help-26

[sonic escarp]

now sort the terms for reals only and imaginary only.

[fathom flint]

Is it
3a+sqrt{a^2+b^2}=-3bi-i
?

[sonic escarp]

no.

real part: a(sqrt(a^2+b^2)) + 2a

try imaginary part again.

[fathom flint]

-bi-i?

[sonic escarp]

you have

• ib(sqrt(a^2+b^2) from the first term (Z|Z|)
• 2bi from the second term (2Z)
• i from the third term (i)

[fathom flint]

Ohhh, so you split (a+bi)sqrt{a^2+b^2} as a(sqrt{a^2+b^2}) and b(sqrt{a^2+b^2})

So -bi(sqrt{a^2+b^2})-2bi-i?

[sonic escarp]

what do you expect? the task was now sort the terms for reals only and imaginary only.

now you have something like real part + i (imaginary part) = 0

so you know real part = 0 and imaginary part (without i) = 0

now you can solve for a and b (hint: you will get different cases).

[fathom flint]

Oki, thanks you

[sonic escarp]

youre welcome

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[toxic aspen]

"Determine the value d such that the plane
$x+y+z=d$ tangents hyperbolid $x^2+y^2-z^2=1$"

[thorny flame]

Merineth 🇸🇪

[toxic aspen]

So far i understand that the gradient of the paraboloid should be parallel to the planes normla vector

i.e i'm looking first for the green arrow?

$\grad f(x,y,z) = (2x,2y,-2z)$

[thorny flame]

Merineth 🇸🇪

[toxic aspen]

normalvector : $(1,1,1)$

[thorny flame]

Merineth 🇸🇪

[toxic aspen]

$(2x,2y,-2z) || (1,1,1)$

[thorny flame]

Merineth 🇸🇪

[toxic aspen]

If i were to solve this for x,y and z

$2x = k\2y=k\-2z = k$

[thorny flame]

Merineth 🇸🇪

[merry mason]

is that parallel to symbol? btw

[toxic aspen]

$x = k/2\y=k/2\z=-2/k$

[thorny flame]

Merineth 🇸🇪

[toxic aspen]

If i input these x,y, and z into the equation of my paraboloid

Wouldn't that result in the dashed line?

Or rather

The result i got was $k = \pm 2$

[thorny flame]

Merineth 🇸🇪

[toxic aspen]

What does this mean?

(2x,2y,-2z) = k(1,1,1)
the k here is a scalar?

k = 2 : (x,y,z) = (1,1,-1)
k = -2: (x,y,z) = (-1,-1,1)

[crimson kayak]

From this you have the values of x, y and z in terms of k
And you have a relation x+y+z = d
You can obtain d

[toxic aspen]

That these two points are where they tangent each other?

[crimson kayak]

Yes k is scalar

[toxic aspen]

aaah

So if i input these x,y and z into the equation for the plane :
x+y+z = d
1+1-1 = d = 1
-1-1+1 = d = -1

[crimson kayak]

Yeah

[toxic aspen]

So we get two total points of tangency, one at d = 1 and d = -1

[crimson kayak]

Yeah

[toxic aspen]

Goddamn

Was it that easy?

[crimson kayak]

Yes

[toxic aspen]

The fact that i have to chatgpt is so sad

:c

[crimson kayak]

Yeah

[toxic aspen]

What made you realize it's a scalar?

[crimson kayak]

If two vectors are parallel
Then we can say

Vector1 = k*Vector2

Where k is a scalar
k is a scalar therefore Vector1 and Vector2 remain parallel, the scalar k only increases or decreases the magnitude of Vector2 and has no effect on the direction of Vector2

[toxic aspen]

Ok and the magnitude being the length of the vector?

[crimson kayak]

Yeah

They're same thing

[toxic aspen]

I think i get what you mean

I'm trying to visualize it

Do you know how i can visualize it better?

[crimson kayak]

Umm I can try wait

[crimson kayak]

You can see
<1,1> and <3,3> have the same direction

We can say that <3,3> is a three times scaled <1,1>

Or vice versa

So basically 'k' only scales the vector manipulating a vector's magnitude
Maybe that's why they're called scaler

[toxic aspen]

(2x,2y,-2z) = k(1,1,1)
So this just means that when we scale the normal vector with k. And solve for x,y,z we get a point?

[crimson kayak]

Yeah

[toxic aspen]

Why exactly do they become tangent points tho?

[crimson kayak]

You can also say it as
(2x,2y,-2z) = (k,k,k)

[crimson kayak]

Okay that is going to be a bigger concept tbh
If you want to completely understand it
Id recommend you watch MIT 18.02 ( available on YouTube)

Umm see
Finding the gradient of a function gives the 'components' of the normal vector of the plane which is parallel to the tangent plane

And here yk x+y+z = d is a tangent plane
So you can say the components of the normal vector to this plane is (1,1,1)

And this normal vector is parallel to the vector (2x,2y,-2z)

[toxic aspen]

could it be

something like this?

[crimson kayak]

Exactly

[toxic aspen]

yeah that does make senseeee

tysm!

finally someone clear !!

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[south yew]

a^2+2bc=x^2+2yz
b^2+2ac=y^2+2xz
c^2+2ab=z^2+2xy

Prove that a^2+b^2+c^2=x^2+y^2+z^2

I know that it is permutation and than you can prove that
ab+ac+bc=xy+xz+yz => a^2+b^2+c^2=x^2+y^2+z^2
but I have no idea how to prove that it is permutation

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[south yew]

<@&286206848099549185>

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[slow rivet]

hi guys, i need help for my calc

[slow rivet]

i’ve taken my final and just got grades back, i was doing perfect

here are the results

is there any error in the grading

historically this teacher has made errors but i don’t grasp these enough to be able to tell if they are truly correct or not

[outer delta]

For question 1, on the interval of [0, 1] it asks about F(1), is the area under the graph from [0, 1] positive or negative?

