help-39

[shy rapids]

wait

[pearl pond]

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[shy rapids]

i still need help lol what do i make the trig sub equal to?

[shy rapids]

currently im trying x=(1/√3)tan theta but im having a hard time

wait holy crap chatgpt is so useful

.close

[pearl pond]

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[acoustic path]

real

[pearl pond]

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[golden orbit]

I barely understand the question, but I tried everything

[golden orbit]

What does it mean by two positive equal roots?

Maximum amount of roots are 3

[fallow nacelle]

b²-4ac=0

[golden orbit]

Quadratic in a quintic?

[fallow nacelle]

you got 3 roots right

[golden orbit]

I don't have them

That's just the max

I don't get what 2 equal roots mean

[pearl pond]

@golden orbit Has your question been resolved?

[golden orbit]

New discovery

k has to be pos

And it has to have 2 pos roots and a negative

[golden orbit]

.close

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[pearl pond]

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[next epoch]

hello

[next epoch]

ill just send the problem in a second

now i dont know what the formula is and im kinda lost

<@&286206848099549185>

[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.

[next epoch]

the problem is it dues in 53 mins so i was a little worried, mb

nope, hw

just its 11:09 pm here and its due next day

so at 12 you cant submit it

i didnt really understand the lesson when the teacher told us but i understand pretty much everything except this

[next epoch]

<@&286206848099549185>

[knotty dew]

it's thales theorem that you should use

[next epoch]

i did it but i think there was a mistake when i did it, so can you try and apply it

[knotty dew]

ok

so

25/(25+10) = x/42 = 35/(35+y)

and you solve for y and x each one at a time

just like a regular equation

[knotty dew]

wdym, a mistake while solving for x and for y or you didn't apply the theorem properly?

[next epoch]

i think i didnt apply it correctly

but also a quick question it says why are the two triangles similar, what did you use to determine this

these two questions are required so can you check it out if you dont mind

<@&286206848099549185>

[knotty dew]

oh of course

[next epoch]

okay

[knotty dew]

i study maths in french so the terms are quite different

i gotta look them up before i answer

[next epoch]

take ur time

[tepid narwhal]

Can I just ask what the question was?

[next epoch]

this

[tepid narwhal]

Okay I see it now. Just one sec.

Thank you

[next epoch]

alright

your welcome.

[knotty dew]

last one is corresponding angles

[next epoch]

oh great

thanks

[tepid narwhal]

Oh it seems like you got an answer already

[knotty dew]

but can you please answer the first one

because i study maths in french

so when terms are technical i gotta translate

[next epoch]

@knotty dew

25
25
+
10

�
/
42
25+10
25
​
=x/42

First, simplify the expression on the left-hand side:

25
35

�
/
42
35
25
​
=x/42

Now, cross-multiply to solve for
�
x:

25
×
42

35
×
�
25×42=35×x

1050

35
�
1050=35x

Divide both sides by 35 to find
�
x:

�

1050
35
x=
35
1050
​

�

30
x=30

35
35
+
�
35+y
35
​

Cross-multiply to solve for
�
y:

25
×
(
35
+
�
)

35
×
42
25×(35+y)=35×42

Distribute on the left side:

875
+
25
�

1470
875+25y=1470

Subtract 875 from both sides:

25
�

595
25y=595

Divide both sides by 25 to find
�
y:

�

595
25
y=
25
595
​

�

23.8
y=23.8

So, the solutions are
�

30
x=30 and
�

23.8
y=23.8.

ignore the question marks cause i did it in another word processor

just take a quick look

[tepid narwhal]

Okay, I thought this was an algebraic problem but this looks like geometry which to be honest with you I am struggled in, a lot, so I don't want to give you the wrong answers.

[next epoch]

go for it, i dont mind

i mean for the first one which says why are the two triangles similar maybe you can help in

or should we get another helper?

[knotty dew]

you are taking the long run just calculate it no need to get to numbers like 1050

but the answers are correct

[next epoch]

oh great

[tepid narwhal]

Okay I can do my best to help you out, but I prefer to get another helper, just so you don't get the wrong answer if this is homework.

[next epoch]

oh okay, idk any specific helper here since im kinda new so maybe u can try

but just incase let me tag them

<@&286206848099549185>

[tepid narwhal]

Okay I will give it my best, but can I ask you do you know what corresponding angles, arem knowing those terms would make a problem like this much more simpler to solve.

[knotty dew]

y = 14

[next epoch]

x is 30 tho right?

[knotty dew]

yh

[next epoch]

of course

[knotty dew]

the others are corresponding angles are congruent

and corresponding angles

[knotty dew]

.

[next epoch]

aight let me submit it and tell you the mark

wish me luck

oh great full mark

Thanks. @knotty dew , @tepid narwhal you guys are the best

best helpers fr

.close

[pearl pond]

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[tepid narwhal]

It is two pairs of corresponding angles by the way, I just got the answer.

[next epoch]

thanks bro

[pearl pond]

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[wintry sable]

not sure how to even start this question, after the first answer.

[ivory swallow]

you may want to graph f

[pearl pond]

@wintry sable Has your question been resolved?

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[unborn abyss]

you can probably write some kind of recursive formula for this

[pearl pond]

@marble needle Has your question been resolved?

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[pearl pond]

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[supple kraken]

i dont even know where to begin im a dumb 8th grader

[blissful cloak]

It's an isoceles trapezoid, that's important for figuring out the rest of the angles

[supple kraken]

this can be closed now dw i managed to figure it out

[pearl pond]

@supple kraken Has your question been resolved?

