help-19

[umbral obsidian]

is there a way to solve the left integral without using the formula on the right (when i complete the square i get the integral on the right

[umbral obsidian]

!status

[odd edge]

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

[umbral obsidian]

2

[cold swift]

just let u=x-1.5 and a=1/2

[elfin zodiac]

Show your work then

[cold swift]

oh wait you didnt want to use this way?

[umbral obsidian]

thing is i didnt do any working all i did was complete the square

and i just subbed into the formula

which is the correct answer

[cold swift]

mhm

[umbral obsidian]

but my teacher said you can't do that in the exam

and i put it into symbolab

and they used 3 substitutions

[cold swift]

what you used up there is really called euler's substitution

specifically the first substitution

thats how the formula is derived

[umbral obsidian]

ye but that means we gotta derive it in an exam i gues

so how would i do it without the substitution

[elfin zodiac]

Sorry i am a bit confused. Why is this?

[umbral obsidian]

honestly idk bro

its not in our

reference sheet

and if its not in our reference sheet we cant use it

unless we prove it

[elfin zodiac]

Ok so you want to do with partial fractions

[umbral obsidian]

like 2 square roots?

on the denominator

[elfin zodiac]

Have u ever done partial fractions

[umbral obsidian]

and rationalise?

[elfin zodiac]

Oh wait

Forget about it

I forgot something

I got crazy with denominator

[cold swift]

you can use trig sub here actually

i think

x=sec(theta) maybe?

[umbral obsidian]

whats that get you tho

like sqrt(3/2) sec theta ?

[cold swift]

you can get rid of the square root in the denominator

[umbral obsidian]

or just x = sec theta

[cold swift]

oh well like in this case itll be x-3/2=sec(theta) sorry

[umbral obsidian]

i sub sec theta into this >? or the complete the square one

[cold swift]

the complete the square one

[umbral obsidian]

hmm ill have a look

[cold swift]

yeah i just did it, it should work

your answer will be off by a constant tho but itll be fine since its indefinite

[umbral obsidian]

did you sub in x = sec theta + 3/2

[cold swift]

compared to the formula one ofc

well i basically derived the formula up there

like the integral of 1/sqrt(x^2-a^2)

honestly might be more clean if you just take u=x-3/2 and put it all in terms of u

[clever fjord]

y0shi

[umbral obsidian]

oh yea

sub asec theta

i forgot about that

i did that once to prove it i think

took me 2 pages

but i get it

hanks

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[clever fjord]

Hamdy Hisham

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[lament sage]

hello

[lament sage]

i’m trying to convert this into integral notation

and i’m having trouble figuring out the a and b value.

have i done this right?

when i convert it into integral notation, would i have to subtract 1 from both the numerator and denominator so that it matches with the summation notation?

like this?

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[outer wadi]

I'm having trouble solving this with 3 variables

[outer wadi]

I am able to do these questions with two variables but I am a bit confused with three

What I tried doing was setting all the partial for f(x,y,z) to 0 and found no critical points

but I'm confused on how to deal with the constraint

I tried solving for z and then substituing and then taking the partial derivaitve of x and y and setting those to 0 and finding where they both equal 0 but I couldn't get that to quite work

so I wasn't sure if that was the right method

I can't use lagrange multipliers

[honest quest]

Suppose we want the maximum. If there is slack in the constraint this means we can improve the objective function. Therefore we set an equal sign for the constraint

Why can't we use langrangian multipliers I was going suggest that

[outer wadi]

this is before we learn lagrange multipliers

[upper ice]

How to perform La Grange multipliers

[outer wadi]

you'd set the gradient of f equal to lamda times the gradient of the constraint function and do some substitutions

but I'm trying to figure out how to do it without that

so far the only two critical points I've gotten is

(root(2), root(2)) and (-root(2), -root(2))

but im not sure if those are right

theres also probably more

[odd edge]

@outer wadi Has your question been resolved?

[tulip spire]

You do the exact same thing as for 2 variable, find critical points. We see that f is continous so by Weierstrass’ theorem minimum/maximum will be either in one of the points you found or somewhere on the edge of this ball(the constraint)

because i beliebe this ball is a closed set

also im not sure if what i mean is called ball in english so correct me

[outer wadi]

hmm ill try again

I think your looking for the word sphere

[tulip spire]

And by the edge of the ball i mean you need to check all points such that x^2 + y^2 + z^2 = 2

[outer wadi]

but a ball is a sphere so its basically the same

[tulip spire]

ohhh right, lmao then why is the symbol for it B like its a ball damn

polish naming always stupid

Anyway, just use Weierstrass’ theorem,
do you know it?

[outer wadi]

oh no we havent learned that

but I used an online max and min calculator to look at the answer and it seems I have the right equation I just am not sure how to solve it

my algebra seems to be lacking

Solving this should give 8/sqrt(33) for x and y

let me try solving it again

[tulip spire]

maybe you did but without the name

necause without that its pretty much impossible to do it

[outer wadi]

somehow my algebra got me 8/sqrt(31)

oh I know why

[outer wadi]

ok I was able to solve this for x and y and got 8/sqrt(33)

got to move things to the other side square it and solve for one variable and substitute

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[barren dawn]

Where is my false part?