[slow rivet]

the area is positive

and increasing

i understand how that was wrong

i thought they meant the function of the line instead of the integral

[outer delta]

Question 2 has variable(s) in the bounds of the integral and requires the chain rule

[slow rivet]

im just wondering if the grading is justified

i had 90s and 100s every exam but i messed up on the final

[outer delta]

I don't see how question 8 lost 2 points, it could be due to the seemingly random f(x). All it's asking is which part(s) of the function are odd and which are even, it's not asking about the overall function right?

Wait do they line up, I'm confused by the notation of points

[slow rivet]

but everything else is correct?

[outer delta]

Sorry I misinterpreted the notation of the website(?)

The blue letters tell me what question to look at

I mean yeah it seems like fair grading

Question 3. There's no work

Multiple choice questions can be ruthless, you either get all the points or none of the points

Unless it has part a. Part b

Question 10, you didn't simplify your answer, were you running out of time?

[slow rivet]

he said we didn’t need to

[outer delta]

You also didn't show all of your work

[slow rivet]

was the answer for 10 correct tho

[outer delta]

No?

[slow rivet]

so it seems graded fairly yea

[outer delta]

Yeah

[slow rivet]

damn man

ngl this hurts so much lmao

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[trail hull]

what;s the general method of factoring

$Ax+By=Cxy$
into
$(x+m)(y+n)=k$?

for example, how does one factor $2b+ 3c= 5bc$ into $(5b−3)(5c−2) = 6$

[thorny flame]

matrix of ones
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gilded plume]

oh here

$Ax+By=Cxy$

[thorny flame]

Arnavutköy

[gilded plume]

$Cxy-Ax-By=0$

[thorny flame]

Arnavutköy

[trail hull]

yeah

[gilded plume]

$C(xy-\frac{A}{C}x-\frac{B}{C}y)=0$

[thorny flame]

Arnavutköy

[gilded plume]

$C((x-\frac{B}{C})(y-\frac{A}{C})-\frac{A}{C}\frac{B}{C})=0$

[thorny flame]

Arnavutköy

[gilded plume]

$C(x-\frac{B}{C})(y-\frac{A}{C})=\frac{AB}{C}$

[thorny flame]

Arnavutköy

[gilded plume]

$(x-\frac{B}{C})(y-\frac{A}{C})=\frac{AB}{C^2}$

[thorny flame]

Arnavutköy

[gilded plume]

or

$(Cx-B)(Cy-A)=AB$

[thorny flame]

Arnavutköy

[gilded plume]

you will usually have to use these last 2 equations

[trail hull]

okay that's nice

thank ypoui

.close

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[agile meadow]

I got OCX=90 but my friend got 45, can someone tell whos correct

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[next rain]

Do we always just make the pi disappear when finding angle?

[next rain]

What was the logic behind canceling out the pi?

Am I tweaking or can we not cancel out the pi because there's 3 on the bottom and 5 on the top?

Or we can because it's multiplication and it's a common factor

[cedar wagon]

Thats the purpose of translating from rad to degrees

[cedar wagon]

We can and its a common factor

[next rain]

Yeah right, I'm rusty at math

Ty

.solve

.solved

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[surreal mural]

[acoustic pecan]

what is going on here

[neon iron]

prove what lol

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[agile harness]

prove all positive integers can be written as

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[rotund yoke]

that any natural number

can be written as

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[rotund yoke]

@neon iron prove that any natural number can be written as one of equasion

[odd pagoda]

whatever you wrote there is hard to read, so it would help if you wrote it down on a piece of paper and made a photo or wrote it in latex

[rotund yoke]

ok sure

just give me a minute

done it

so?

[rotund yoke]

@odd pagoda what do you think?

anyone?

@everyone

[thorny plinth]

wdym by "all the integers can be written as one of the expressions"?

(also @ Helpers not @ everyone...)

[rotund yoke]

ok sorry

[rotund yoke]

I meant like every integrer

can be written as one of these expressions

i mean like

1 is 2^0

2 is 2^1

3 is 2^2 -3^0

and so on

how can I prove i?

<@&286206848099549185>

[thorny plinth]

What are the a,b,c here?

[rotund yoke]

integrers

[thorny plinth]

Yes but what are they for 3?

[rotund yoke]

its a =2

2^2 -1 =3

[thorny plinth]

Wait yeah ok, your expressions are a bit confusing though.. so you have as many 2^something as you want, then at some random point you start putting 3^something in as well?

[rotund yoke]

no

its like this

first ust 2 do the a power

then it minus one

then we multiply it by 2^b

then we just -3

[neon iron]

Hi can someone reccomend me some places where i could practice yr 7 mathematics

[rotund yoke]

times 2^c

[neon iron]

Like websites wtc

Etc

[rotund yoke]

idk

[neon iron]

Ok

[rotund yoke]

then -9

times 2^d

and so on

[thorny plinth]

okk

[rotund yoke]

sdo

what is the answer?

??