[pearl pond]

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[latent spire]

Using the Pythagorean Theorem, I got the Tree's height to be 32

If that is: The tree =b

Last time I checked the Hypotenuse is the "c" in the Pythagorean Theorean

[last summit]

a^2+b^2=c^2

the hypotenuse is c

[latent spire]

yeah

the ladder is the hypotunese right?

[last summit]

yes

[latent spire]

so

the equation would be

4+b=36 right??

[last summit]

no

[latent spire]

what?

[last summit]

2^(2) + b^(2) = 6^(2)

4+b^(2)=36

[latent spire]

yeah

OH!!

I forgot that b is still squared

so then what??

[last summit]

subtract 4 from both sides

take the square root

[latent spire]

oh

so sqr root of 32=b?

wait

the answer is weired though

like

oh nm

the question says to round

to the nearest tenth

wait

lemme check if the answer is right

[latent spire]

YAY

thanks!!

.close

[pearl pond]

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[vale turtle]

I didn't understand the reason

[vale turtle]

They mentioned that log is always positive so it will be divergent

.close

[pearl pond]

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[wooden mantle]

Can I have help with this equation please:

f(x) = ax^2 + bx + c
f(x-1) + f(x) + f(x+1) = x^2 + 1

What is f(2)?

[wooden mantle]

My math teacher gave it to us as a challenge and while it’s not required I have solved every other math challenge she threw at us with little struggle, this one however has stumped me and I would like to know how to solve it

[unborn abyss]

try taking that second equation and applying the definition of f(x)

and then see if you can figure out a, b, and c

[wooden mantle]

I did and it simplified down to 14a + 6b + 3c = x^2 + 1
I don’t know if you can put in 2 as x on the other side of the equals sign but if you can it’s 5

[unborn abyss]

where'd all the x's go

[wooden mantle]

But the answer shouldn’t have and variables because when rounded it’ll be the answer to a safe

They were put in as 2 should they not have?

[unborn abyss]

take this equation
f(x-1) + f(x) + f(x+1) = x^2 + 1

and apply this equation
f(x) = ax^2 + bx + c

[wooden mantle]

Right so (a(x-1)^2 + b(x-1) + c) + (a(x)^2 + b(x) + c) + (a(x+1)^2 + b(x+1) + c) = x^2 + 1

[unborn abyss]

yep

[wooden mantle]

Is that simplified
3(ax^2) + 2a + 3(bx) + 3c = x^2 +1 ?

Or did I do something wrong

And where do I go next

[cosmic charm]

compare coefficients

[wooden mantle]

Because if I’m trying to find f(2) wouldn’t I just replace x with 2

How do I do that

[cosmic charm]

you see on the left

whats your coefficient of x^2

[wooden mantle]

a?

[cosmic charm]

no...

[wooden mantle]

3?

[cosmic charm]

no..

[wooden mantle]

Then what’s a coefficient

[cosmic charm]

the number infront of the x^2

[cosmic charm]

you have 3ax^2

so your coefficient is 3a

[wooden mantle]

Ok

[cosmic charm]

and on the right

whats the coefficient of x^2

[wooden mantle]

1

[cosmic charm]

so what can you conclude form that

[wooden mantle]

That it would be 2ax^2

Or 3a(2x^2)

[cosmic charm]

no..?

[wooden mantle]

I’m confused

[wooden mantle]

What do you mean

[cosmic charm]

on the left your coefficient of x^2 is 3a

and on the right its 1

[wooden mantle]

Yes

[cosmic charm]

and since they're supposed to be equal to each other

3a = 1

[wooden mantle]

So a = 1/3?

[cosmic charm]

yes

[wooden mantle]

Wait but why should 3ax^2 equal x^2

[cosmic charm]

because the left function is equal to the right function

[wooden mantle]

But why isn’t the rest of that side included on this
We are just leaving out 2a + 3bx + 3c

[cosmic charm]

we're comparing the coefficients individually

[wooden mantle]

Why are we able to do that

[cosmic charm]

ok here we'll go simpler

lets say we have ax + b = 7x + 4

what is a and b

[wooden mantle]

Yeah

A is 7 and b is 4?

[cosmic charm]

yes

now apply the same logic to the original question

[wooden mantle]

But couldn’t it also be like a is one and b is 6x + 4

[cosmic charm]

a and b are constants

meaning they are numbers

they are not associated with x

[wooden mantle]

So they aren’t variables?

[cosmic charm]

no

they are numbers

[wooden mantle]

Ok that makes more sense

I was thinking of them as variables

So back to the original equation
A is 1/3
And the rest it
2/3 + 3bx + 3c = 1

[cosmic charm]

you still havent determined b and c

[wooden mantle]

Right

[cosmic charm]

what is the coefficient of x on the left

[wooden mantle]

3b

[cosmic charm]

on the right?

[wooden mantle]

0?

[cosmic charm]

yes

[wooden mantle]

So b is 0

How do I find c

[cosmic charm]

do the same thing

[wooden mantle]

But c isn’t a coefficient of x

[cosmic charm]

you have a constant on the right

what is that constant

[wooden mantle]

1

[cosmic charm]

and your constant on the left?

[wooden mantle]

3c

[cosmic charm]

only 3c?

[wooden mantle]

And 2a

[cosmic charm]

and what was a?

[wooden mantle]

1/3

[cosmic charm]

now can you find c?

[wooden mantle]

So c is 1/6?