[barren dawn]

Need some help

[odd edge]

@barren dawn Has your question been resolved?

[barren dawn]

No

[shrewd sapphire]

For #1, because the rate of change is positive, does that mean the number of abnormal cells will be increasing or decreasing?

@barren dawn

[barren dawn]

yeah

Gonna be incerease

right?

[shrewd sapphire]

Yep, so picking false was right
Next, the proporition of abnormal cells is the number of abnormal cells divided by the total number of cells

[barren dawn]

yeah, so this is right for sure

[shrewd sapphire]

Mhmm

So are you good on 1 & 2?

[barren dawn]

yeah?

[shrewd sapphire]

Is the question mark there since you're still unsure about something or because you'd like me to continue?

[barren dawn]

I am ok

Thanks!

[shrewd sapphire]

Since you now know what #1 and #2 are you know that #3 must be false

[barren dawn]

yeah

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[weary tusk]

want someone to check this for me : If A is a square matrix of dimension mxm and c is a constant ,then : $|cA|= c^{m} |A|$

[clever fjord]

Hamdy Hisham

[zinc glacier]

yes

you can think of cA as (cI)(A)

so |(cI)(A)|=|(cI)||A|=c^m|A|

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[fervent solstice]

good afternoon, i would like some help

[fervent solstice]

-(2m-2)/2 = -m+1=3, -m=-3-1, -m=-4, m = 4?

is it right?

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[fresh sphinx]

Hi I only have one attempt remaining for this question and would like to verify what im doing wrong

[fresh sphinx]

this is the wrong answer, can anyone show me where i went wrong

wait i totally did ibp wrong

can anyone please double check this?

waut

[errant epoch]

u went wrong in the first step

[fresh sphinx]

holy cow yeah

derivative of x is mostly likely not f(x)

[errant epoch]

yuppp

[fresh sphinx]

well

i will be back in 2 minutes

@errant epoch does this look correct lol

[errant epoch]

yuppp

[fresh sphinx]

yayy thanks!

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[tribal basin]

Did I do this math wrong?

[tribal basin]

1st part was trickiest for me.

Hm

[undone mauve]

give me a bit

[odd edge]

@tribal basin Has your question been resolved?

[tribal basin]

👍 thx

[undone mauve]

sorry

which part do you need help with?

@tribal basin ^

[tribal basin]

The first part I just want to know if I did the math correct, if not where I could change.

Might as well check b and C if you could too I would appreciate it.

[undone mauve]

I'm not really aware of the formula you have used, but confirm whether this is what you did: you added the areas of the 2 rectangles and the triangle

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[frank tide]

So y= log ,a’ (x) (the base is a, if someone can write this in latex ill give them a video of some1 placing a sunflower on a duck’s head)

[frank tide]

My teacher says that

[meager juniper]

$y = \log_a(x)$

[clever fjord]

OmnipotentEntity

[tawdry cave]

Where is the video, man? I am waiting here.

[meager juniper]

same

[frank tide]

Can i atleast send yhe question first

[tawdry cave]

What? Such betrayal.

[frank tide]

Fine

[tawdry cave]

But yeah, send the question. Lol.

[frank tide]

Here

[frank tide]

So my teacher said that A cant be a negative number and- A can be a negative number when x is 0 and is y is 1… also, he said that x cant be 1 which is weird bc x can be 1 when A X and Y are all 1

Which now im confused

[tawdry cave]

$y = \log_a(x)$

[clever fjord]

Anonynmous

[tawdry cave]

x can't be 0. Are you sure that your teacher said that?

[frank tide]

Wait wait

[meager juniper]

So $y = log_a(x)$ is equivalent to solving the expression $a^y = x$. So we have a bit of a problem when $a = 1$ or when $a$ is a negative number.

[clever fjord]

OmnipotentEntity

[frank tide]

Yeah i meant that A is 1 x can be anything and y will be 1 so

[meager juniper]

So, if you have $1^x = 2$ what is x?

[clever fjord]

OmnipotentEntity

[frank tide]

I do understand yhe problems with a=1 but it is solvable eith certain values of y and x. Which contraricts what mu teacher said “a CANT be 1”

And a can be a negative number as long as x is inbetween 0 and 1

[meager juniper]

how do you figure?

[frank tide]

Wdym?

[meager juniper]

ok, so, coming back to the definition of log.

[frank tide]

You can just tell me that my teacher is inaccurate or im wrong then explain bc i dont wanna annoy you with my slow brain

[meager juniper]

if we have $y = \log_a(x)$ as being defined as the value of y that makes the expression $a^y = x$ a true statement. And I put a 1 in for $a$, then there's only one possible value of $x$ for which this statement has any true values for $y$, do we agree here?

[clever fjord]

OmnipotentEntity

[frank tide]

Yes

Which is 1

[meager juniper]

ok

now here's the question

we agree that $log_1(x)$ has no value, is undefined, unless $x = 1$. But what is its value when $x = 1$?

