help-28

[steel mist]

f(a) = f(-1)

[next hawk]

go find that tangent equation now

[steel mist]

so u go back to ur f(x) and plug a -1

$f(x) = -x^2 + 4x \implies f(-1) = -(-1)^2 + 4(-1)$

[glossy valve]

Aza

[steel mist]

= -5

so f(-1) = -5

f'(-1)?

we got f'(x)

$f'(x) = -2x +4 \implies f'(-1) = -2(-1)+4$

[glossy valve]

Aza

[steel mist]

= 6

x is x, as in y = mx+n

that x represent ur input

and a = -1

so we got the line

[glad pine]

so to get the slop you take the x and you take the derivative and do f'(x) and that gives you the slope?

[steel mist]

$y - (-5) = 6 (x - (-1))$

this is ur tangent

[glossy valve]

Aza

[steel mist]

$y+5 = 6(x+1)$

[glossy valve]

Aza

[glad pine]

and subtract 5 from both sides

y=6x+1

[steel mist]

thats up to u

[glad pine]

ok thank you

[steel mist]

but that is the equation of the tangent line to ur f(x) at the point (-1, -5)

and actually, ive just realised

ofc f(a) = -5

XD

because the point is -1, -5

[glad pine]

yes lol

[steel mist]

and f(a) is the y coordinate of the point

indeed -5

so now u do the same with the other exercise

[glad pine]

ok

[steel mist]

is easy once u do it a few times

probably u will learn inmediate derivatives soon

so u safe the time doing the limit

but is rlly good to do it that way at least a few times

[glad pine]

I heard of a way to convert to derivative form by doing x^e turns into e^x-1

is that true?

[steel mist]

then unheard that

and stick with what u know so far

[glad pine]

ok

[steel mist]

i mean, u can try tho

nothing stops u from doing

[glad pine]

so to get the derivative

[steel mist]

$f'(x) = \lim_{h\to0} \frac{e^{x+h}-e^x}{h}$

u can do this, and see what u get

[glad pine]

you used this formula?

[glossy valve]

Aza

[steel mist]

this is not a formula

this is the definition of the derivative

and it comes from the slope (the triangle i showed u)

[glad pine]

ohh

thank you

[steel mist]

$f'(x) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$

[glossy valve]

Aza

[steel mist]

now if ur f(x) = e^x

well, just replace

[steel mist]

u will get this

and u can try to solve this limit to find that is the derivative of e^x

and u will see if what u heard is truth or not

$f'(x) = \lim_{h\to0} \frac{f(x+h)-f(x)}{x+h-x}$

[glossy valve]

Aza

[steel mist]

maybe this way u see better the relation between the triangle and the derivative

[glad pine]

thank you for the help

[steel mist]

x+h-x is just h

[glad pine]

this makes a lot more sense now

[glad pine]

I have to go now, but thank you for everything

[steel mist]

it is literally this

exactly this

[glad pine]

I have saved the pictures

thank you for your help!

have a good day

or night

.close

[full forum]

Channel closed

Closed by @glad pine

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[steel mist]

see ya, good luck

[glad pine]

thank you

[full forum]

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[brave heath]

HELP

ME

[full forum]

Please don't occupy multiple help channels.

[full forum]

Please don't occupy multiple help channels.

[brave heath]

hi

?

HELLO

[quaint prawn]

Stick to one channel @brave heath I closed your other one

[brave heath]

ok

help me with this 1 thing

in life

[upper anvil]

is that for an assessment 💀

[brave heath]

YES

EXACTLY

HELP ME NOW

[quaint prawn]

<@&268886789983436800>

Test

[brave heath]

ok

it was a joke

its homework

[quaint prawn]

It clearly isn’t

[brave heath]

who has a test on 11:48

pm

WDYM?

ahem

[quaint prawn]

No calculators allowed, there’s a test code at the top, and you said it’s a test

[brave heath]

very real test

[upper anvil]

have you tried watching the vido

[quaint prawn]

This has your full name in it

[brave heath]

its a bookmark thing

[quaint prawn]

You probably don’t want to share that

[brave heath]

shi

not even my account..

same thing

can u help me plz

[upper anvil]

the coverup is crazy

you should check the video provided first

[brave heath]

i did

i did 4 times

it doesnt make sence

[brave heath]

very real test tho

[gritty spindle]

This is not helping your case

[calm nacelle]

It doesn't seem like it's a test, but you're not gonna get help if you just keep spamming "help"

[brave heath]

huh

then

can you plz help me

last question in life

then i can go to bed

plz 🙂

[nimble crane]

can you describe what you're confused about?