[cosmic charm]

really?

[wooden mantle]

1/9?

Hold on lemme grab a calculator

[wooden mantle]

Yeah 1/9 right?

[cosmic charm]

yeah

[wooden mantle]

Ok

Well doesn’t that all go to x^2 +1 = x^2 + 1

[cosmic charm]

thats the point

you got a b c

put it back into f(x)

put x = 2 inside f(x)

get answer

[wooden mantle]

Oh ok so 1.444444…

Right?

[cosmic charm]

yeah

[wooden mantle]

Let’s go

Thanks

.close

[pearl pond]

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[pearl pond]

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[limpid roost]

Proof that gcd(a,b) = gcd(b, r)

[limpid roost]

Where r = a mod b

I don't get it

a = bq + r
a - bq = r
gcd(a,b) divides a-bq, therefore must also divide r?

What am I missing? <@&286206848099549185>

[pearl pond]

@limpid roost Has your question been resolved?

[radiant terrace]

gcd(a,b) divides b, r based on what you said, so it is a common divisor of b and r. But gcd(b,r) is the greatest common divisor of b and r.

This gives you a useful inequality.

To get the reverse inequality try and apply similar reasoning to show gcd(b,r) is a common divisor of a, b

Recall if X<=Y and Y<=X then Y=X

[limpid roost]

Ok now I see it

I just did a number theory!

[radiant terrace]

You had all the right ideas fwiw. Make sure to keep an eye on how your problems and definitions connect.

[limpid roost]

Seems pretty simple now.

Now I want to prove Bezout's lemma

If gcd(a,b) = 1 then there exist x and y st ax + by = 1

[pearl pond]

@limpid roost Has your question been resolved?

[pearl pond]

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[pearl pond]

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[cosmic torrent]

You can take 4 common and sum r 3 to 3o meaning 3+4+5+... +40 and multiple it by 4

n ??

What is n

l,a??

You can use simple add formula for natural number

1+2+3+...+n =n(n+1)/2

No you just need to add r 3 to 30 not whole function

Let me e

Let me explain

Your series is
4(3)+7+4(4)+7+...+4(30)+7

[cosmic torrent]

Right 👍

Now you can see how many time you need to add 7

And you can get 4 common

Like
4(3+4+5+...+30) +7+7...+7

[pearl pond]

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[grizzled hamlet]

i havent got none of these anyone wanna help me

[lavish portal]

Uhhh

Do you know soh cah toa

Or 30 60 90 triangles

@grizzled hamlet

[grizzled hamlet]

30 60 90 triangles?

like sin and crap?

not rlly

and 30 60 90 ues

[pearl pond]

@grizzled hamlet Has your question been resolved?

[runic shale]

this problem can only be solved with sohcahtoa tho…

,tex .sohcahtoa

[jolly parrot]

onion

[runic shale]

if that helps

[pearl pond]

@grizzled hamlet Has your question been resolved?

[pearl pond]

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[pearl pond]

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[real moon]

"State the slope of the function in each interval and change in slope for each interval"
a) 6x+1/2x-4 b) 1/x^2-9

[real moon]

How do I find them?

[pearl pond]

@real moon Has your question been resolved?

[real moon]

<@&286206848099549185>

[fresh night]

Let f(x) equals (6x+1)/(2x-4)

We can calculate f’(x) for its slope

By quotient rule

f’(x)= -(13/2(x-2)^2)

When x =2, the denominator becomes 0, which leaves the function undefined at x =2

To see if the slope is going up in an interval or not, we can either sub in some values to check or calculate its second derivative

[real moon]

Ok that's helpful

I'm just confused where did this come from? -(13/2(x-2)^2)

[pearl pond]

@real moon Has your question been resolved?

[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

[potent patrol]

are these the same?

[pearl pond]

@potent patrol Has your question been resolved?

[iron path]

!1c

[pearl pond]

Please stick to your channel.

[iron path]

also

they are

they're the same thing

[pearl pond]

@potent patrol Has your question been resolved?

[potent patrol]

ok thanks

[pearl pond]

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[frosty badge]

geometric sequences a1+a2+a3 equals 91 if we add 25 to a1 27 to a2 and 1 to a3 we will get 3 numbers that make a aretmatic sequence we need to find the geometric sequences 7th number

[iron path]

i dmed

[pearl pond]

@frosty badge Has your question been resolved?

[pearl pond]

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[opal kiln]

idk how to solve

[summer imp]

Separate the integral in two pieces so that you can integrate the parts as they are defined.

So, you would break it up into [-1,1] and [1,4]

[opal kiln]

Ohh ok

@summer imp thank you!

I solved it

-close

[summer imp]

The command is .close

And you're welcome

[opal kiln]

.close

[pearl pond]

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[lost sigil]

why does it have no critical numbers?

i derivated it

and its -1/x^2

doesnt that mean the critical number should be 0?

[buoyant ridge]

but 0 is not in the domain of the original function

[lost sigil]

how could i know the domain?

[buoyant ridge]

well you can know for sure that 0 is not in the domain

functions stated with a formula are usually assumed to take the largest possible domain on which the formula is defined

but the formula 1/x is not defined at x = 0

[lost sigil]

ohhh okay thanks lots

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[dusk otter]

Can anyone help with this please/

22b)

This is what I did for part a) if that helps

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[vernal pine]

Is this right so far?

[pearl pond]

@vernal pine Has your question been resolved?