[thorny plinth]

idk man.. feels like a weird question.. also can you do 5? Bc looking mod 3 i dont think you can do anything that's 2 mod 3

[thorny plinth]

This channel is occupied

#❓how-to-get-help

[neon iron]

What does that mean

Ok

[rotund yoke]

It has own problem

[neon iron]

Oh ok

[rotund yoke]

you need to get cahnnel with out a problem

like help 12

[rotund yoke]

5?

[thorny plinth]

5 = 2^.. ?

[rotund yoke]

16-8-3

a=1 b=3

2^a+b -2^b-3

[thorny plinth]

Oh right 1-2=-1 I'm dumb

[rotund yoke]

nah

you are smart

and you know how to solve this problem

I believe you do

[thorny plinth]

Uhh i don't think so..

I need to go now as well.. so um good luck with this

[rotund yoke]

Do you have any idea wha should I try?

[rotund yoke]

hae a nice day]

[thorny plinth]

You can rewrite it without the +, as in 2^a - 2^b - 3·2^c - 9·2^d...

[rotund yoke]

yeap

[thorny plinth]

But yeah i dont know what else to do

[rotund yoke]

ok thx for your time an help

have a nice day

<@&286206848099549185>

[rotund yoke]

how do I solve it?

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[tame loom]

what is an identity, conditional equation, or an inconsistent equation?

[agile harness]

identity : always true
conditional equation : may be true depends on the value of x
inconsistent equation: never true

[tame loom]

how do i find out the equation is one of those

i solved 15

x=6

but

[agile harness]

but what

[tame loom]

its positive, so would it be an identity?

[agile harness]

🤔

what does it being positive have to do with anything

[tame loom]

i misread

woops

[agile harness]

geometrically there is one intersection of the line y = 2x-5 and y = 7

does 2x-5 always = 7

or does it depend on the value of x?

[tame loom]

does it depend on the value of x?

[agile harness]

i’m asking you

is it true that for all x, 2x-5 = 7?

meaning regardless of the value of x, we will have a true statement

[tame loom]

i think so

[sullen locust]

no

it is not true

[agile harness]

thank you walk down

we didn’t know

[sullen locust]

ofc

it would depend on the value of x

[agile harness]

!help

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[sullen locust]

what you have to do is try out every way to find out what it would be

i'm talking about rearranging the numbers\

[agile harness]

brother

[sullen locust]

mb

ima just go..

[tame loom]

LOL??

[agile harness]

it is not true, consider a counter example like x = 0

2(0) - 5 =-5 ≠7

[tame loom]

so could i try that for each problem?

[sullen locust]

that's what i meant by trying different methods...

yes

[agile harness]

sure

so because the truth value of this statement depends on the value of x

this is conditional

it is not always true

nor is it always false

[tame loom]

i seee

i see

walkdown bro

[sullen locust]

what happened?

i apoligize

[tame loom]

anywats

thank you both

[sullen locust]

NOW CLOSE THE GOD DAMN CHANNEL

[tame loom]

i WAS GOING TO

[agile harness]

.close

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[sullen locust]

.reopen

oh...

[agile harness]

.reopen

[topaz sinew]

✅

[agile harness]

.close

[sullen locust]

HOW

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[tame loom]

CLOSE TS!!

[sullen locust]

how do i get permission?

[tame loom]

.close

[agile harness]

get better bro

[sullen locust]

..

[agile harness]

already closed it

[tame loom]

😕

[agile harness]

wild

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[lunar grail]

i need help

[wooden osprey]

send your question

[coarse herald]

[wooden osprey]

offtopic, but i love your pfp

[coarse herald]

thanks

[lunar grail]

okay bett

its like this project actually

idk whats going on in this class fr

its geometry

<@&286206848099549185>

[coarse herald]

i think its just similarity

[lunar grail]

no idea what that means

i need someone that knows how to do that math

[coarse herald]

well how did you solve the first one then

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@lunar grail Has your question been resolved?

[lunar grail]

my friend did it

w me in class

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[dim tangle]

id watch a video on it first to understand

https://youtu.be/4TVeYKgLmN8?si=nA-0dB1XKytfkSQB

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[surreal river]

Help

[surreal river]

Can anyone explain this

[half edge]

@surreal river still need help?

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[burnt swift]

guys wtf do I do?

[lean pebble]

you know what EBC is right?

[flint stump]

Find the sum of angles of ABC and ABD

[burnt swift]

how

[flint stump]

sum of angles of the two triangles

[burnt swift]

I dont know

[burnt swift]

ok let me do a drawing

[burnt swift]

one equation and three variables

[mystic jay]

no

use isosceles condition

[burnt swift]

a+b=90

[flint stump]

Now consider triangle ABE

[coarse tusk]

exactly

so you are done

[burnt swift]

2(c+d)=270

,calc 270/2

[thorny flame]

Result:

135

[burnt swift]

okay I appreciate the help, ty

.solved

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[sleek bay]

I started by splitting the integral.

[lucid junco]

did the substitution for the previous integral work

[sleek bay]

No I did a diff method

A(denominator) + B(derivative od denominator) = numerator

Find a and b then sub and cancel stuff ans came out easily

This one I think I gotta use ur sub method

T formula

[lucid junco]

so after splitting what did u get?

[sleek bay]

2/ sinx (1+cos x) + 3/1+cos x

[lucid junco]

yes

[sleek bay]

Dx and bounds

[thorny flame]

water beam

[sleek bay]

Yes

[lucid junco]

okay lets use this

lets tackle the first part first

2/sinx(1+cosx)

[topaz sinew]

@sleek bay Has your question been resolved?

