help-19

[pastel orbit]

θ_2 comes from here

[hearty cape]

ooh shitt

da,mn

[pastel orbit]

just remember that n_1 must be a greater number than n_2

because total internal reflection cant happen from less dense to more dense

only vice versa

so n_1 must be 1.52 and n_2 must be 1.34

θ_2 is 90 because we want critical angle

and then you solve

[hearty cape]

i see now

thank you

[pastel orbit]

no problem

[hearty cape]

would u have any idea why the mark scheme has the solutio nas this

they both give the same answer

[pastel orbit]

ngl, i hate this notation lol

[pastel orbit]

it's basically the same thing, but i prefer to use snell's law

since imo, it's clearer what's going on

this is kinda ugly imo, since they dont leave everything in variable form

also, i actually didnt know that you could calculate the refractive index between two materials by taking the ratio of their refractive indicies wrt air/vacuum

i suppose that's what you were confused about earlier

my apologies

but it really is the same as snell's law, and i highly doubt your instructor would take marks off if you used snell's law instead

[hearty cape]

yeah no worries

do u know where this line comes from?

[pastel orbit]

again, it's the same as snell's law

they derived it from there

n_1sinθ_1 = n_2sinθ_2 becomes n_2/n_1 = sinθ_1/sinθ_2

n_2/n_1 = 0.8816 and sinθ_2 = sin90

they let θ_1 = C

again though, i highly advise leaving everything in variable form if you are comfortable with it

it makes solutions much clearer, and easier to follow

plus, you're less likely to make mistakes, and more likely to understand the principles better if you do it consistently

does that help? @hearty cape

[hearty cape]

alriighy yess

very much so

thank u kind sir

[pastel orbit]

if you're done here, don't forget to close the channel

[hearty cape]

how i do that

[pastel orbit]

.close

just type that

[hearty cape]

aight let me just copy some text so i remember

aight thanks

.close

[odd edge]

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[hearty cape]

.reopen

[odd edge]

✅

[hearty cape]

actually i have anthoer question 💀

[pastel orbit]

go ahead

[hearty cape]

im conflicted what they want me to shade out

idk how to choose where to shade

[pastel orbit]

ah, this one may be out of my area of expertise lol

i'm not too familiar with complex numbers

[hearty cape]

its just like normal loci

[pastel orbit]

okay, we have two conditions

[hearty cape]

the first thing is a circle with radius less than or equal to 4 and centre 0,2

[pastel orbit]

yes

[hearty cape]

and the 2nd thing is a perpendiculr bisector through the middle of the points

between -4, 1 and 0, -1

with it being less than or equal to -4, 1 ?

idk what that means

[pastel orbit]

hmm

"the distance from the origin to z + 4 - i must be less than or equal to the distance from the origin to z + i" is what i'm getting

but please dont take my word for it lol

perhaps someone else can help with this one?

[hearty cape]

@everyone

imagine

[pastel orbit]

does @ everyone actually ping everyone

[hearty cape]

na lol

[pastel orbit]

i see lol

perhaps wait for someone to pass by and help?

[hearty cape]

aighty

[pastel orbit]

if not, you can ping @ Helpers (no space) after ~15m have passed

[hearty cape]

could it be the distance from the point to the perpendicular bisector is closer to z + 4 - i than to z + i

<@&286206848099549185>

[hearty cape]

.

[odd edge]

@hearty cape Has your question been resolved?

[odd edge]

@hearty cape Has your question been resolved?

[odd edge]

@hearty cape Has your question been resolved?

[odd edge]

@hearty cape Has your question been resolved?

[odd edge]

@hearty cape Has your question been resolved?

[warped grove]

So it seemed like you already figured out the first equation is a circle and the second equation is one side of a perpendicular bisector. That's correct.

You can read the second equation as saying "The distance from z to (-4, 1) is less than the distance from z to (0, -1)".

So shade in all of the z where both z is inside the circle and also that's true.

Does that make sense, or are you confused by something?

[neon sonnet]

<@&286206848099549185> can someone help me please

[warped grove]

This is someone else's channel, please don't interrupt it

[warped grove]

Please read #❓how-to-get-help

[neon sonnet]

oki

[odd edge]

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[cinder elk]

You have to use logarithms, you familiar with them yet?