[quick yacht]

each entry in the table is exclusive

"and" and "or" would be out of the total number of people

so, a is right
b so far is right
c and d are wrong

think about how many people ordered pizza or wings

this would be W and P, W and -P, -W and P

which is how many total?

[vernal pine]

56

[quick yacht]

good

out of how many customers total?

[vernal pine]

65?

56 isnt an option tho

@quick yacht

[quick yacht]

[quick yacht]

this is for the numerator of c?

[vernal pine]

these are all the options for any box

[quick yacht]

then | must not mean or

oh wait

given that

[quick yacht]

forget everything i said, i apparently dont know what im talking about, the answers are fine

[vernal pine]

all of them ?

[quick yacht]

yeah

sorry about that

i forgot my statistics notation there

[vernal pine]

Whats this one

[quick yacht]

wings and pizza is out of everybody

so, simply the total number of customers

[vernal pine]

65 ?

[quick yacht]

yeah

[vernal pine]

Thank you!

[quick yacht]

np

[pearl pond]

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[uneven summit]

Can someone help me with this?

[uneven summit]

I actually have the answer, but I am not sure if it is correct

[merry carbon]

What do you doubt about the solution? Do you have an idea of how you'd solve the question so far?

[uneven summit]

Yes

I solved it

Already

O just wanna confirm answers so

[hard crystal]

Send ur work

[uneven summit]

Okay

There

I wouldn’t lie about this, I have a final tomorrow and this is all practice. Nothing to worry about, this isn’t homework I’m trying to copy lol

But anyways

290.25 is the height and 2.375 is the time

I used vertex form to get this

[hard crystal]

Yep its correct

[uneven summit]

Okay

Thanks

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[waxen smelt]

in spherical coordinates grad derivation

they came to this and said that they basically used chain rule but i dont understand how and on what they used the chian rule to come to that. can someone please explain

[pearl pond]

@waxen smelt Has your question been resolved?

[pearl pond]

@waxen smelt Has your question been resolved?

[humble lintel]

r, theta, and phi are each functions of x,y, z. I'm not sure what part of the chain rule you don't understand.

[waxen smelt]

I know it's fairly obvious but I just can't see how
Del/Delr = delx/delr...

[humble lintel]

If you have a function f(r,theta,z) and you take the partial derivative of f with respect to r, and r is itself a function of x,y,z, the chain rule tells you to differentiate r with respect to x, y, and z.

[pearl pond]

@waxen smelt Has your question been resolved?

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[gaunt ravine]

i didn't get the marked part here and how they solved the final equation

[gaunt ravine]

The 4th line of the marked part to be exact

I didn't get what he did there

[versed remnant]

small rearrangement , pretty sure

[gaunt ravine]

Yeah has to be but i can't see it lol

[versed remnant]

13 = 65 - 117 + (65 * 1)
13 = 65 +65 - 117
13 = 65(1+1) +117(-1)
13 = 65(2) +117(-1)

[gaunt ravine]

Owhh i see

Wait look at the next step after that

Its the samw as before

Why do that then

[versed remnant]

yeah , idk what they are doing

[gaunt ravine]

Hm i see

[versed remnant]

must be a mistake , just dont look at that step

[gaunt ravine]

And then how to solve that equation

13 = 65m + 117n

[versed remnant]

13 = 65m + 117n
13 = 65(2) + 117(-1) from before
m = 2 , n = -1

[gaunt ravine]

Ohhh i see he used that here

Thank u so much for the help !

[versed remnant]

np

[pearl pond]

@gaunt ravine Has your question been resolved?

[gaunt ravine]

@versed remnant similar problem as before i almost got it but got stuck again TT

Got that so far but idk where the other 81 went

[versed remnant]

-26(81) - 12(81)
81(-26-12)
81(-38)
-38(81)

[gaunt ravine]

Owh yeah lmao 😭

I legit can't see the obvious stuff in math

Sorry for bothering thank you

[versed remnant]

np

[gaunt ravine]

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[static nacelle]

If we have a linear function f: R^2 -> R^3 and we know that f((1,3)) = (2,-3,-1) and f((3,2)) = (-4,6,0) can we know the dim() of the kernel and image? If so, what are they

[grizzled dust]

so we know f is mapping 2 linearly independent vectors from the domain to 2 linearly independent vectors in the codomain right?

since (1,3), (3,2) are a basis for the domain, they will mapped to a basis of the image (the 2 R3 vectors)

@static nacelle are you still here?

[static nacelle]

I am I am

[grizzled dust]

does that make sense so far?

[static nacelle]

yes

[grizzled dust]

so what is the rank then?

[static nacelle]

rank as in R^2 has rank 2, R^n rank n?

not english, we dont use that term, sorry

[grizzled dust]

what is dim (image) ?

that's all that rank is

[static nacelle]

yes, that! would be 2

[grizzled dust]

yep, so what would the dim(kernel) be?

[static nacelle]

does that imply that the function is injective? since then dim(kernel) would be 0

[grizzled dust]

yes

exactly

[static nacelle]

but not surjective because for that it would need to have rank 3, correct?