[brave heath]

ok

so basically

its (,)

rigt

right

then when i do it

the input is wrong

[quaint prawn]

What

[brave heath]

and then i watch the video

yea exactly

its confusing

[nimble crane]

no

what are you saying

you're being incoherent rn

[quaint prawn]

“It” isn’t confusing, you are being confusing

[nimble crane]

what does this mean?

[nimble crane]

or this?

be a bit clearer with your words

[brave heath]

its like this

[quaint prawn]

Yeah? That’s asking for the answer

[brave heath]

i just need help

[nimble crane]

yes, but what is your issue with the question?

[quaint prawn]

That’s where you put the ordered pair that makes it a rectangle

[brave heath]

ok so i do not understand VERTICES

very confusing

[quaint prawn]

That’s not much of a question

[brave heath]

and then when i put the 4th point -5 -3

it says its WRONG

[quaint prawn]

Because that doesn’t make a rectangle

[brave heath]

it odes

does

what does then

[nimble crane]

it doesnt...

[brave heath]

but it should!

it has the math on it

its shaped

exactly

like a triangle

[nimble crane]

???

[brave heath]

i mean

[quaint prawn]

Rectangles aren’t triangles

[brave heath]

rectangle

[quaint prawn]

Are you trolling

[brave heath]

no

plz

acctually dude

[brave heath]

e

[stone aspen]

please be respectful when asking for help

[brave heath]

i said please like 5 times....

[brave heath]

😮

[quaint prawn]

Do you think this is a rectangle?

[brave heath]

ty

[nimble crane]

ah austin beat me to it

[brave heath]

-5 -3

thats exactly what i wrote

yea

[quaint prawn]

Your drawing is better

How does that look like a rectangle at all??????

[brave heath]

look

https://tenor.com/view/box-highlight-kyocera-rectangle-shape-gif-17757040

like that

but upwaards

[quaint prawn]

That looks nothing like the picture we just sent

[brave heath]

man im just confused now :

[nimble crane]

(-5, -3) is the bottom left vertex of the shape we drew

[brave heath]

YEA

thats exactly

what i said

[nimble crane]

does that point complete the rectangle?

[brave heath]

and it said its wrong

yea

[quaint prawn]

No it doesn’t

[brave heath]

bro plz help

[quaint prawn]

It doesn’t make a rectangle

[brave heath]

a poor soul

but it does look

[quaint prawn]

Do you know what a rectangle is?

[brave heath]

e

yea

its like 4 dots

then just connect them

to make a rectangle

[quaint prawn]

That’s not what a rectangle is

[nimble crane]

HOW IS THAT A RECTANGLE

[brave heath]

https://tenor.com/view/math-geometry-polyhedra-prism-quadrangular-prism-gif-18002691

this?

uh

[quaint prawn]

A rectangle has 4 sides, and every opposite side has even length. The angles inside a rectangle are all right angles

[brave heath]

https://tenor.com/view/animation-rectangle-gif-24785950

[brave heath]

but one of the lines

or dots they did

is different

from the other side

[nimble crane]

who is they??

[brave heath]

the people who made this thing

[quaint prawn]

Are you on your phone right now?

[brave heath]

no

macbook

[quaint prawn]

Imagine you had your phone

[brave heath]

ok

[quaint prawn]

It’s a rectangle

Tilt it

[brave heath]

omg

yea

[quaint prawn]

One corner of your phone will be higher up

[calm nacelle]

Vegito are you older than 13

[quaint prawn]

But it’s still a rectangle

[brave heath]

i am 13

[calm nacelle]

Okay

[brave heath]

im not good with maths ok

[brave heath]

hey can u help?

omg its the next day...

00:00 😮

...

<@&286206848099549185>

[stone aspen]

Do you want help or an answer

[brave heath]

help

[stone aspen]

Alright, so what can you conclude from the given points and austin’s information about tilts

[brave heath]

that its a rectangle

thing

[stone aspen]

a rectangle is still a rectangle even when tilted

[brave heath]

and tilt my phone

[stone aspen]

correct?

[brave heath]

i guess

[stone aspen]

yes not compare that to your graph

[brave heath]

but one of the dots

or crosses

is above

the thing

so its not the same

righ

right?

[stone aspen]

the tilt only changes the view aka the points

[brave heath]

E

no no

[stone aspen]

yes

[brave heath]

look at the points

[stone aspen]

hence a tilt

[brave heath]

one is different

from all others

eh?