[clever fjord]

OmnipotentEntity

[frank tide]

1

Or wait

No

[meager juniper]

🙂

You see the problem now?

[frank tide]

When a is 1 and x is 1 then y can be anything

[meager juniper]

Yes, but a function requires only a single value for a function

if any value will work, then what you have is not a function.

[frank tide]

Oh yes

Hmmm okay

[meager juniper]

Next, let's talk about negative values for a.

So this one is a little bit more wibbly, your math teacher is correct, about real numbers.

Because in general for non-integer exponents and negative bases, a^x is not defined.

As you point out, if we restrict ourselves to only certain values of x, yes, we can find some solutions.

But realistically speaking, there's actually quite a bit going on here.

if we consider complex numbers.

[frank tide]

But this is not hoe a function work, yeah?

[frank tide]

How a function works*

[meager juniper]

No, you can define a function that is mostly undefined, with only a few isolated defined values

that's perfectly valid.

it's just not useful

[frank tide]

So whats the prob-

Oh

[meager juniper]

and it kinda gives you some bad intuition

because later on, when you learn about the complex logarithm, you'll see that it's not the entire story.

so, for simplicity, we generally don't teach negative base logarithms.

[frank tide]

Ooooh… that sounds kinda exicting

Wait

Complex logs exist?

Why

I mean

[meager juniper]

Yes.

So you know a + bi is a complex number?

[frank tide]

Sure theyre useful in some firlds (complex numbers) but r they reallu that useful in advanced math

[frank tide]

Yes

Its also abi which is a pretty name

Sorry

[meager juniper]

There's another way of writing a complex number that is equivalent, called the polar form, it's r e^(it), where r is the length of the complex number and t is the angle.

[frank tide]

I shouldve zipped it

[meager juniper]

the key to realize is that if you add 2pi to t, you get back the same number.

which means that the representation of the number changes, but the number itself doesn't.

[frank tide]

Okay yeah ty for everything

[meager juniper]

like 1/2 and 0.5 are the same number.

[frank tide]

Aye i understood that

[meager juniper]

now, if we take the natural logarithm of this number, we have ln(r) + i(t + 2pi k) where k is an integer.

which means that the complex logarithm is not single valued, like we would expect out of a function.

now, the way that exponents are defined for real and complex numbers is actually built on exp and ln, so this is quite important.

[frank tide]

Oh- i mean yes i see.

I understand the second part atleast

[meager juniper]

which means that z^w for general complex numbers z and w is also not single valued.

So you can see this is actually quite a conundrum that touches a lot of areas of complex numbers.

we have to talk about the "principal" exponent, and so on.

and that's a bit of a can of worms to open on the heads of poor, unsuspecting high schoolers just learning about logarithms.

so we say negative bases aren't allowed. 🙂

[frank tide]

Daaaammn-

Ok nice

Thats hella interesting

Abt to leave mu studies and make research abt it

.leave

[meager juniper]

!done

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[frank tide]

Oh 😭

[frank tide]

Ikkk but its been a while

.close

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[tribal basin]

What would the formula be

I thought I could try to answer it my own without help clearly not 🥲😂

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@tribal basin Has your question been resolved?

[hollow quarry]

Lets start with some assumptions.

What can you say about the front face (the cross-section) of the feeder? What shape is it?

[tribal basin]

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[untold blade]

where did i mess up?

[untold blade]

y = lnx +5x

lny = ln(lnx) + ln5x

y'/y = 1/xlnx + 5/5x

y' = y(1/xlnx + 1/x)

[dawn tiger]

if y = ln(x) + ln(e^(x^5)), then you can't simplify ln(e^(x^5)) to 5x

[untold blade]

ohhh

[dawn tiger]

the 5 is the exponent of just the x, and the whole x^5 is the exponent of e

so it's ln(e^(x^5)), not ln((e^x)^5)

5x would be a correct simplification if this was the second case

[untold blade]

so ln(e^5x)

thats correct right

i just realized its not even asking for the derivative 💀

[dawn tiger]

y is ln(x) + x^5

because you get ln(e^(x^5)) which is x^5 * ln(e) = x^5

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[untold blade]

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[tribal basin]

Is this correct ?

[odd edge]

@tribal basin Has your question been resolved?

[tribal basin]

@jolly lantern

[tribal basin]

<@&286206848099549185>

[twilit garnet]

@tyson Your solution seems to be correct.

Are you instructed to evaluate $\pi$ as 3.14?

If not, please just keep it as the exact value.

[clever fjord]

Your LaTeX Helper

[tribal basin]

Ok thank you

[twilit garnet]

If that answers your question, please type .close to allow other people to use the help channels.

[tribal basin]

.close

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[mystic saffron]

im having trouble finding reference number to a terminal point

[mystic saffron]

like i can find it but i need to know if the point is in first quadrant or second or third or fourth

how can i do that?

given a terminal point in radian

how to know which quadrant

[latent scaffold]

If it's not already, make sure the point is within [0, 2pi) by adding or subtracting multiples of 2pi from the angle you have

Then it should be easier to tell which quadrant it is from useful angles on [0,2pi).