[grizzled dust]

it's not onto because dim R3 > dim R2 but it is injective

yes exactly

[static nacelle]

okay helped a lot man! thanks

have a nice day

.close

[grizzled dust]

you too 👍

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[midnight haven]

\dm
Suppose that the shelf life, in years, of a certain perishable food product packaged in cardboard containers is a random variable whose probability density function is given by
\[
\m fx = \env{cases}{
e^{-x}, \q x >0, \\ 
0, \tqx{elsewhere}
}
\]
Let $X_1, \, X_2, \,$ and $X_3$ repesent the shelf lives for three of these containers selected independently. Find $\m\P{X_1 < 2, 1 < X_2 < 3, X_3 > 2}$

[midnight haven]

im a bit confused about something

what exactly is f(x) describing? the probability of what? and how do X_1, X_2, X_3 relate to it

[sonic spire]

fx is a PDF

[midnight haven]

yes, i get that

[sonic spire]

in this case its the exponential distribution

parameter = 1

[midnight haven]

yes, i get that

[unborn abyss]

you were working with these yesterday

[sonic spire]

its describing the likelyhood of an event given its value

[midnight haven]

thats not my question. I am asking how do the extra random variables specified relate to the random variable that f(x) is describing the probability of

[sonic spire]

$X_1, , X_2, ,$ and $X_3$ follow the distribution that fx describes

they should have really said $X_1, X_2, X_3 \sim\text{Exp}(1)$

[unborn abyss]

\sim

[midnight haven]

\dm So do we have $\m\P{X =x} = \m fx$ or like, $\m\P{X_1 = x, X_2 = x, X_3 = x} = \m fx$

[midnight haven]

or something

[unborn abyss]

the first one (technically it only describes intervals not exact values because it's continuous but. the first one, not the second)

[jolly parrot]

Spoon

[sonic spire]

replace the = in P with \leq

and we good

[unborn abyss]

no it's not a cdf

[midnight haven]

ok so like

what im meant to be doing is

,, \dm \m f{x_1, x_2, x_3} = \m f{x_1}\m f{x_2} \m f{x_3} = e^{-x_1-x_2-x_3}

[midnight haven]

and now like

,, \dm \m\P{X_1 < 2, 1 < X_2 < 3, X_3 > 2} = \int_{2}^\infty\int_1^3\int_0^2 e^{-x_1-x_2-x_3}\dd{x_1}\dd{x_2}\dd{x_3}

[rustic gate]

bro is using \dm now

[midnight haven]

i cant stand anything light

[unborn abyss]

i mean you could do it that way

or you could just use the fact that they're independent

and split $\mathscr{P}(X_1 < 2, 1<X_2<3, X_3>2)$ into $\mathscr{P}(X_1<2) \cdot \mathscr{P}(1<X_2<3) \cdot \mathscr{P}(X_3>2)$

[jolly parrot]

ヘイリー

[midnight haven]

oh hmm

i guess u can do that

[rustic gate]

oh my god

\mathscr

[unborn abyss]

i'm so fancy u already know

[midnight haven]

so like

,, \dm\int_0^2 e^{-x_1} \dd{x_1} \cd \int_1^3 e^{-x_2}\dd{x_2} \cd \int_2^\infty e^{-x_3}\dd{x_3} = \int_{2}^\infty\int_1^3\int_0^2 e^{-x_1-x_2-x_3}\dd{x_1}\dd{x_2}\dd{x_3}

woops uhm

[rustic gate]

why are you putting braces around x_1

[sonic spire]

let him cook

[rustic gate]

this is right

[midnight haven]

[unborn abyss]

you need braces after \dd or the spacing is wrong

[midnight haven]

,, \dd x_2

[rustic gate]

not for lex

[midnight haven]

oh i didnt know that im so used to bracing everything that has any subscripts

or superscripts

[rustic gate]

\dd doesnt take an argument

[midnight haven]

[unborn abyss]

lex can have little a brace, as a treat

[midnight haven]

is this true doe

[rustic gate]

yes

[unborn abyss]

yes the integral is separable

[rustic gate]

thats just factorising the integrand

[unborn abyss]

or whatever you call it

[midnight haven]

OH YEAH

[rustic gate]

the calculus thing where [ \iint \map fx \map gy \dd x \dd y= \int \map fx \dd x \int \map g y \dd y]

[midnight haven]

you called me out on this snow

like a month ago

[rustic gate]

what

[rustic gate]

did i

[midnight haven]

ye like

u can split the integral if like the bounds are like not

[rustic gate]

oh i might have

[midnight haven]

varying

[rustic gate]

i rember

[midnight haven]

anyways bye thanks hayley and spoon

[rustic gate]

your brain should go factorise when you see independence anyway

[midnight haven]

[rustic gate]

independence and factorisation are synonymous almost

[midnight haven]

gotchu king

.ckose

.ckose

.close

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[midnight haven]

bruh

[rustic gate]

ckose-lounge moment

[midnight haven]

[rustic gate]

[pearl pond]

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[storm harness]

where did this t go here?

or why

[pearl pond]

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[storm harness]

.close

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[quaint crown]

Why i can`t do 3^^-2? How to solve 3^^0.5? I mean tetration
(yes, ik samples like 3^^3 is 3^3^3=3^27)

[quaint crown]

So, how to type tetration in WolframAlpha?

[woven cliff]

I didn't get ur question

[quaint crown]

^^ is tetration

3^^2 or ²3

3^^3=³3=3^(3^3)=3^27

Like this

[inland lantern]

,w tetration 3 2

[quaint crown]

hmm, wat if i^^i

[quaint crown]

Because, in 3^1/2 it`s √3(bruh, why 1/2 is √?), 3^^1/2 is a? Wat

How to understand this?

[pearl pond]

@quaint crown Has your question been resolved?

[pearl pond]

@quaint crown Has your question been resolved?

[pearl pond]

@quaint crown Has your question been resolved?