[sleek bay]

Got it

Thanks @lucid junco

I used that exactly and got it

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[frigid swan]

find radius

and length of QE

[topaz sinew]

@frigid swan Has your question been resolved?

[wooden moon]

well, PA * PB = PC * PD

Do you know this property of intersecting chords?

also, AB is perpendicular to CD

[fading owl]

@notakshar do you know power of a point?

Did he leave the server?

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[zenith sonnet]

Quick question, is sin²(x) = (sin(x))² ?

[wooden osprey]

yes, they are the same

[zenith sonnet]

okay thank you, and is it also the same as sin(x)²?

[wooden osprey]

ehhh

it's ambiguous

if you write it like that i'll assume you are referring to sin(x^2)

[zenith sonnet]

Okay, then I will always put in brackets

Thank you^^

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[coarse tusk]

I've usually seen this to mean sin^2(x) as well

f(x)^2 usually means (f(x))^2

[wooden osprey]

that isn't really a good analogy

[coarse tusk]

Notation often depends on context

[coarse tusk]

What do you mean?

[wooden osprey]

f(x)^2 is not as ambiguous as sin(x)^2 imo

i wouldn't immediately default to f(x^2) if you say the former

[coarse tusk]

Yeah you are right

It's just bad notation imo

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[wispy gulch]

i have no idea how to solve

[topaz sinew]

@wispy gulch Has your question been resolved?

[smoky sparrow]

ah, so the axis of symmetry of y^2 = 4x is just y = 0

when reflected across y = -x - 2

try drawing it out, the axis of symmetry of the image must be x = -2

[smoky sparrow]

yeah then you just need to reflect (0, 0) on y^2 = 4x across the line

then that point will be on the new curve

[wispy gulch]

oh

lemme try it out

in getting -2,-2

as the new vertex

so now is it a shifted curve with new coordinates?

@smoky sparrow

[smoky sparrow]

yeah that gives you enough information

from this it's C

[wispy gulch]

it wud be option C

ye

[smoky sparrow]

since it's a reflection it has to be a translated version of $x^2 = -4y$

$x + y + 2$ is parallel to $y = -x$, so that's why the minus sign appears and it's not like $x^2 = 4y$

[thorny flame]

south

[smoky sparrow]

so then you just follow the vertex, a translation of (-2, -2) down

$(x - (-2))^2 = -4(y - (-2))$ and bingo

[thorny flame]

south

[wispy gulch]

ohhk

so if its a reflection, x and y will interchange and there is a possiblity of one going -ve

i have another question similar to this

ill solve it and then close this channel

got it brotha thanks

.close

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[half tulip]

hi

[half tulip]

I have a quick question about refuting uniform continuity. You have to select an x ​​and x_0 so that the distance between them is less than delta, but for exactly these x_0 and x the epsilon criterion should go wrong. My question is how to select or find this epsilon

[keen raptor]

If a function is continuous but not uniformly continuous, it probably gets steeper and steeper without limit in some sense

Any epsilon will probably work as long as you choose your x at a part that's steep enough (based on delta)

For some graphs like sin(x^2) you'll need an epsilon value small enough that you can escape (y-ε, y+ε) though

Anything <2 will work

[half tulip]

😅 rn im trying it with sin(x^2)

for x i choose x_0-Delta:2

[keen raptor]

ok then how do you choose x_0

[half tulip]

i didnt change x_0

but if i would then most likely pi

[half tulip]

or wait can i say that delta has to be bigger then pi/2

if yes then i could take epsilon=1 and it wouldnt count

[keen raptor]

No you can't choose delta

And you need to choose x_0 depending on delta

if your choice of x_0 does not depend on delta, then you're disproving ordinary continuity

sin(x^2) is continuous, so that's impossible

[half tulip]

ok then if x_0= Delta/2 + pi or Delta/2 + pi/2

and x ist then pi or pi/2

the delta kritria would work

ok doenst make sense

https://www.youtube.com/watch?v=iFdyD0EcWNI&ab_channel=SBM i saw this vid rn and in this example x_0 isnt delta related

<@&286206848099549185>

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[runic sail]

help

[runic sail]

how i should solve this?

[stuck tusk]

if $A=OS$ where $O$ is orthogonal and $S$ is symmetric, then $A^T=SO^T$, so $A^TA=S^2$. Now try diagonliazing $A^TA$, and you can find a square root of $A^TA$, because this matrix is symmetric positive

[thorny flame]

TimourX

[stuck tusk]

And then, you can let $S$ be a square root of $A^TA$, and let $O=AS^{-1}$, if $S$ is invertible (I hope so)

[thorny flame]

TimourX

[runic sail]

okay thank you i will try that ^^

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[deep forum]

in regular pentagon ABCDE, F and G are midpoints of AE and CD. If AB∩DE=H, AE∩BC=J, DF∩GH=K and DE∩JK=L, prove that BG⊥GL

[olive ridge]

What does AB∩DE=H mean?

[deep forum]

AB and DE intersect at the point H

[olive ridge]

OK

[topaz sinew]

@deep forum Has your question been resolved?

[deep forum]

<@&286206848099549185>

[wooden moon]

technically there is a way to brute force this using sine and cosine laws, I don't see a easier way personally

[deep forum]

i'm fine with any way honestly

just let's say that i only have one page to fit the solution in

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[topaz sinew]

@deep forum Has your question been resolved?