[south plume]

hint $\frac{1}{2\sqrt{2}} = \frac{1}{2^{3/2}} = 2^{-3/2}$

[clever fjord]

artemetra

[cinder elk]

I think if you put both to the log_4 it should give you 3x - 2 = log_4(1/2sqrt(2)

[plain ruin]

what

Why can't you just make the same basements

[wooden python]

it's fine, just roundabout

[wooden python]

basements?

did you mean bases

[plain ruin]

Like 2^x=2^1 and you get x=1

[cinder elk]

This would be a little bit more useful if they had the same base

[wooden python]

yeah you meant bases

[south plume]

they can get it

$4^{3x-2} = 2^{2(3x-2)}$

[clever fjord]

artemetra

[plain ruin]

👍

Number in some root equal to the same number with another root if only their roots are equal

√2/4 make it base 2 somehow

√2

What it is

It is 2^(1/2)

There is a rule

Lemme find

like this

[charred veldt]

yes

[plain ruin]

so you can say that 4 is 2^2 isn't it

so do 2^(1/2)/2^2

Do division

Sorry

I wasn't here lol

When you divide number with same bases

What you do

Nah, never use calculator

we need algebra

Do you know

a^b/a^c=a^(b-c)

What you'll get

nah

b is 1/2

c is 2

Yess

Yes

Do it

Yes

1/2-2

Is it hard

Yes

Number in some root equal to the same number with another root if only their roots are equal

Like 2^x=2^1 and you get x=1

Can you understand

So what's next

write your new equation

[plain ruin]

.

[plain ruin]

.

Can you do it and understand?

Yes it's good

So

Then the "powers" have to be equal

So it's linear equation

I hope you can do it

Hm

Yes

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[mystic saffron]

sin a= 30

as in the angle opposite side a is 30?

[stable night]

2

[wooden python]

!xy

[odd edge]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[stable night]

if the angle is opposite the shorter side than there are always 2 triangles

[stable night]

he said the question

[wooden python]

he said it in a strange way

[stable night]

ah okay

kind of like this

yeah its 2

[wooden python]

ok that makes a lot more sense

"sin a=30°" was fucked up

[stable night]

oh lol didnt see that

[stable night]

like this

you can change the position of the shorter side (a) and still have a triangle with those 3 conditions

[mystic saffron]

no

[stable night]

you said sine of a is 30

[mystic saffron]

sin is to do w the ratio of sides so idk how it is equal to an angle

[wooden python]

in the future always send a picture of the question

[plain badge]

which board?

@remote dawn

ooh

no lol

high school

lol

use sin rule

so how many triangles possible?

i gtg

srry

just use sin rule

[stable night]

there is 2

[proud vale]

Send the question

Can u draw the triangle for me

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[wide jackal]

So equation of a line passing thorugh P which meets the given conditions

so according to the tutor, the answer to first is y=-3,z = 4

but trying it on desmos 3D and geogebra , does not give anything close

Am i writing the equations wrong?

the comma dosent work

i am new to 3d desmos so maybe i am missing something

this is eating me alive ffs, am i being taught wrong, is the grapher wrong

and how on earth

do i do that

let me google

not yet

i mean

from what i was taught

correct madam

[wide jackal]

the guy said d2,y,d1 is the eqn of the line

actually, just take the entire pic

YES

has this something to do with y=z= lambda ?

AH i see what u did

smart

i like this

whats wrong with the thing i wrote tho

and why cant i just add a comma

ah i see

is there a calc which works with what i wrote

i can see why

oh

that interesting

thank you so much for helping, wait i have another doubt

why does this work then

i get a line ||el to x axis and passing through the point

actually

wait

dont answer it

i will find it myself

didnt see

thanks

you were so helpful

.close

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[dull light]

Could balanced base 3i work?

[shy smelt]

Maybe if you have 9 digits

like 0, 1, -1, i, -i, 1 + i, etc..

[dull light]

Hmm

[shy smelt]

Just 0, 1, -1, i, -i doesn't seem to do it

[dull light]

Wait why would the digits not all be real

[shy smelt]

wait, (3i)^n is complex of course

Do you just want to represent integers?

or extend it to the complex integers as well

[dull light]

I mean I only need integers

But complex integers is objectively better

[odd edge]

@dull light Has your question been resolved?

[shy smelt]

Then you probably need all 9 of them

[dull light]

This might be one of the worse ways possible to count

[odd edge]

@dull light Has your question been resolved?

[dull light]

Do the integers have unique representations

[shy smelt]

Yeah, I think so

https://en.wikipedia.org/wiki/Quater-imaginary_base

Actually, you might need 17 digits

[dull light]

17?