[quaint crown]

Okay

.close

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[frosty badge]

I got 2 chords that come from the same point the big one is 12cm and the smaller one is 10cm. The distance betwwen the small ones midpoint to the big chord is 4cm need to find the radius of the circle

[versed remnant]

any diagrams?

[frosty badge]

@versed remnant

Should bd someting like this

[versed remnant]

is it specified at which point of big chord does the line join?

[frosty badge]

No just its from the midpoint of the small one

[versed remnant]

i mean the only thing i can think is using pythagoras theorm but i doubt if it solves it

as we dont know if its a right triangle or if big chord crosses the origin of circle

[frosty badge]

Sooo

What do we need to know to solve it

[versed remnant]

if big chord goes through the origin i suppose

whats the answer as per answer key?

[frosty badge]

The answer is 6.25

[versed remnant]

i mean im getting 6.40 but i suppose the way im doing has lots of errors

[frosty badge]

So can you solve it??

.close

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[subtle skiff]

I apologies in advance for my messy writing haha. The area of a rectangle is 20a^4b^4 − 8a^3^b + 6a^4b^3
. The length of one side of the rectangle is 2a2b . What is the length of the other side of the rectangle if a = 3 and b = 2 ? I feel like I did it way more complicated then it need to be.

[zealous shore]

I get 816 as well when I follow along your work. I don't see a quick way to do the problem, it's just a lot of work. You can save a bit of time if you look at (20a^4b^4 − 8a^3b + 6a^4b^3)/(2a^2b) and notice that the 2a^2b can cancel with everything in the numerator, but there's still a lot of calculations to do.

[subtle skiff]

I appreciate your help @zealous shore and feed back!

.close

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[unkempt rapids]

can someone help idk how to solve these questions

[tall flint]

if you're allowed to use l'hopital's rule, that is probably the path of least resistance

if not, then things perhaps get harder

[light helm]

factorisation
limit identity involving sin(k)/k

[pearl pond]

@unkempt rapids Has your question been resolved?

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[young widget]

hii can someone help me w the function

i also tried -tan(2pi x)

[hot stone]

So the number that multiplies with x in trig functions is called "frequency"

Example:

Sin(3x)

Frequency = 3

Period = 2pi/frequency

[young widget]

even in tan graphs ?

[hot stone]

Oh sorry

[young widget]

i thought that was only for sin and cos

oh

[hot stone]

Pi/freq

[young widget]

so pi/2 ?

[hot stone]

Because period of tanx = pi

Yes but youre in degree

So

180 degree/360 degree = pi/period

= freq

So the number you get from that is 1/2

Therefore the number you'd multiply to x is 1/2

Not 2

The period is twice as long as "normal", so the frequency is 1/2

[young widget]

wait i don't understand why it's 1/2 and not 2

[midnight haven]

a good way to think about it is like

recognise that tan(x) is asymptomatic at pi/2 *k

because cos(x) = 0 at pi/2

so, from your graph you can see that it's asymptomatic at pi's instead

so you want to multiply by a 1/2 for that pi/2 to appear at an instance of x = pi

[young widget]

ohhh wait i get it

[hot stone]

I think of it this way:

The normal period is 180 degree

You have a 360 degree period

The period is twice as long, which means it takes twice as long for the x values to go through the period

Therefore you can imagine that the x values are being slowed down

Hence why you multiply by something less than 1

[young widget]

okay i think i get it

[hot stone]

If you had a 90 degree period, you'd multiply by 2

Because you go through the period twice as fast

[young widget]

okk

so for the graph

1/2 x ?

[hot stone]

Yea that would agree with a tan curve that has a 360.degree period

[young widget]

ok let me try it

[hot stone]

You find that number by doing

"Normal Period" / Actual Period

[young widget]

ok tysm

it's right

.close

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[azure mist]

can anyone help me with this

[azure mist]

<@&286206848099549185>

[robust oar]

we don't help in quizzes/tests

[azure mist]

ight

.close

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[robust oar]

same question?

you need to find two points into the graph and then plug these numbers into the formula

[midnight haven]

how do i find the graphs

[robust oar]

look at the graph

[midnight haven]

so whats the equation

[robust oar]

here's an example of this

[midnight haven]

i got 60

as my answer

is that correct

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

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[hybrid hawk]

given that the radius of the inscribed circle is 1, how to find the distance from (0,0) e.g. center of the circle to one of the vertices of the polygon?

[vital estuary]

use trig using one of the tangency points

[hybrid hawk]

what if i dont have the tangency points

are there any properties of pentagon that can be used?

[vital estuary]

you know the angles in the pentagon

[hybrid hawk]

yep its a regular pentagon

[hybrid hawk]

alright cool

thanks

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[pearl pond]

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[versed venture]

Hello. I have the following problem

d/dx[(x+1)*(3_sqrt((x+1)/(x-1))], find f'(2)

I am fine with what to do with the function to find f'(2) once I have found the derivative; however, I am having trouble getting the derivative.

I just do the product rule and then the quotient rule when I take the derivative of the second term, correct?

[versed venture]

or is there an easier way to go about it?

[light helm]

can you show your attempt

[unborn abyss]

$(x+1)\cdot\sqrt[3]{\frac{x+1}{x-1}}$ is that right?

[light helm]

was 3_sqrt supposed to represent cube root?

[unborn abyss]

oh that would make more sense

[jolly parrot]

ヘイリー

[versed venture]

yes that is the problem

[unborn abyss]

how did you handle the cube root?