[mystic saffron]

oh yeah

this makes sense

im able to do the problems

thank you

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[pure fossil]

My conclusions aren't equivalent. I'm not sure where I went wrong. Apologies for messy handwriting it's almost 1 am for me

[mystic saffron]

I can’t read the handwriting.

Can you type it out; perchance?

@mystic saffron You have good eyes, can you read this?

Um no the words Are too skinny and it's setting an Unrealistic standard for me.

[pure fossil]

Ah sorry

There are 2 equations that should be equivalent

1/2(x+1)² - 1/4(2x+1)

-1/4(x-1) + 1/4x(2x+3)

You're supposed to simplify to prove their equivalent but my answers don't match

For the first equation

1/2(x²+2x+1) - 1/4(2x+1)
1/2x² + x + 1/2 -1/2x - 1/4
1/2x² + 1/2x + 1/4

The second equation

-1/4(x-1) + 1/4x(2x+3)
-1/4x + 1/4 + 1/2x² + 3/4
1/2x² - 1/4 + 1

I'm not sure where the mistake is lol

[latent scaffold]

This wasn't expanded properly

[pure fossil]

Oops

Thanks

Ok I have another question

NEVERMIND GUYS I SOLVED IT YAYYYYYYY

.close

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[pastel orbit]

does this proof work?
#1260424130822541403 message

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[green elm]

why oh why

[pastel orbit]

Artichoke said it worked

you can check yourself ig

maybe he missed smth

[green elm]

no i meant why oh why point to some thread instead of just asking here

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[crimson heron]

Hello, im using this website to help me optimize the launch angle of a projectile - https://conversationofmomentum.wordpress.com/2014/07/31/optimal-launch-angle/
When i try find the optimal launch angles with my values i get a undefined answer

[crimson heron]

Solved it .close

.close

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[loud grotto]

Explain the distinguishing feature of the distance between the foci and a point on the ellipse.

[loud grotto]

what is the distinguishing feature?

[runic swift]

what do you think it is

you can kind of get a hint from the standard form of an ellipse

[loud grotto]

well

my best guess

which is probably wrong

there are 2 foci

that doesnt make sense nvmd

[runic swift]

well i said you can get a hint from the standard form of an ellipse so heres a first hint

the right side of the standard form is just a number

[loud grotto]

ok let me look smth up

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[elder vault]

hello

[elder vault]

I recall there's a channel specifically about linear algebra

but I cannot find it anymore

is it because the role that I selected?

[leaden karma]

#linear-algebra

this is not accessible to you?

[elder vault]

Until now.

It seems like only people with early university as their role would be able to access it.

How to reselect my role

I want to see all the channel.

[leaden karma]

I dont have the early university role

[elder vault]

because you joined the server on 2022

[leaden karma]

can you not see this whole section?

[elder vault]

i cannot

[pliant latch]

Select your role corresponding to whichever level of course material you’re studying. Linear algebra is conventionally taught in University. If you want to change your roles, click channels & roles on the channel bar

[elder vault]

but I didn't see there's an option that allows me to change my role into early university.

ohh I see it now

[pliant latch]

You can also change yours roles in #info

[spiral basalt]

So that's what being a helper is like
Why?

[leaden karma]

what?

[spiral basalt]

So many pings

[leaden karma]

oh, yeah lol

a lot of them are from other servers though

[spiral basalt]

I don't understand why this role isn't deprecated yet

[leaden karma]

i used to have like 50k pings at one point, cleaned out a lot of useless servers

[spiral basalt]

And why people hold on to it

[leaden karma]

true it doesn't matter at this point

[spiral basalt]

The only thing I let pile up is the unread posts in the help forum
Unfortunately I occasionally click on it by accident so it never reached 800, only 700

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[fallen remnant]

If x+y=1
Minimum value of √(xy)+1/√(xy)
x and y are positive reals

[fallen remnant]

I tried am gm but didn't get the correct answer
Got ≥2
But the answer is 2.5

Ik we can get the answer by writing the entire equation in terms of x and differentiating it but want to know if there's any better method
Like am gm cauchy or something like that

[pulsar elbow]

@fallen remnant in case you’re away (there was text here but i misremembered the C-S inequality..)

disregard what i said, sorry, 1 sec :P

[fallen remnant]

🫠

[pulsar elbow]

ugh okay sorry don’t you actually want to maximize this function

you can get an upper bound of 2.5 pretty easily when you try to maximize

and of course it’s not true when you try to say “it’s greater than or equal to 2.5”

in which case all the text i deleted before was actually right and i will feel vindicated

and i’ll retype it for you

because the function squared is just less than (x+1/x)(y+1/y) by cauchy schwartz, with equality holding if and only if x=y=1/2

where the 1/2 part comes from your constraint

now you get a maximum of 2.5 just from plugging in

@fallen remnant sorry

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[candid tangle]

Is this the correct approach to the question?

[odd edge]

@candid tangle Has your question been resolved?

[odd edge]

@candid tangle Has your question been resolved?

[odd edge]

@candid tangle Has your question been resolved?