[shy smelt]

No, that's only if you use purely imaginary and purely real ones

[dull light]

Even that I don’t understand

Oh wait thats 9+9-1

I see

[shy smelt]

Yeah, you have the even powers of (3i) and the odd ones

[dull light]

Congratulations you found a worse way to count

[shy smelt]

but it seems like you can cover the complex plane with just 9 digits if you include complex ones

$$15 + i = (i) \cdot (3i)^0 + (i) \cdot (3i)^1 + (1) \cdot (3i)^2 + (-1) \cdot (3i)^3$$

[clever fjord]

Jelle

[shy smelt]

As an example

[dull light]

Do all reals have a unique form?

[shy smelt]

Yeah, wait maybe the thing is that all the digits are supposed to be real though

[dull light]

Iirc they don’t in balanced base 3

[shy smelt]

They do, right?

[dull light]

I think it was something like 1.TTTTT…. = 0.11111…

[shy smelt]

Oh, but that's in every base I think

0.99999999... = 1

[dull light]

Yea but that’s not what being written

I could be wrong but that might of been 1/2 if I remember correctly

[shy smelt]

Oooh, btw if you only use real digits you still need 9 digits. This was incorrect

[dull light]

Wait then you can use just 3 digits in base sqrt(3)i

[shy smelt]

No, because only the imaginary part has integer values

[odd edge]

@dull light Has your question been resolved?

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[valid river]

Gaussian integral?

[graceful berry]

whats that

[valid river]

its the integral of the bell curve

see this: https://quantummechanics.ucsd.edu/ph130a/130_notes/node87.html

[graceful berry]

thanks

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[minor ice]

How to derive the formula for the sum of this series? I can differentiate and integrate, but I don't understand how to derive formulas for sums.

[minor ice]

Also, if anyone knows any good sources on finding the sum of series, I would like to ask for a link to it.

(books, videos etc.)

I didn't find anything in my native language 😦

[quasi sparrow]

https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-10-15/v/differentiating-power-series

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@minor ice Has your question been resolved?

[minor ice]

Can we find the sum of each series by differentiating and integrating? Are there any series whose sum cannot be calculated?

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@minor ice Has your question been resolved?

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[mellow socket]

I have to show it is Darboux integrable

[cold urchin]

What's the issue?

What have you tried and where are you stuck?

[mellow socket]

I want to split it into two intervals

[a,c) and (c,b]

but I don't know how to show that it will be integrable on those two intervals

if it was [a,c] and [b,c] I could do it😅

am I approaching it wrong?

[odd edge]

@mellow socket Has your question been resolved?

[mellow socket]

<@&286206848099549185>

[mellow socket]

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[mystic saffron]

do you know about completing the square

i tried to watch videos and understand but i dont really get the image

i can try and explain

if you wanna

yes

go ahead

okay so

[
(a+b)^2 = a^2 + 2ab + b^2
]

[mystic saffron]

do you know of this

expanding the bracket?

yeah pretty much. I just want you to keep it in mind

so like

a quadratic is usually given as $ax^2 + bx + c$ right?

[mystic saffron]

yes

alright so we let's make the assumption that a = 1

we now have $x^2 + bx +c$

[mystic saffron]

so the idea behind completing the square is that we want to get SOMETHING that will get us something similar to this

so its just factorising?

not really. I'll get to that

so like

okay so

lets ignore the c for now (its still there but we dont want to look at it now.

alright

we have [
\c b{x^2} + \c g{bx} ]
and our square above was [
(a+b)^2 = \c b{a^2} + \c r 2 \c g{ab} + \c o{b^2}
]

[mystic saffron]

now compare the first one with the second

what do you think is missing in the first one?

that the second one has

colour

they look exactly the same

[mystic saffron]

wait these two?

yeah

I colour coded the similar items

hm

like the orange b^2 is not there for one, right?

and the red 2 is not really clear

yeah

a^2 is kinda there with the x^2

so okay great

we know we want to get something similar to (a+b)^2 but the problem is that we are lacking an apparent b^2 and the red 2

whats an apparent

now this is the crucial part and i need you to concentrate

[mystic saffron]

take it to mean clear

ahh

[mystic saffron]

okay

[mystic saffron]

so lets deel with the 2 issue

in our equation [
x^2 +bx ]
there should have been a $2bx$ instead of a $bx$, but it's gone

[mystic saffron]

what do you think turns 2bx into bx?

like, what do you do to go from 2bx to bx algebraically?