[versed venture]

i change it to powers

[unborn abyss]

ok cool

what you describe will work but ummmm I'd recommend combining that (x+1) in front into the numerator

[versed venture]

right, so quotient rule on ((x+1)(x+1)^1/3))/(x-1)

so the algebra will just kind of be annoying?

[unborn abyss]

combine them more

$(x+1)¹(x+1)^{\sfrac13}$

[jolly parrot]

ヘイリー

[unborn abyss]

can combine those into (x+1)^4/3

[versed venture]

oh yeah ofc

okay thanks

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[versed venture]

.reopen

[pearl pond]

✅

[versed venture]

just to clarify... so now i apply d/dx[(p'q-pq')/q^2] where in this case the p' and q' require the chain rule to be applied, yeah?

[unborn abyss]

errrr the pqqp stuff is taking the derivative so the d/dx would go away atp

but yeah quotient rule will work for this

(or you can do what i tend to prefer and use negative exponents and the product rule)

[versed venture]

and so rearrange it to (x+1)^4/3 * (x-1)^-1/3?

[versed venture]

ohh okay thanks for telling me that

[unborn abyss]

yeah

[versed venture]

yeah that is a nicer way to solve it. thank you

[unborn abyss]

join the quotient rule hating team

[versed venture]

fr

okay thank you

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[pearl pond]

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[tall prawn]

I've been really scratching my head at this, can someone help me before 11:59?
The questions is only worth a point but it's all I have left and I want to get it

[pearl pond]

@tall prawn Has your question been resolved?

[tall prawn]

omg guys I can't believe I actually got it!! AHHHH
I had the right concept, just should've used a constant in place of C instead, for the first part it should've been 28-ln(7) rather than 4x-ln(x)
and I feel A M A Z I N G now, I CAN FINALLY SLEEEP–
aight, I'mma go sleep now, I wish everyone else here luck with their math work, gn everyone and sweet dreams

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[pearl pond]

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[regal fern]

can i get help on this question

[karmic fern]

let's try not to cheat on your timed assessments

[regal fern]

bruh please

[prime bramble]

It’s in #rules

[regal fern]

okay

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[pearl pond]

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[prisma basin]

Is it LinReg equation y=ax+b?

*a+bx

that’s what I got when I plugged it in

[plush bramble]

Doesn't look like you rounded according to the instructions

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

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[pearl pond]

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[signal lynx]

i just wanted some clarifications on this question. i have an idea on how to solve it but i wanted to know if its correct and if it could be cleaned up a bit

[signal lynx]

my current working theory is proof by contrapositive. we prove if x<0 then x^3 +5x <= x^2 +1

if x<0 and x is a rational number, then x is negative

a negative number cubed is negative. a positive int times a negative int is also negative. thus, x^3 +5x is negative

a negative number squared is positive so x^2 is positive

with the addition to 1 the righthand side of tghe inequality is positive

making it negative number x <= positive number x

which is true

which should prove the original question?

[pearl pond]

@signal lynx Has your question been resolved?

[pearl pond]

@signal lynx Has your question been resolved?

[glass meadow]

Yes it's correct, though the wording is a little awkward

$$x<0 \implies [ x^3 + 5x < 0 \text{ and } x^2 + 1 > 0 ] \implies x^3 + 5x \le x^2 + 1$$

That's all you need really

[jolly parrot]

Nel

[signal lynx]

yeah im not really good at translating my like mental logic to formal notation 😭

[signal lynx]

i dont have to prove that x^2 +5x is less than 0?

[glass meadow]

You can if you want, I just think it's obvious

[glass meadow]

Actually this isn't quite the contrapositive

$$x \le 0 \implies [ x^3 + 5x \le 0 \text{ and } x^2 + 1 > 0 ] \implies x^3 + 5x < x^2 + 1$$

[jolly parrot]

Nel

[glass meadow]

This is the contraposition of " x^3 + 5x >= x^2 + 1 implies x > 0 "

[signal lynx]

ah

that makes sense thank you

would you mind helping me w this next one as well? my current logic for it is to suppose n is an even integer. this means n=2j for some j in Z. n=4k is the same as n=2(2k). k is an integer so 2k is basically any integer in Z which proves n=4k is even. similarly, n=4k+2 is equal to n=2(2k+1) where 2k+1 is also a valid int. both n values fit the definition of even so if n is even then it is also equal to n=4k or n=4k+2 for some int k

[pearl pond]

@signal lynx Has your question been resolved?

[split aurora]

this only proves that if n=4k or n=4k+2 then n is even

[signal lynx]

how come? if n is even its equal to 2k for some arbitrary int k. n=4k and 4k+2 are also even for their respective arbitrary k’s. doesn’t than mean they’re equal?

how else would you prove it?

[split aurora]

Maybe there is some even number not of the form 4k+2 or 4k but still even

For example 6k and 6k+2 are both even but not all even numbers are of that form

They can also be of the form 6k+4

[split aurora]

Do you know the division theorem

[signal lynx]

uh

i don’t think so

[signal lynx]

ah

that makes sense shit

[split aurora]

It says that for any two integers n and q there exists two unique integers k and r where 0<=r<|q| such that n=qk+r

It's relevant in this case because it means every integer is of one of four forms:

• 4k
• 4k+1
• 4k+2
• 4k+3
  You prove that 4k+1 and 4k+3 are impossible if n is even, and then you just have to show that 4k and 4k+2 are both possible.

[signal lynx]

how might you go about proving it without the use of the division theorem?