[deep forum]

.close

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[deep forum]

.reopen

[topaz sinew]

✅

[rustic garden]

Man idk

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[astral hazel]

Can i have help in my math test

[astral hazel]

Im 1 point away from A-

[placid merlin]

we cant help with tests, but we can help you study

[smoky sparrow]

you're a goated helper btw

[astral hazel]

ok can u explain

[hollow drum]

Are you taking a test right now?

[astral hazel]

uh

so our teacher let us do the test home he says we can use notes and everything but i dont understand some part

im on big ideas

math

its geometry level honors

[hollow drum]

Open notes/everything doesn't mean you can use discord

[pearl fog]

vouchh

[astral hazel]

no he lets use outside sources

bro i dont want the answer i just want an explanation so ik how to do it

[smoky sparrow]

still though asking other people to do your test for you is not allowed

that's like the one thing you can't do

[astral hazel]

good point

[pearl fog]

why does teachers allow to do testing at home outside of the pandemic a while back

[astral hazel]

bro u wanna see

well i use u guys as a source

[mortal hull]

fr lmao

[pearl fog]

use chatgpt

[mortal hull]

we're better than that

[astral hazel]

.

[mortal hull]

i mean

[astral hazel]

chat just help me out

Pls

[mortal hull]

you're better than that

[pearl fog]

if were better we dont help :3

[dark swan]

i think its safe to assume you cannot ask for help from people on this test

if it's not clearly stated

[astral hazel]

It is

[empty spoke]

your course probably has a precise definition of what this means, I don't think direct one on one assistance falls into "sources"

[smoky sparrow]

yes check your school's academic honesty policy as well

[empty spoke]

in any case the help channels are meant to be about specific math questions (e.g. an exercise or problem), so I'm gonna go ahead and close this

.close

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[neon iron]

uhm...how?

[sweet shard]

Plot the two of them

Circle and ellipse

[topaz sinew]

@neon iron Has your question been resolved?

[pseudo horizon]

the ellipse has major axis 12 and minor axis 4

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[prisma fulcrum]

help with part c

[smoky sparrow]

so remember that line l is perpendicular to AB, so that is the perpendicular height

you first need to find the length of AB using the distance formula
that is the base length

then you should have height = 5

but then for every 3 units to the right and 4 units up, the length is 5 units by Pythagoras

(if you have height = 10 scale up, so 3 * 2 and 4 * 2 in that case)

so it's just M + (3, 4) and M - (3, 4)

[smoky sparrow]

we choose this cause the gradient of l is 4/3

[prisma fulcrum]

hold up

i don't understand fully

can you be a little more rigorous?

[smoky sparrow]

which part don't you understand?

[prisma fulcrum]

the M + (3,4), M - (3,4)

[smoky sparrow]

ahhhhh

this diagram should explain

[prisma fulcrum]

i alr have this diagram drawn

if the first coordinates are x1, y1

and the second x2,y2

[smoky sparrow]

basically you can go in the opposite direction from M and get another triangle with the same area

[prisma fulcrum]

ye

[smoky sparrow]

that's why it's like this then

[prisma fulcrum]

i was tryna do something like

$\frac{6.5-y_1}{2-x_1} = \frac{4}{3}$

[thorny flame]

hoax

[prisma fulcrum]

in this aren't you assuming the units moved in the x direction and then the y direction are integers?

[smoky sparrow]

ah that doesn't work actually

that just tells you that (x1, y1) is on the perpendicular line

[prisma fulcrum]

a^2 + b^2 = 5^2

[smoky sparrow]

you'd also need the equation of the circle with centre M and radius 5

then it's a big algebraic pain

[prisma fulcrum]

if we were restricted to integers a and b only

then yes we can say a,b = 3,4

[smoky sparrow]

you could choose the direction vector (0.3, 0.4) too

[prisma fulcrum]

but otherwise, infinitely many solutions :/

[smoky sparrow]

you just need a vector which has the same gradient as the perpendicular line

we're lucky because the scale factor for (3, 4) is 1
that already has length 5

for these questions they will give you convenient numbers

[prisma fulcrum]

now an approach i tried was

first finding y1 in terms of x1 by using the gradient formula

y2-y1/x2-x1

then i decided to input all that into the equation i get when determining the length of MC in terms of x1 and y1

[prisma fulcrum]

and then u get a quadratic equation in terms of x1 only

but unfortunately it ends up reducing to one with no real solutions

[smoky sparrow]

if you do it correctly you should get two solutions for x1

[smoky sparrow]

but like I said, huge pain

[prisma fulcrum]

i think i did it pretty much correctly

but no real roots :/

perhaps i should take the effort of writing it all in latex

wait

for $m = \frac{y_2 - y_1}{x_2 - x_1} \text{Does it matter if}y_2<y_1$

[thorny flame]

hoax

[prisma fulcrum]

and $x_2<x_1$?