divide by 2

yeah exactly

[mystic saffron]

so again, notice how we have x^2 correspond with a^2 right?

so that means x = a, right?

yeah

yeah but what about b?

what do u think it should be

remember like

it must be 2ab, but it transformed into bx

so hmmm

[
2ab \to bx
]

[mystic saffron]

what do you think the second 'b' should be for this to make sense

2

nope! i think its an issue with them being the same letter

let me redo this for you

we have [
\c b{x^2} + \c g{bx} ]
and our square above was [
(c+d)^2 = \c b{c^2} + \c r 2 \c g{cd} + \c o{d^2}
]

[mystic saffron]

this might be better now

so now

[
2cd \to bx
]

[mystic saffron]

what do you think b should be for this to make sense?

remember c = x

and we have x on both sides

so...

b^2

nope

hmmm how do i explain this

okay another approach

[mystic saffron]

we know that c = x here

yes they are the same

we have [
\c b{x^2} + \c g{bx} ]
and our square above was [
(d+d)^2 = \c b{x^2} + \c r 2 \c g{xd} + \c o{d^2}
]

[mystic saffron]

so now

bx and 2xd

the first one we have a 2 missing

what do we divide with to make 2xd be xd

2

yeah

so 2xd/2 = xd

so after we divided by 2

we get bx and xd which means b = d

but like

we had to divide by 2 for this, so there is something we need to compensate for as well

another way to look at it is this

[
x^2 +bx = x^2 + 2x\f b2
]

[mystic saffron]

right? @mystic saffron

yes

so like

c^2 + 2cd
x^2 + 2xb/2

what is c equal to

and what is d equal to

d=b

c=x

nuh uh

like okay

c^2 = x^2

so that's cool

so we are left with

2cd
2xb/2

we know that c = x

so like

2xd
2xb/2

so like, what do u think it should be now

what is d equal to

2d

no

wait

d = 2

my brain is fried

yeah i think you need to take a break

give this a look later

if you want

yeah i think thats a good idea

thank you for trying to explain it to me in 5 different ways when i didnt get it😭

have a good day

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[dusk rivet]

I am trying to understanding laplace transformation of heaviside function. What's bother me why we including terms from the intervals when t < a into when t > a. For example, in this example, we're adding and subtracting t why?

[dusk rivet]

reference: https://tutorial.math.lamar.edu/classes/de/StepFunctions.aspx#mjx-eqn-eqeq1

for example here why we're taking g(t) to have t+5 as input rather than t-5?

[odd edge]

@dusk rivet Has your question been resolved?

[odd edge]

@dusk rivet Has your question been resolved?

[odd edge]

@dusk rivet Has your question been resolved?

[upper gale]

they include those terms cause they want a single expression for f(t), thats why they use heaviside functions

the include a -t term so it cancels with t when t>=6

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[warped grove]

The three sides of the triangle are the adjacent, opposite, and hypotenuse, right?

Which one would the 50 cm side be?

Adjacent, opposite, or hypotenuse?

Please don't give out answers

(User got banned anyways)

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[lusty garden]

.close

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[hoary marsh]

?

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[wispy elm]

hi

[wispy elm]

can anyone help me

[quasi sparrow]

[wispy elm]

how do I solve the second part of this

?

[warped grove]

I'm assuming you've already proved that the two triangles are similar. What does that imply about the side lengths of the triangles?

[wispy elm]

The sides are proportional

[warped grove]

Yup. Which sides, in this case?

[wispy elm]

be and ec

[warped grove]

Not quite

EC isn't one of the sides of triangle ABC

[wispy elm]

oh

be and bc

[warped grove]

What's the side of triangle ABC that corresponds to the side BE of DBE?

Yup exactly

And then what's the side of triangle ABC that corresponds to the side ED of DBE?

[wispy elm]

AC

Correct?

[warped grove]

Yep! perfect

so the relationship between AC and ED is the same as the relationship between BC and BE.

[wispy elm]

ah

[warped grove]

Can you write an equation describing this fact?

[wispy elm]

ya

so its something like

x/y = z/r

[warped grove]

yeah something like that

[wispy elm]

alr

48/32 = E/28

[warped grove]

what is E?

[wispy elm]

E=42

[warped grove]

Yeah, what do you mean by the letter E though?

[wispy elm]

Be

BE

ah thank you

[warped grove]

No, BE is 28

[wispy elm]

oh whoops\

so

42 - 28

=

14

EC = 14

so

BC = 42

units

[warped grove]

yup! perfect :)

[wispy elm]

Thank You So Much

[warped grove]

no problem

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[odd edge]

@midnight jasper Has your question been resolved?