[dusty wasp]

kena tambah dengan 2C2 x 9C2 rasanya

sebab boleh jadi dia nak dua2 N tu sekali kan

[candid tangle]

tapi soalan tak sebut hanya boleh wujud sekali?

[dusty wasp]

maksudnya boleh wujud dua lebih dari sekali la kan?

[candid tangle]

ye

[dusty wasp]

so faham ke

(jawapan awak) + (2C2 x 9C2)

[candid tangle]

ooo faham faham, terima kasih

[dusty wasp]

sama2

for b tau ke?

[candid tangle]

bagi b jawapan saya 6c4 betul ke?

[dusty wasp]

hmm betul

tkde skema ke?

[candid tangle]

tak

[dusty wasp]

ohh ok

[candid tangle]

mcm tu je la , terima kasih atas bantuan

.close

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[marsh hound]

Why is a bounded

[leaden karma]

it's the inside of a circle

[marsh hound]

Ah

And the others are not because they can just go to infinity? Or neg infinity?

[leaden karma]

more rigorously, in the real numbers $x^2\ge 0$, so from the definition of (a) you get $x_1^2\le 1$ and $x_2^2\le 1$

[clever fjord]

kheerii

[marsh hound]

.solved

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[plain hound]

Hey

[plain hound]

A line passes through the point (1, -4) and is perpendicular to the line 3x + 2y – 6 = 0. Determine a scalar equation for the line.

How do I do this?

[rough rivet]

first find the gradient of that line by

m1 * m2 = -1

from then solve the rest

aply (1-4)

ull get what u want

[plain hound]

Can I use Ax+By+Cz + D = 0 for this

A = 3, B = 2, x = 1 and y = -4

[rough rivet]

huh whats that..

[plain hound]

A formula

[rough rivet]

idk bt that..

[plain hound]

Oh

Well idk what you are talking about as well

[rough rivet]

then we r drawn >:(

[plain hound]

Thanks for the help tho

[rough rivet]

wai i can reexplain things

[shrewd coral]

Given an equation $Ax + By + C = 0$, do you know how to find the slope/gradient of the line?

[clever fjord]

PowerUp

[plain hound]

All i know is that the final answer is 2x-3y-14=0

[rough rivet]

too bad 💀

change 3x + 2x - 6 = 0 to a form of y = ax + b

try it

ur a is the gradient of that line

[plain hound]

No need I will figure it out

Thanks tho

.close

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[hoary estuary]

How do I solve this?

[hoary estuary]

<@&286206848099549185>

[warped grove]

Please only ping Helpers once after 15 minutes

[hoary estuary]

oops sorry

im new so i didnt know lol

.close

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[hidden gale]

this is false right

[frosty oyster]

No

Because they have 2 sides in common

along with one angle which is equivalent

they are congruent by SSA law side side angle

[hidden gale]

that doesnt exist

isnt it SAS or ASS

[frosty oyster]

bro same thing

[hidden gale]

no its not

[frosty oyster]

I mean google says it exists sooo but im pretty sure it is congruent by that law

[hidden gale]

💀

[frosty oyster]

hold on

Ohhhh ok mb i didnt knwo what i was talking abt

so the angle between the two common lines must be equal

[hidden gale]

Only using SAS and ASS

they are not congruent right

[frosty oyster]

ok in that case yeah

[hidden gale]

thanks

[frosty oyster]

they arent congruent or not enough

[hidden gale]

ok

ty

i didnt mean to be rude with the skull btw

[frosty oyster]

nah its good

[hidden gale]

bruh i dont get it

do u know

@frosty oyster \

nvm

i know what i did wrong

[sullen ferry]

the diagram need not be drawn to scale

[hidden gale]

the signs are a different way

[sullen ferry]

(think of a kite)

[sullen ferry]

largest side corresponds to the largest angle

[hidden gale]

@pseudo widget

@sullen ferry

is it QS

[sullen ferry]

no

[hidden gale]

its not?

[odd edge]

@hidden gale Has your question been resolved?

[sullen ferry]

it is not.

you have to remember its not drawn to scale

[hidden gale]

how

but the biggest angle is R

in the whole diagram

in QRS, it is 105, 30, and 45

in QPS it is 70, 65 and 45

[sullen ferry]

this only works for triangles. you are doing it for as a whole

think about the case of triangle QRS and QPS separately

[hidden gale]

i am

[sullen ferry]

note that angle 45 corresponds to side QS

in triangle QPS

[hidden gale]

in QPS from least to greatest it is QS<PQ<PS

[sullen ferry]

so you have a side which is greater than QS

[hidden gale]

so what is the longest side in the figure

in the whole figure

[sullen ferry]

well, what are the other triangles sides

in comparisson to QS

triangle QRS

the largest angle in QRS is angle R, which corresponds to side QS

[hidden gale]

uh from least to greatest it is qr<RS<QS

[sullen ferry]

so you can combine these two inequalities

QR<RS<QS=QS<PQ<PS

what is the largest side then?

[hidden gale]

PS?

[sullen ferry]

mhm

[hidden gale]

ok

[odd edge]

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[velvet axle]

how would i go about answering this question

[olive needle]

Parallel lines have the same slope.