[signal lynx]

.close

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[pearl pond]

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[vital estuary]

cool photo

ask a question

[abstract fox]

How do I find the equation?

[tender tartan]

Start by finding the equation for velocity as a function of time

to do so, you take the antiderivative of the acceleration such that v(0) = 1

[abstract fox]

1/4t^2 + 4t + C?

[tender tartan]

yes, and substitute C so that v(0) = 1

[abstract fox]

Set the equation equal to 1?

@tender tartan or 0?

[tender tartan]

at t=0, you want the equation to equal 1

[abstract fox]

Ohh ok

[tender tartan]

Then do the same thing again to get the position

[abstract fox]

So find the integral again?

Ohh ok ok I got it thank you

@tender tartan may you help me with one more?

[tender tartan]

sure, what is it ?

[abstract fox]

The derivative of f(x)=(e^x)(lnx)

[tender tartan]

use the product rule : the derivative of u(x)v(x) is u'(x)v(x) + u(x)v'(x)

[abstract fox]

I did

I got e^x+e^xlnx/x, but it's wrong

[tender tartan]

what's the derivative of e^x, and what's the derivative of ln(x)

[abstract fox]

e^x and 1/x

[tender tartan]

then (e^x)'*ln(x) and e^x(ln(x))' are ?

[abstract fox]

Isn't it just e^x?

[abstract fox]

e^x(lnx)+e^x(1/x)?

[tender tartan]

yep

[abstract fox]

But the answer is e^x(1+xlnx)/x

[tender tartan]

ln(x) = xln(x)/x

[abstract fox]

Wait it is?

[tender tartan]

you multiply ln(x) by x/x (which is 1)

[abstract fox]

Why x/x?

[tender tartan]

you want to factor out 1/x, but you can only multiply ln(x) by one without changing the function, so you multiply by one making 1/x appear

so x/x

[abstract fox]

I'm confused sorry

Right after this what do I do?

[tender tartan]

you want to factor out a 1/x

on the right it's easy

on the left, you have to make it appear

[abstract fox]

Like this?

[tender tartan]

so you say that 1 = x/x, making the left part e^xln(x)x/x

[tender tartan]

no, you have e^x/x + e^xlnx = e^x/x + e^xlnx * 1 = e^x/x + e^xlnx * x/x = e^x/x(1 + xlnx)

[abstract fox]

Ohh ok

I get it now

Thank you so much

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[pearl pond]

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[tawdry wolf]

is desmos broken

[tawdry wolf]

geogebra gets it right

is it possible to get negative values with this equation?

[solid pier]

-1^(-1) = -1

you get x^x negative for all values of the form x= -1/k where k is odd

[tawdry wolf]

wait then geogebra is wrong

[tawdry wolf]

is it something like "it oscillates so fast between positive and negative" that it doesnt display on the -x side

[solid pier]

well x^x is weird at negative values

[solid pier]

yeah something like this

also, what is x^x for a negative irrational value? well it’s really not easy to tell what it should be

so maybe geogebra says “uh hello? what is this weirdness going on?” and doesn’t graph it there

[sharp vigil]

if you have a negative number to a non-integer power, you get an complex number. so a graph that only shows real outputs doesn't fully capture it

,w plot z^z

[jolly parrot]

Results provided by WolframAlpha

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Requested by cloudedhorizons

[tawdry wolf]

interesting

[solid pier]

not entirely true no? (-1/3)^(-1/3) is -3^(1/3)

it depends on the branch cut ig

[sharp vigil]

wolframs plots the principal root, which is complex, whereas desmos just plots it implicitly, so anywhere where the real-valued root exists is plotted

[pearl pond]

@tawdry wolf Has your question been resolved?

[pearl pond]

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[blissful orbit]

1. (8-1)! = 7! = 5,040

2: 4!4! = 576

3. 4! = 24

Correct me if I'm wrong

[pearl pond]

@blissful orbit Has your question been resolved?

[blissful orbit]

No

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[hexed spoke]

in a vector field $ F = x^3i + y^3j + z^3k $ we are calculating flux of hemisphere with using divergence teorem and spherical coordinates we have property of $ z = 1 $ and $z = sqrt( 1 - x^2 - y^2 ) $ what is the flux going through that hemisphere ?

[hexed spoke]

I used divergence teorem than find 3(x^2 + y^2 + z^2) and used triple integrals, also since it is a hemispherical shape used spherical coordinates to evaluate

[pearl pond]

@hexed spoke Has your question been resolved?

[pearl pond]

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[teal tendon]

i’m not sure how to find the value of x here
i tried to find the total area but i’m not sure where that brought me or what to do with what i have
this question is in a packet about quadratics so i know it’s related to that but i’m not sure how

[earnest stratus]

Now you've gotten area as a function of x.

Which is a quadratic.

Do you know when a quadratic attains max/min value?

[teal tendon]

the vertex right?

[earnest stratus]

Yes.

[teal tendon]

so do -b/2a?

[earnest stratus]

yes.

[teal tendon]

and that’s the x value that i’m looking for?

[earnest stratus]

yes.

[teal tendon]

oh alr

thanks

i’ll try it

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[craggy spoke]

I'm trying to integrate sin(lnx)dx, im doing the u-substitution
but the part that I get stuck on is. I set u = lnx, so du= 1/x dx. so to replace dx would I leave it as x*du ( dx= xdu) or do I just pull out the 1/x ?

[craggy spoke]

[glacial sequoia]

write x as e^u

[craggy spoke]

I just had to process that. and that made sense to me. Thank you

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