[midnight jasper]

<@&286206848099549185>

[odd edge]

@midnight jasper Has your question been resolved?

[fluid tundra]

what progress have you made on this problem so far?

[midnight jasper]

AHhh someone finally.. I really havent made any progress thus far. I think im mostly stuck on how to begin

[fluid tundra]

essentially, what this problem wants you to do is take the two given approximately normal distributions and subtract one from the other, resulting in another normal distribution

does this concept sound familiar to you?

[midnight jasper]

so like subtracting 1 mean from the other?

[fluid tundra]

yes that's part of it

[midnight jasper]

Sorry for taking a while to respond.. can we also subtract the standard deviations?

[fluid tundra]

no worries

but no you cannot

the procedure for combining standard deviations is sqrt(A^2 + C^2)

[midnight jasper]

i got a mean of d is 100 and std dev of d is 120

Is it still normal?

[midnight jasper]

I got .7976

[fluid tundra]

this is what i get, and yes it is still a normal distribution

[fluid tundra]

that's also what i get so i believe this is correct (just make sure to round when inputting your answer)

[fluid tundra]

very nice

[midnight jasper]

Thank you for being patient and helping me through out the problem! I understand the concept a lot better now. Hope u have a great rest of ur week

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[terse pike]

can i rewrite -log5 as log5^-1?

[dark kraken]

Log(5^-1) yes log(5)^-1 no

[wooden python]

$-\log(5)$ as $\log(5^{-1})$?

[clever fjord]

Ann

[terse pike]

wait

okok

hold on

correc?

wait i think im doing it wrong

[mystic saffron]

i think its fine and then you csn solve for x

,w 9^(3x-2) = 5^(7x-1)

[clever fjord]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by qylooo

[mystic saffron]

yeah its correct

[terse pike]

OH

YAY

ok but rewrote log9^3 - 7log5 as log((9^3)/(5^7))

and then did (log(81/5))/log((9^3)/(5^7))

[mystic saffron]

thats fine

[terse pike]

and got -0.59580186

am i doing somehting wrong on my calculator or soemthing

[mystic saffron]

,calc (log(81/5))/(log(729)-7log(5))

[clever fjord]

Result:

-0.59580185995785

[mystic saffron]

that matches

its correct

[terse pike]

YAYAYAYAYA

TTHANK YOU

[mystic saffron]

no problemm

[terse pike]

:3

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[spice spindle]

Left is my work and right is answer key. I checked my answers with online calculator and I think the key is wrong. Is there something I'm missing? (I'm doing bad in the class so I'm very unsure of myself 😭 🙏 )

[dark kraken]

left is correct on both

[spice spindle]

alright thanks so much ❤️

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[delicate orbit]

homework help

[mystic saffron]

!show

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[wheat raptor]

revisiting vectors; came across this notation

[wheat raptor]

what does it mean to 'dot product by 'i' '

[lyric marlin]

'Take the first coordinate'

As i is (1,0,0)

[wheat raptor]

right, so i should interpret i in vector calculations as (1, 0, 0 ...), j as (0, 1, 0 ...)?

makes sense

[lyric marlin]

Yes

i,j,k with hats are notation for the standard basis in R^3

dunno if that applies to F^3 in general

[wheat raptor]

thanks, i'll give this a re-read with that in mind

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[mystic saffron]

2 questions did i do this right, and is that a quadratic function?(running late on my school project ahahaha

[elfin zodiac]

That doesn’t look correct

-2x-2x is not 0

Slso, (-2)(-2) is not -4

Do all the steps again

[mystic saffron]

yeah i forgot bro

[pearl patio]

btw you can use difference of squares

y = a^2 - b^2 = (a-b)(a+b)

the a here is x-2

and the b here is x

[mystic saffron]

ty ill try that

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[sick sigil]

Sorry for this very stupid question but where did I go wrong here

[sick sigil]

the derivative of the first one is a multiplication, so then we use the f(x)'*g(x)+f(x)*g'(x) no?

which should give us 1cos(x)+x*-sin(x)

doesnt that mean we end up with cos(x)-xsin(x), aka cos(x)+ (x*(-sin(x)))

[harsh vapor]

it seems correct

[sick sigil]

but wtf

[harsh vapor]

i looked up to the 3rd derivative

[harsh vapor]

seems all fine

[sick sigil]

they want it to be this

[harsh vapor]

ohh thats another thing

[sick sigil]

but if i have 3 -cos(x)

that would be -3

and then you have f'''(0)x^3/3!