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[velvet axle]

ty

[odd edge]

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[velvet axle]

How do I go about answering this question?

[mystic saffron]

whats the gradient of a horizontal line

and a tangent is just the gradient at any given point

what can we do to the function to determine the gradient at any given point

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[violet nacelle]

Could use help with this problem:

[fresh pine]

Build a graph

Y as V and X as t

It will be a parabolic one

[mystic saffron]

long

[fresh pine]

Or

[mystic saffron]

just integrate

[fresh pine]

🗣️🫰🏿

Or js

Count

Time it moves backwards

And then forward

And count it like that

It's 8 th grade physics

It will go backwards till t=5

After wards it will go forward

Find distance

And subtract for first question

[mystic saffron]

yeah then calculate the area under the curve for distance

[fresh pine]

Yeh

[mystic saffron]

so integrate

get integrating bud @violet nacelle

[fresh pine]

🙌🏿🙌🏿🗣️🗣️

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[analog pewter]

I could use the help of someone who speaks Dutch to teach me formulas and solutions to some problems for an exam

[analog pewter]

If possible, DM

[ember oak]

when's the exam?

[analog pewter]

Tomorrow

[ember oak]

I don't speak dutch, but I can try to help if your english is okay

[analog pewter]

So yeah, it might be a bit late

Ah, not sure if you'll recognize but I might send you a document or something of a previous exam

I can try translating

I'll go eat something and come back later

[odd edge]

@analog pewter Has your question been resolved?

[analog pewter]

Back

Found it

I think just the formulas will suffice

[ember oak]

There really aren't "formulas" to solve all of these problems

Not everything can be put into a calculator so easily

these problems are designed to test understanding, not the fact that a formula has been memorized

Give me one specific problem you are stuck on here and let's see what we can do about it

[analog pewter]

Ok

Do you happen to know the rules?

[ember oak]

which rules?

[analog pewter]

I don't know the names needed for some of these

You got time?

I'll go study this and return later for questions

[odd edge]

@analog pewter Has your question been resolved?

[odd edge]

@analog pewter Has your question been resolved?

[odd edge]

@analog pewter Has your question been resolved?

[analog pewter]

3
6
11
12
14
16
17
21
23
24
25
26
27
28
29
31

<@&286206848099549185>

Can anyone explain?

[warped grove]

What are these numbers...? Are they the problem numbers you want explained?

[keen latch]

those series have any rules?

[analog pewter]

Yeah...

[analog pewter]

I lost the pages so I lack the materials

Unless you mean something else

[warped grove]

I can't help rn, but I suggest if you want ppl to help you, start by asking one question and showing what progress you've made on it so far

Once you finish that question, you can ask another question

Ppl aren't gonna wanna help if you have like a whole list of questions

[analog pewter]

True

But I'm tired

I think this it for me

How do I stop occupying?

[odd edge]

If you are done with this channel, please mark your problem as solved by typing .close

[analog pewter]

.close

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[ivory flame]

hii could someone help me verify whether these answers for a, b, d are right

[odd edge]

@ivory flame Has your question been resolved?

[hollow quarry]

(a) I would include the base case, n = 1
(b) + I believe you meant 5q^2 = 2p^2 rather than 2q^2 = 5p^2
+ You go from p^2 = 5k to p = sqrt(5k) yet you make no statements on k.
What if k is a multiple of 5 also?
If p^2 is a multiple of 5 and p is a positive integer, p must be also a multiple of 5

For (d), while trivial, you should consider n = 2k as well. Also 6k is always even as well, 6 is a multiple of 2 after all

[odd edge]

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[wide drift]

I know how to do this but im really out of practice

I forget the first step

Dont I need to like integrate sin(x/2) between 3pi and 0 or something

[spiral basalt]

0 to 3pi would be a better start

[wide drift]

But the method is correct?

[spiral basalt]

that's not what I said

[wide drift]

I havent done these in so long 😭

[spiral basalt]

does the integral compute the area ?

[wide drift]

I think so? I forget the method

[spiral basalt]

what's the area enclosed by the x-axis and f(x) = -2 for x in (0, 2) ?

[wide drift]

4

[spiral basalt]

is that the integral of f ?

[wide drift]

Ah

Wait yeah it was 2

🤦‍♂️

[spiral basalt]

the area was 2 ?

[wide drift]

yes

[spiral basalt]

it's kinda a 2x2 rectangle though

[wide drift]

is it not (2^2)/2 - (0^2)/2

This is what my original question was about I dont remember the method

I think I will just watch a video on it again

[spiral basalt]

Congratulations
I wanted you to spot the problem and deduce the solution
And you revealed you forgot how integrating constants works

[wide drift]

I cant tell if youre being sarcastic or not 🥲

[spiral basalt]

well that's not how integrating constants works

and you have yet to find what I wanted you to notice

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[kindred nest]

claim

[kindred nest]

!status

[odd edge]

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

[kindred nest]

1

I honestly don't even understand the wording

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[daring vector]

i have done a, wondering if what i have started for b is correct and how to continue

[daring vector]

<@&268886789983436800>

i would really appreciate any help

[sinful grove]

1. looks correct and you can see this via continuity as well

For 2) what are ur thoughts?