[thorny flame]

hoax

[prisma fulcrum]

$\frac{6.5-y}{2-x} = \frac{4}{3}
y = \frac{4}{3}x - \frac{23}{6}$

[thorny flame]

hoax

[prisma fulcrum]

$\frac{6.5-y}{2-x} = \frac{4}{3}//
y = \frac{4}{3}x - \frac{23}{6}$

[thorny flame]

hoax

[prisma fulcrum]

what was the command for new line i forgot

[smoky sparrow]

try using $$ ... $$

[thorny flame]

south

[smoky sparrow]

then you can type all at once and it will automatically generate new lines

[prisma fulcrum]

it was \ i think

[smoky sparrow]

that's like for a space

[prisma fulcrum]

$4 \
4$

[thorny flame]

hoax

[smoky sparrow]

yeah you can do that if you use $...$ then

[thorny flame]

south

[prisma fulcrum]

$4 ...
4$

[thorny flame]

hoax

[prisma fulcrum]

$\frac{6.5-y}{2-x} = \frac{4}{3} \
y = \frac{4}{3}x - \frac{23}{6}$

[thorny flame]

hoax

[prisma fulcrum]

$\sqrt{(6.5-y)^2 - (2-x)^2} = 5

[smoky sparrow]

have fun I guess solving it? it is possible ofc

[prisma fulcrum]

$\sqrt{(6.5-y)^2 - (2-x)^2} = 5 \
(6.5-y)^ - (2-x)^2 = 25 \
\frac{169}{4} - 13y + y^2 -(4-4x + x^2) = 25 \
\frac{169}{4} - 13y + y^2 -4 + 16x -x^2 = 25 \
y^2 - 13y -x^2 + 16x + \frac{169}{4} = 25 \
(\frac{4x}{3} - \frac{23}{6})^2 - 13(\frac{4x}{3}) - x^2 + 16x = \frac{-69}{4} \
\frac{16x^2}{9} - \frac{92x}{9} + \frac{529}{36} - \frac{52x}{3} - x^2 + 16x =\frac{-69}{4}$

[thorny flame]

hoax

[prisma fulcrum]

$\frac{7x^2}{9} -\frac{104x}{9} = -\frac{575}{18} \
14x^2 -208x = -575 \
14x^2 - 208x + 575 = 0 \$

[thorny flame]

hoax

[prisma fulcrum]

these are the roots :/

@smoky sparrow

[smoky sparrow]

yikes

[smoky sparrow]

oh wait it should be $\sqrt{(6.5-y)^2 + (2-x)^2} = 5$

[prisma fulcrum]

ahh fuck

[thorny flame]

south

[prisma fulcrum]

lemme solve it in my notebook rq

no real roots 😭

@smoky sparrow

i need some help fr

[topaz sinew]

@prisma fulcrum Has your question been resolved?

[exotic summit]

The trick is that your perpendicular bisector (line l) has the same slope as the hypotenuse of a 345 triangle

Try drawing a 345 triangle with 1 corner on the midpoint and another somewhere else on the line

[topaz sinew]

@prisma fulcrum Has your question been resolved?

[prisma fulcrum]

The hypotenuse of a 345 triangle is 5

The slope is 4/3

[exotic summit]

Yes

I'm talking about the slope of that hypotenuse though

It's the same as your line

In other words, your line is like an extended hypotenuse of a 345 triangle

Which you can leverage to solve the problem

[prisma fulcrum]

who said anything about the base and the perpendicular having integer lengths 😭😭

[exotic summit]

The length of the base can be found by taking $\sqrt{((-4)-8)^2+(11-2)^2}=\sqrt{(-12)^2+(9)^2}=\sqrt{144+81}=\sqrt{225}=15$

[thorny flame]

Cyktard_

[prisma fulcrum]

i alr know that

AB = 15

[exotic summit]

Then triangle ABC has base AB=15 and height MC which we'll call h

Applying the formula for the area of a triangle we get $15h/2=37.5$, which we can rearrange into $h=75/15=5$

[thorny flame]

Cyktard_

[exotic summit]

Thus the base and height are both integers

Specifically 15 and 5

From there just apply the 345 triangle

[topaz sinew]

@prisma fulcrum Has your question been resolved?

[topaz sinew]

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[cedar wagon]

Is it correct ?

[cedar wagon]

Phi is the bijective application in C^1 and his jacobian below

And third page im using change of variable theorem

[topaz sinew]

@cedar wagon Has your question been resolved?

[topaz sinew]

@cedar wagon Has your question been resolved?

[topaz sinew]

@cedar wagon Has your question been resolved?

[golden blade]

Let's say I got the same in cartesian coordinates

,calc 9/16

[thorny flame]

Result:

0.5625

[cedar wagon]

Ok ty for the result !

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[quiet dock]

i have a surface area of a leaf which I calculated to be 21.073 units ^2 on geogebra, I would like to convert this to cm. How can I do that?

[quiet dock]

this was my leaf

[whole geode]

ok. Well, what's the units correspond to in cm?

1 unit = x cm. If you can tell us that, then we can tell you that 1 unit^2 = x^2 cm^2

[quiet dock]

im not sure I can't find any information on the website

[whole geode]

the unit doesn't come from the website

[quiet dock]

oh

[whole geode]

it comes from your picture

[quiet dock]

so I js have to measure it

?

[whole geode]

yes

and then compare

So you have the leaf going out to 5.5 units.

[quiet dock]

but if the picture isn't shrinked / enlarged at all is there any way to calculate it

[whole geode]

if you measure the leave to be 11 cm, then that's 1 unit = 2 cm

because 11/5.5 = 2

[whole geode]

No, because you can take a picture of a leaf with that camera close up, but you can also take a picture of an entire cruise ship with that camera if it's a little farther away, right?

It depends on the field of view, the distance to the object, whether or not you're taking a picture at an angle or head on, and so on.

[quiet dock]

oh true

ok well that is simple enough

thank you

oh also

for a lab do you think id have to explain the steps im doing here in more detail?

do you think it should be explaining the stepes of integration?

[whole geode]

it looks like power law stuff. I would probably omit it as it's simple, maybe quote the exact coefficients in an appendix just for reproducibility

[quiet dock]

okay thanks

.close

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[distant rapids]

I've got some geometry questions.
First one:
The height of a regular quadrangular truncated pyramid is 4 cm, the sides of the bases are 2 cm and 8 cm. Find the area of ​​the diagonal section.