[harsh vapor]

that is with taylor expansion probably

[sick sigil]

so we get -3x^3/3!

nvm

i see the issue

[harsh vapor]

do you know the taylor expansion?

[sick sigil]

my brain is fried to the point where i took the 2 to the nominator instead of the denominator

well ish

f(x) = f(0) + f'(0)x + f''(0)x^2/2! etc

but with taylor expansion i think you do

f(x-a)

and replace all the x's with a instead

[harsh vapor]

cos can be represented as 1-x^2/2 etc

you can also directly input a

without the x

no sorry

you set the a as zero in this case

and you just get x

[harsh vapor]

and you get this

then multiplying by the other x you get your photo

and you can derivate that to get an approx of the derivative

but your method is cleaner

[sick sigil]

but i feel like im not getting the right answer here q.q

[harsh vapor]

might be faster to use taylor if you have many derivatives of derivatives though

[harsh vapor]

so try that

ohhh

listen

[harsh vapor]

divide it into pieves

pieces

[sick sigil]

that's what I've tried to do

but the x^3

[harsh vapor]

hm

[sick sigil]

that is always going to be 0 no?

[harsh vapor]

you know identities?

[sick sigil]

what kind of identities?

[harsh vapor]

limit identities

[sick sigil]

like trig identities?

uhhh

[harsh vapor]

like sinx/x as x goes to 0 is 1

[sick sigil]

i possibly do but you'd have to elaborate

ah i know a few of those

[harsh vapor]

thats the key here

with some manipulation

i'll try it aswell

are you allowed to use l'hopital

[sick sigil]

i'll say no, because for some of the exams they do let us but sometimes they dont q.q

[harsh vapor]

okok

ok i solved it with just identities

a bit of a silly solve but it works

it equals 1 right?

[sick sigil]

5/6

;x

[harsh vapor]

huh

i might have done an error myself

huh im sure its one

let me check on wolfram alpha

it is 5/6

id try to use taylor then

if you are allowed obviously

[sick sigil]

yep

do you know why this works? q.q

this is the part i struggle with I think

[harsh vapor]

oh thats easy

let me write it

[sick sigil]

ahhhhh

yeah ok that makes sense

thank you!

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[mystic saffron]

The formula circled in red is what I need to show is the green circled one

I would like to get my work checked so far

[wooden python]

!msgdel

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[urban atlas]

https://i.imgur.com/2rlrw43.png

[urban atlas]

for 10(a) i had to find how to split it into 2 right triangles by testing the 3 configurations with the pythagorean theorem, in the end I got to A = 420 which is right but it took a good amount of calculation

is there a trick I'm missing or do you really have to try all 3 configs

or is there something completely different/easier I'm missing other than splitting it into 2 right triangles

[stone cipher]

trigonometry to find an angle and then use area = 0.5(AB)(AC)sin(BAC)

but there might be an easier way

i guess it's just like

splitting it into two right triangles isn't even that easy but yeah

[odd edge]

@urban atlas Has your question been resolved?

[fluid tundra]

there is not much calculation actually because it is all pythagorean triple ratios

[fluid tundra]

the easiest way to find the area of these triangles aside from splitting them into right triangles would be to apply heron's formula

but the right triangle method is fun and actually takes less time if you use it well

[urban atlas]

I don't get the pythagorean triple ratios thing

I know what you mean but first it's not really clear to me which configuration of the split I'd go with and then even if I somehow knew, (now that I do) I still don't see how ratios give me the answer without doing the math

I know one of the right triangles has hypotenuse 56 and the other hypotenuse 39

[quartz tinsel]

what about cosine rule to get one angle then the 1/2 * sin a * bc

[fluid tundra]

👀 that is extremely not fun and will take much more algebra than you want or need

[fluid tundra]

yes you'll probably have to play around with it a bit

[urban atlas]

well I guess the problem is just more tedious than it seems on the surface 🤷‍♂️ was just making sure I wasn't missing anything

[quartz tinsel]

i hv no clue abt heron’s formula thats why

[fluid tundra]

the second one 10(b) is a lot more obvious in my opinion on what you're supposed to do

[fluid tundra]

no it is not tedious

but yes it doesn't seem you missed anything 👍

[quartz tinsel]

can u help me!