[daring vector]

well for +φo we get the numerator approaching 2φ and the denominator approches 0

but this is undefined

inside a Ln

so im not too sure what to say about it, and for -φo we get x(φ) approaching ln(0/-2φ)

and ln(0)is undefined

its a bit confusing then what to do

@sinful grove

ok

.close

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[silent comet]

Hi i just wanna know the name of this subject so i can search it on yt

[silent comet]

I know its integration but i dont know what type of thing this is

[indigo crater]

I think it is midpoint rule? I'm not sure but this is more towards numerical integration

[odd edge]

@silent comet Has your question been resolved?

[silent comet]

I searched it up i dont think its the exact same way of solving it

im confused on the delta x

its weird

my text book does it in a kinda sheet style

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[mystic saffron]

Correct reasoning?

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@mystic saffron Has your question been resolved?

[mystic saffron]

.close

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[silent jay]

Am I correct?

a. ( -ab ) is positive.
b. ( abc ) is positive.
c. ( b - ac ) cannot be definitively determined.
d. ( ab + bc ) is negative.
e. ( a^3c ) is positive.

[clever fjord]

clikz

[rough rivet]

yeh u are

[silent jay]

Thanks

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@silent jay Has your question been resolved?

[silent jay]

Even for c?

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[rich night]

i cant tell if my tension is right

i worked out the reaction force for ii, but I think it's wrong

<@&286206848099549185>

[odd edge]

@rich night Has your question been resolved?

[rich night]

<@&286206848099549185>

[keen latch]

you can use moment equation

so you solved question 1.

right?

are you there?

HI

[rich night]

yes]

hallow

i dont think part 2 is right

i got part 1 yes

[rich night]

im hereeee

[keen latch]

in the question 2,

there are 2 kinda forces

one is the horizontal part of T

other is the sum of 3 forces

[rich night]

what do h

you mean

[keen latch]

weights of two beams and vertical part of T

[lunar tree]

While solving question 1, why did you ignore torque due to gravity?

Which acts at the centre?

[rich night]

if youre talking about the mass

i thought the centre of mass was one of the given forces

im wronh

thank you for that

[keen latch]

Oh, yes

you missed the weight of the beam

[rich night]

Okay I wrote it out again

[lunar tree]

Since the value of T changed, shouldn't Tcos theta also change?

[rich night]

yep its 3083.7N

however, I still don't think thats the answer

[lunar tree]

Okay, give me some time to solve it myself.

[rich night]

in class we did horizontal and vertical moments, but i dont know how to do that with this question

[lunar tree]

I think you've done a calculation mistake here

Probably while calculating sin(theta)

Maybe forgot to set the angle to degree while calculating or the other.

[rich night]

nuh uk

nuh uh

ive done the calculation again

theta equals artan0.6

feed that directly to sin function

[rich night]

you didnt multiply t by 5

for the moment

[lunar tree]

Oh my bad

[rich night]

🥳 I'M NOT TOTALLY USELESS

[lunar tree]

Well it's 1 AM here

I'm in a weird state of mind rn

I'm sleepy but don't wanna go to sleep lmao

[rich night]

nah jkjk

[rich night]

whys that so relateable

[lunar tree]

🙂💀

[rich night]

@lunar tree

<@&286206848099549185>

[lunar tree]

Umm why do you exactly think that ur final answer is incorrect?

[rich night]

.

[lunar tree]

I don't think I've heard about horizontal and vertical moments.

I've just passed high school, uni's gonna start next month.

[odd edge]

@rich night Has your question been resolved?

[rich night]

same

ill finish highschool next year

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[reef wigeon]

Two Problems, A and B. This is an Algebra 2 Application Problem.

A) According to a reliable source, 15 bags of compostable trash will yield 75 cubic feet of compost. Use the equation calculating compost from trash to solve for s (99% sure that is (s-1)3 because Volume of bag is s3 and you lose 1 cubic foot of compost by the time it's done composting)(and I believe the general equation would be (Compost Gained - Compost Lost)^3, but I’m also pretty sure that’s wrong).

B) Calculate the interior volume of the composing bin (4 feet by 4 feet by 4 feet) to find the amount of trash you can put in it. Using the value for s, how many bags of trash (t) can fit in it, and how much mulch can you expect from that trash?

I’m kinda stuck on all of it, so any help would be amazing. If I forgot an important detail I can find it. Thanks, and have a great day!

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@reef wigeon Has your question been resolved?

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@reef wigeon Has your question been resolved?

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@reef wigeon Has your question been resolved?

[reef wigeon]

<@&286206848099549185>

[reef wigeon]

Bro. Fuck this server.

