok so 6 times x

Will divide

Y

?

Hold up

Is B the y intercept?

[split socket]

Exactly

Yes

[hallow spruce]

..