[fluid tundra]

oh

do you want me to explain heron's formula? or do you want something else

[quartz tinsel]

#help-32

[fluid tundra]

i see

[urban atlas]

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[grim stream]

Can someone help with these?

[wooden python]

this is a question about summation notation / big sigma notation

do you know what that is

[grim stream]

idk much

[wooden python]

surely you were introduced to it in the recent past...?

like, you've even got the fancier version of it here

where you are summing over a set

[grim stream]

I know how to do these types

[wooden python]

ok right.

let's look at the first sum. the one that says "i in [5]" under it

[5] means {1,2,3,4,5} (generally, [n] is the set of all natural numbers up to and including n)

so $\sum_{i\in [5]}$ is synonymous with $\sum_{i=1}^5$

[clever fjord]

Ann

[grim stream]

ah so the first one would be 30 because 2 + 4 + 6 + 8 + 10?

[wooden python]

$\sum_{i\in{3,6,9}}$ means there are three terms: one for $i=3$, one for $i=6$ and one for $i=9$ (and no others, only specifically these)

[clever fjord]

Ann

[wooden python]

yes

[grim stream]

then this would be just 3 + 6 + 9 so 18?

[wooden python]

yes

[grim stream]

Oh the answer is 12 ig

oh its not that bad

the last one is a bit confusing tho, how would u go about it?

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[onyx nacelle]

Is there a semiprime between each square number? Starting at a large enough square?

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[fathom bolt]

I have to factor this term

I understand the first step in which I have to convert x(y-z) to xy-xz, but unfortunately I don't understand the second step in which I have to convert -(z-y) to (y-z). I would like to know how to convert -(z-y) to (y-z), nothing more. Thanks.

[rotund hawk]

$$x(y-z)-(z-y)=x(y-z)+(-(z-y))$$
$$x(y-z)+(-(z-y))=x(y-z)+(-z-(-y))$$
$$x(y-z)+(-z-(-y))= x(y-z)+(-z+y)$$
$$x(y-z)+(-z+y)=x(y-z)+(y-z)$$

[clever fjord]

FirstNameLastName

[rotund hawk]

@fathom bolt anything unclear?

[fathom bolt]

one moment im looking at it

no that perfectly makes sense

[forest sky]

[(-(z-y) = -1(z-y)] this is essentially by definition of the notation. then you can distribute the multiplication

[fathom bolt]

thank you very much

[clever fjord]

cloud

[fathom bolt]

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[sick sigil]

Could someone please help me find the issue in my calculation

Question is:

Find the primitive function to f(x) = x+10/x^3 + 4x^2+5x

So first off I took out x in the denominator

And then I use partial fraction decomposition to get 2/x + -2x-7/x^2+4x+5

so the primitive to 2x is 2ln(x), so that one is now done

then i find the square for x^2+4x+5, which is (x+2)^2+1

i then split them up into two

and use a variable exchange on the bottom to get t^2+1

and from that i get -2t/t^2+1 which is the derivative of ln(t^2+1)

only one im unsure on is the arctan (is it legal to take out the -7? i think it is since it's a constant), but thats not even the one im having issues with lol

[devout sapphire]

u went wrong when substituting x + 2 = t

[sick sigil]

why/where is it wrong?

[devout sapphire]

-2t = -2x - 4

that means theres only a -3 left

u still put -7

[sick sigil]

but x is t-2

so it's -2(t-2)

-2t+4?

oh

yes

XD

[devout sapphire]

small error lol

[sick sigil]

yeah i see it now

thank you a ton

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[paper moth]

can anyone help me

[paper moth]

im thinking it is 15 can anyone confirm

????

[quasi sparrow]

! show

[odd edge]

Show your work, and if possible, explain where you are stuck.

[paper moth]

ok....

so without leak it takes 30 mins

with leak 45 mins

so an extra 15 mins is being added

so ye

im thinking 15

@quasi sparrow

[lyric marlin]

sanity check on why this is wrong: if it took 15 minutes to empty, that means the leak empties the tank faster than the pipe fills it up, so it couldn't be filled with a leak

[paper moth]

bruh wdym

just tell me if im wrong or right

im wrong im pretty sure

is it 90

?

[latent scaffold]

They pretty much told you you were wrong.

[paper moth]

is it 90

??

@latent scaffold

BRUH WHY DOES EVERYONE JUST LEAVE LIKE THAT😭

[lyric marlin]

it's been 1 minute calm down

[paper moth]

ok sorry

[lyric marlin]

I can see that your answer is wrong, I just need some time to double check that what I'm thinking actually works to give the right answer

[paper moth]

ok.....