[pale bobcat]

-_- i can help

problem A:

Problem B:

something like that

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[stuck wasp]

8 b and c

[stuck wasp]

,rotate

[meager juniper]

@stuck wasp double angle formula

[stuck wasp]

Ya cos^2 -sin ^2but how to apply

[meager juniper]

You know sin x and cos x

Plug them in

[stuck wasp]

So cos ^2 is just 1- sin ^2

[meager juniper]

You know cos from part a

Might make your life easier

But that way is also valid

[stuck wasp]

Oh ok so 4/5^2 - 3/5^2

I got wrong awnser

Cos of 16/25 - Sin of 9/25 i got like .45 but im supposed to get .28

[nimble blaze]

why are you doing cos of 16/25 etc

[stuck wasp]

should i not?

[nimble blaze]

you shouldn't

[stuck wasp]

Cos 2 x is cos^2 -sin ^2

[nimble blaze]

yes

missing x, but yeh

[stuck wasp]

how i do?

[nimble blaze]

you had it before, albeit missing ()

and then you did something completely different

[stuck wasp]

So i do not square cos (4/5)?

[nimble blaze]

no

[nimble blaze]

missing ()
(4/5)^2 - (3/5)^2

just simplify that

[stuck wasp]

16/25-9/25

its 7/25

[nimble blaze]

yes

[stuck wasp]

wait what

HOW?

[nimble blaze]

cos(x) = 4/5,
cos^2(x) = (4/5)^2

same for the sin(x), sin^2(x)

[stuck wasp]

Oh so me figuring out cos (x) =4/5 in a) part, mean i dont have to do cos again?

[nimble blaze]

well, yeh

[stuck wasp]

wait so i do next question

So sin 2x is 2 sin x cos x

So its 2 x 3/5 x 4/5?

24/25

[nimble blaze]

yes

[stuck wasp]

damm so easy

Btw

Could you explain double angle formula

i get how i use Sin 2x and Cos 2x but what does it come from

[nimble blaze]

look up derivation of compound angle identity

[stuck wasp]

Ok got it thanks man

.close

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[torn lodge]

I am stuck on the step on producing a generalised equation for the integral

At the bottom, I found the equation for the first sub interval.

[modern sundial]

Your work is really messy and hard to follow. You should redo it more systematically by starting with defining your partition of [0,1] into a set of numbers x_1, x_2, ..., x_N = 1/N, 2/N, 3/N,... and writing the rule in terms of f(x) = 2^x.

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[neat shadow]

Hi, I'm trying to study matrices and I have a question. If x is an eigenvalue for a matrix B, is B^4 the eigenvalue for a matrix B^4?

[neat shadow]

I'm thinking its yes? But I dont know if I'm right or how to prove it

[zenith jasper]

you mean is x^4 the eigenvalue for B^4?

[neat shadow]

Yes, sorry

[zenith jasper]

yes it is

[neat shadow]

I see thank you

[zenith jasper]

you can use mathematical induction to prove it

A^nx=λ^nx

[neat shadow]

Is a square matrix (m x m) A diagonalizable if and only if it has m distinct eigenvalues?

[zenith jasper]

i think so, but im not entirely sure u can ask other helpers

[neat shadow]

I see, Ill tag then

<@&286206848099549185> if you have a moment please

[zenith jasper]

will this help

[neat shadow]

it might but Im slightly worried about the wording

because that says n distinct eigenvalues leads to diagonalizable

but what about the other way round

[green elm]

if, but not only if

the identity matrix does not have distinct eigenvalues (they are all 1) but it is diagonalizable because it is diagonal

[neat shadow]

So it is the wording then, okay

[signal hare]

Diagonalizable = all Jordan blocks are 1x1 matrix. When you have distinct eigenvalues the Jordan blocks have to be 1x1

[neat shadow]

Can I ask what a jordan block is? I have not learnt that

[signal hare]

U can lookup “Jordan normal form” on wikipedia. It says that every matrix is similar to something looks like this. The Jordan form theory is very useful when talking about similarity

[neat shadow]

I see, thank you for your help

.close

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[weary flame]

If 𝑝 = 2010^2 + 2011^2 𝑎𝑛𝑑 𝑞 = 2021^2 + 2013^2
, then the simple
value of √1 − 2(𝑝 + 𝑞) + 4𝑝𝑞 = ….

I have tried to brute force it but I realised that the numbers would get too large, I am trying to solve this without a calculator. Is there a simpler way to do this?

[tulip spire]

x^4 is eigenvalue of B^4

[steep mantle]

is the question $\sqrt{1-2(p+q)+4pq}$?

[clever fjord]

Skill_Issue

[steep mantle]

dont bruteforce it lmao

@weary flame

hm

intresting question

[odd edge]

@weary flame Has your question been resolved?

[weary flame]

yea I'm figuring out how to not do that

[weary flame]

yes

[steep mantle]

hm

is this an olympiad question?

[weary flame]

yes

I'm practicing for one

[steep mantle]

ok

i noticed it could be factored to (2p-1)(2q-1)

[weary flame]

this?

[steep mantle]

the inside

also is it really 2013?

maybe its like 2012 so it goes 2010 2011 2012

[weary flame]

yes

I am a little bit sus of the practice questions tho

[steep mantle]

damn

[steep mantle]

wdym?

[weary flame]

Because there was a question about number sequences that was kinda wrong so I had to assume some things

let me show it here