[warm turtle]

try finding the diagonals of the top and bottom base first

[distant rapids]

Yes, that is what I am thinking about.
Channel magic has helped me to understand the algorithm just as I posted a question 🙂

However I've got other tasks as well

The lateral edge of a regular quadrangular pyramid with base side 2 is inclined to the plane of the base at an angle of 60°. Calculate the cross-sectional area of ​​the pyramid by the plane passing through the midpoints of the adjacent edges of the base perpendicular to the base.

[warm turtle]

smth like this?

apologies for the poor drawing 😅

[warm turtle]

so the area within that orange cross section

since the angle is 60, therefore it would be a ?

what type of triangle

within that regular 3d square pyramid

[distant rapids]

My mathematical terminology isn't good, I don't study in English, sorry for that.
Though, I would assume that every angle will be equal to 60 deg

[warm turtle]

it's okay dww

but yes u are correct

so you can just think of the problem as finding the area of that triangle

[distant rapids]

Are the faces you have drawn your cross-section across are adjacent? Or am I getting it wrong?

[warm turtle]

OHH omg actually im wrong

i drew them opposite 😭

im sorry wait

the answer is gonna be different

[distant rapids]

Could you please illustrate the drawing?
Is it correct that cross-section will form an AC on the base if we consider naming the base - ABCD?

[distant rapids]

Oh, got you.
I understand how to get the base but not the sides though

[warm turtle]

yes , if the ABCD are the midpoints of each base

[distant rapids]

So basically the base of a triangle is going to be sqrt(1)?

[warm turtle]

yes

to get the area, i think all you need to do is use the 30 60 90 right triangle property to find the height of the pyramid first

or actually, the height of the "section" of the pyramid at the midpoint of the line AC, given that the AC is the base of the triangle

[distant rapids]

Thanks a lot, I think I understand it!

[distant rapids]

Would you have some time to help me with theoretical questions and one other?

[warm turtle]

yay! youre welcome

[warm turtle]

suree

as long as they won't take too much time i'm okay

[distant rapids]

How many planes of symmetry does a regular square pyramid have?

I think it's two but I'm deffinitely not sure

[warm turtle]

i think it's 4

but i'm not sure as well hahaha

because you can slice the square pyramid, if you are looking at the pyramid from the top, vertically, 2 ways diagonally, and horizontally

[distant rapids]

When googling in English it actually gives an answer. And you are right, it is 4

[warm turtle]

ahhh okay

yipee

[distant rapids]

In the pyramid SABC, SD is the height of the face ASB, (SAB)⊥(ABC). What segment is the height of the pyramid?

[warm turtle]

did the question specify what kind of height?

if it was the slant height it would be different from the perpendicular height

but if it's asking for the perpendicular height i think it should be SD

[distant rapids]

Nope, it didn't that is why I am curious

[warm turtle]

hmm

i'm not sure 100% how to figure that out

[distant rapids]

Well, there is no option that represents slant height(if i'm correct)

So I think it's right

[warm turtle]

yep

theres no option

SC would be an edge

as well as SA and SB

[distant rapids]

Last one

The lower base of the cylinder belongs to plane a. Point M is taken on this same plane. MA is tangent to the circle of the lower base of the cylinder. At what point will the straight line MH cross the surface of the cylinder?

1. median
2. height
3. bisector

[topaz sinew]

@distant rapids Has your question been resolved?

[distant rapids]

Nevermind, solved everything I needed. Huge thanks @warm turtle

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[golden blade]

<@&268886789983436800>

[craggy haven]

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[wide forge]

How can I determine the smallest square that can contain this shape?

[keen venture]

Are all angles equal? This is basically an equilateral pentagon?

[wide forge]

yes

it is a 5 pointed star if you will

all angles and lengths are equal

@keen venture I think a 27.58 by 27.58 square would work

but is there a smaller box with some angle trickery?

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@wide forge Has your question been resolved?

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[cyan elk]

Please guys im dumb 😭🙏

[rose sierra]

find the total area of the 1 sections

then divide by total area

[cyan elk]

Ok hold on

Nvm I got it

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[marsh mirage]

I have used the inequality between x^2+y^2+z^2>xy+yz+zx . Considering theta as an acute angle, I have got the answer (b). Is it correct?

[mortal thunder]

Is it given that theta is acute?

[marsh mirage]

No. That's the problem. I can't find any relation if theta is any angle. If you can, then it will be nice.

[mortal thunder]

First of all, you've only sent a part of the question is what I can see. Can you send the full question if there's any data given before that can be used?

[mortal thunder]

Alr. Firstly what can you say about the expression x² + y² + z² - xy - yz - zx ?

Hint: double the expression and try forming perfect squares

[marsh mirage]

Yeah. I have already done it.

[mortal thunder]

So, you must have that x² + y² + z² ≥ xy + yz + zx and since x, y, z are all distinct positives, that gives you (x² + y² + z²)/(xy + yz + zx) ≥ 1

[floral sable]

(Not to interrupt but your writing it beautiful, ok sorry bye)

[marsh mirage]

Yes.

[mortal thunder]

From this, you can check that:

Option a) demands cos theta ≥ 1 so false
Option c) is same
Option d) demands cosec theta ≤ 1 which is, again false

So (b) is correct

[marsh mirage]

Thanks. I have done it in the same way.