[lyric marlin]

In a single minute, the pipe would fill 1/30 of the tank. In the same minute, the leak removes a certain proportion as well.

with the leak, 1/45 of the tank is filled each minute

so the amount the leak removes a minute is the difference between 1/30 and 1/45

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[blazing onyx]

hey i need some help

[blazing onyx]

https://www.youtube.com/watch?v=pGe2uSidWEY

4:16

specifically, why is the last line true

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@blazing onyx Has your question been resolved?

[blazing onyx]

graph theory help please

<@&286206848099549185>

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[sharp oak]

What's the question?

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[mystic saffron]

How do I use a=pert to solve this?

[warm quartz]

well what are you trying to find

you have the amount of money they want to get, the time, and the interest rate

[mystic saffron]

I think P?

[warm quartz]

yeah

[mystic saffron]

So my equation would be $8000=pe^5.052$

[clever fjord]

micock

[mystic saffron]

I don’t kkow

I tried this way and I couldn’t get a good answer

?

[warm quartz]

you can isolate the t right

i mean the p

with t = 5

[mystic saffron]

The problem is that I did e^5.052 and after subtraction it still wasn’t right

[warm quartz]

subtraction?

shouldnt you be dividing

[mystic saffron]

Don’t I multiply 5 and 5.2

[warm quartz]

what is the rate

0.052 right

[mystic saffron]

Yeah

[warm quartz]

because 5.2%

0.052(5)=0.26

[mystic saffron]

Well

When multiplying exponents u add them

[warm quartz]

but this is e^ab

[mystic saffron]

Oh

[warm quartz]

e^a times e^b is e^(a+b)

i believe

but its a product

so you should just multiply the stuff in exponent

[mystic saffron]

So i divide the e^.26 from each

Thanks bro

I was almost cooked

.close

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[pearl patio]

Hi, maybe a dumb question but i don't get the logic they're utilizing here

[pearl patio]

why is operating on b not just n operations?

[sterile blaze]

what's b?

[pearl patio]

Ax = b

where b is the solution vector ig

the context is probably only that picture tbf

[sterile blaze]

for each column we make the leading nonzero a one then we eliminate all rows below it right

[pearl patio]

yes

[sterile blaze]

so that's n row order operations to do it once. but there are n variables in each row as well so that's n^2. but for b it's only n

[pearl patio]

i get that

why is it only b for n

[sterile blaze]

then you repeat it for the second leading nonzero etc, generally you do it n times, so it'll be some factor of n^2

[sterile blaze]

hm?

[pearl patio]

i'm confused, maybe i should just write out what i think and then you could maybe help me point out the flaw with what i have in mind i guess

if b is a n x 1 vector then to change the element in the second row is 1 operation

u repeat that for everything below

so you have a sum of 1's n times which is just n

i don't see what else is there to do with b besides that

[sterile blaze]

oh

it's just because you perform all the normal row operations on the matrix to get rref

and you repeat those on the augmented matrix as well (A|b)

you are just doing the rref on A and repeating the same operations on b. not doing rref for b

if you on the other hand were doing rref for b then it'll simply be 0 or 1 as there's only one column. and yes. also n operation

[pearl patio]

so you mean when i'm eliminating one row then that's n operations on b too?

oh wait, i guess it's more like this:

you mean like if i'm eliminating all the first elements in the first column

then that's already n operations on b right?

then i have to do it with the second elements in the A matrix which is n-1 operations on b and then so on?

and then sum from k = 1 to n of k is approximately n^2

okay i guess that makes sense

[sterile blaze]

idk i didn't really do the math but it's some constant * n^2 probably

[pearl patio]

i'm not doing the math either

[sterile blaze]

yesh

[pearl patio]

hmm okay tyty

not sure why that was hard to undderstand lol

*understand

[sterile blaze]

•⩊•👍

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@pearl patio Has your question been resolved?

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[wooden python]

a bit shoddy but yes

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[outer saffron]

Domain of √5-4x-x^2+ x^2 log(4+x)

[outer saffron]

Here are my steps :

5-4x-x^2>0
x^2+4x-5<0

[-5, 1]

x^2 log(1+x) is main task

[fallen remnant]

≥

[fallen remnant]

Ah ok

[fallen remnant]

4+x > 0

[outer saffron]

Logrithm property

x>-4

(-4,1] my answer

.close

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[outer saffron]

0 to 3 integration √x (dx)^3

[outer saffron]

What is dx^3