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[spare onyx]

What is (4/5) squared

[topaz sinew]

Please don't occupy multiple help channels.

[spare onyx]

Zayn can u help me

[lean owl]

Someone please tell me how 1 - sin²a/cos²a can be written as cos²a - sin²a / cos²a?

[lean owl]

16/25?

[spare onyx]

I’m no math nerd but I think u can just switch the variables

As long as the operations are the same

[lean owl]

Wait I am in the wrong channel

[fading owl]

16/25

[spare onyx]

Wan you explain it

Can*

[sonic escarp]

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[elfin onyx]

what

[smoky sparrow]

why

[elfin onyx]

.reopen

[verbal crater]

#help-23

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[vivid saffron]

Hello! My thought process was :

o / s = 24 / 5 and** s / t = 1 / 12** and o / f = 16 / 1
**o : s : t : f ** = 48 : 10 : 120 : 3
The biggest number is the third one that is equal to 120 and the smallest number is the fourth one that is equal to 3.
The third height is equal to 10.
o = 10( 48 / 120 ) so o = 4
so f = 4( 1 / 16 )
The smallest numer ( the fourth one ) is equal to 0.25

[vivid saffron]

I am pretty sure this is not right, but I don't really know how to do it right

[smoky sparrow]

let the original height be h

then her second height is 5/24 * h

(o : s) = (24 : 5) = (h : 5/24 h), divide by 24 on both sides

then her third height is 5/24 * h * 12

can you continue?

yeah then you have to work backwards a bit, get the original height first

[vivid saffron]

wouldn't it be 24/5 ?

[smoky sparrow]

and then divide by 16 to find the shortest height

[smoky sparrow]

no, the 2nd height has to be shorter than the original

so the ratio is less than 1

[vivid saffron]

so what is o?

[smoky sparrow]

original

s is second

[vivid saffron]

but h is the original height, no?

[smoky sparrow]

yeah

[vivid saffron]

sorry mate, i think you are trolling

[vivid saffron]

so o = h

[smoky sparrow]

yeah

they used o so I just followed them

[vivid saffron]

i don't understand very well.
h / s = 24 / 5

if i want h to be in the another side

the right side will be multiplied by -1? what happens when i want to get rid of a numerator?

[smoky sparrow]

that's actually not what you want

you can multiply by s on both sides, do the same thing to both sides

you get h = 24/5 s

that doesn't tell you what the 2nd height is, in terms of the original

[smoky sparrow]

but you can flip both fractions to get $\frac{s}{h} = \frac{5}{24}$

[thorny flame]

south

[smoky sparrow]

now you have $s = \frac{5}{24} h$, multiplying by $h$ on both sides

[thorny flame]

south

[smoky sparrow]

it's a consequence of how ratios are defined

if h:s = 24:5, reverse the order, s:h = 5:24

[vivid saffron]

oh, that makes sense so if x / y = 1 / 2, y / x = 2 /1

[smoky sparrow]

yeah

[vivid saffron]

alright, i think i can continue from here, lemme give it a try

[topaz sinew]

@vivid saffron Has your question been resolved?

[vivid saffron]

actually i don't know, h / f = 16 / 1, so f = 1 / 16 * h

f = 24s / 80

my assumption that the third one the the biggest and the fourth one is the smallest is correct?

or it is not possible to know yet

[smoky sparrow]

that is indeed correct

5/24 * 12 = 5/2 so larger than original

and the second and fourth are smaller than the original

basically what you do is to reverse the ratios

so 10 * 1/12 * 24/5 is the original height

[vivid saffron]

what i know is that
h = 24/5 * s
s = 5/24 * h
t = 5/24 * hs
f = 24/80 * s

[smoky sparrow]

yeah I have no idea how you did it this way but 0.25 is correct

[vivid saffron]

lol

[vivid saffron]

alright then, thank you

have a great day

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[smoky sparrow]

you too!

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[sullen scarab]

this is the expression how to do this one?

my idea was to break the -1 to -1/3 and give to the trio but what after that?

[sinful flint]

!status

[topaz sinew]

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

[sullen scarab]

maybe 2

[sinful flint]

!show

[topaz sinew]

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

[sullen scarab]

something like this

@sinful flint

[sinful flint]

and how would that help?

[sullen scarab]

well i did b'coz i though some sort of symmetery will appear

[sinful flint]

try making all the fractions' denominators equal

[sullen scarab]

and also -1 was tempting to break to -1/3

[sullen scarab]

give me a moment

something like this?

[sinful flint]

,rotate

[sullen scarab]

@sinful flint

i forget to do - 1

[sinful flint]

no

[sullen scarab]

what's wrong?

[sinful flint]

remember that $(c-a)=-(a-c)$

[thorny flame]

Bonk

[sullen scarab]

yeah

i did that

in the second one

[sinful flint]

but why did you add (a-c) in the first one?

you already have (c-a)

[sullen scarab]

in the numerator?

[sinful flint]

denominator

if all the denominators are equal, then you can simply add the numerators

[sullen scarab]

yes i multiplied a-b

is the expression wrong?

[strange whale]

You've seen lagrange interpolation or no, neon ?

[sullen scarab]

no

whats that?

[strange whale]

It's fine, it's a way to make a polynomial which passes through points you specify

These (x-a)(x-b)/(c-a)(c-b) are super reminiscent of that

You should try looking at what happens to the formula for x=a x=b and x=c if you want a big hint

[sullen scarab]

individual should i check na?

well it's zero for each case

[strange whale]

Yeah

[sullen scarab]

so it means it have three roots as