[stone aspen]

yeah, okay now define a rectangle

[brave heath]

something thats equal

on all 4 sides

[stone aspen]

not all

[brave heath]

but one of the points IS not equal

[stone aspen]

thats a square

[brave heath]

what

no

a rectangle

has bigger

or longer

you could say

square

[stone aspen]

a square is a special kind of rectangle that lies in the same definition

[brave heath]

k

[stone aspen]

think of it like this, only one side of the shape is shifted (tilt)

[stone aspen]

and if a rectangle is shifted, then the one point moves with the other in correlation

[brave heath]

uhmm

im not following

[stone aspen]

the missing point is the same as the other but flipped.

[quaint prawn]

If it looked like this

Could you make it a rectangle?

[brave heath]

maybe

could be a triangle

[quaint prawn]

(Notice that all I did, was tilt the image)

[quaint prawn]

The whole entire point is to add a fourth vertex to make it a rectangle.

So why are you saying that?

[brave heath]

u asked

tho

[quaint prawn]

I didn’t

[brave heath]

e

[quaint prawn]

I asked if you could make it a rectangle

[brave heath]

no

thats 3

not

4

[nimble crane]

.

[stone aspen]

oh my

[quaint prawn]

There’s no way you’re being serious

[nimble crane]

Austin is asking you if you can complete the rectangle

[brave heath]

yea

-5 -3

[nimble crane]

not if those three points constitute a rectangle

[brave heath]

but it says its wrong .

[nimble crane]

no way man

[quaint prawn]

It is wrong

[brave heath]

-4 -3?

[nimble crane]

are you still doing this

[brave heath]

i 1 upped it

like the other one

[nimble crane]

why??

[quaint prawn]

You need to like ask your teacher or parents for help with this

[brave heath]

:

[quaint prawn]

I don’t think anyone here is going to be able to help you

[brave heath]

its 12 am...

i started at 9...

[quaint prawn]

Go watch the video again then

[brave heath]

i did

it says right angle

[nimble crane]

then why are you still trying -5 -3

[quaint prawn]

https://youtu.be/itgj03p5igM?si=gTOJkMCdsIhoDJkA

Here’s a different video you can watch

[nimble crane]

and why are you trying - 4 -3

[brave heath]

shoot

i got it worng

they gave me a whole new question

[quaint prawn]

That’s the exact same type of question

[nimble crane]

^^

[brave heath]

no its different

[nimble crane]

no its not

its the same type of question

[brave heath]

...

who knows the coords

[stone aspen]

different question, same type of question. the issue is still the same

[quaint prawn]

!noans

[full forum]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[brave heath]

!noans

[full forum]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[quaint prawn]

What?

[brave heath]

idek

i tried ur thing

!help

[quaint prawn]

Are you still trying to solve the problem or have you just given up and messing around now?

[brave heath]

im tryna solve it

OM

i got it i think

-4 positive 3

and im wrong

[stone aspen]

…?

[brave heath]

god dang it

[nimble crane]

how did you get that

[brave heath]

becuase of the thing

but it reset

im so done :

!close

?

!done

[full forum]

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

[brave heath]

.close

[full forum]

Channel closed

Closed by @brave heath

Use .reopen if this was a mistake.

[brave heath]

i got it thanks for your help

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[tight peak]

I want to find the maximum error of approximation program but all input (64bit integer) is too big to bruteforce.
Since I make 2 32bit block out of it and multiply them together at the end, calculating max error in each block and somehow calculate the total max error is enough.

The question is, what should the maximum error be of a*b when each has 2bits of error at max? Is it simply 2+2? How can I describe how I calculated it?

[tight peak]

"How I calculated it" as in I need to sort of prove it's accurate enough to users in README or an article and alike

[full forum]

@tight peak Has your question been resolved?

[full forum]

@tight peak Has your question been resolved?

[tight peak]

<@&286206848099549185> hi :)

[full forum]

@tight peak Has your question been resolved?

[full forum]

@tight peak Has your question been resolved?

[full forum]

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[thorny spire]

how am I suppose to solve this

[random coral]

do you have the formular for revolving around x axis?

[thorny spire]

Is this right?

[random coral]

isnt that the one for finding the length of the curve of the function?

if you go search on google

[thorny spire]

I think it's for surface area

length of a curve would just be the square root

I think

[random coral]

well your original seems correctly formulated then, what are you stuck on?

[thorny spire]

how to solve it, I don't know where to start, it's intimidating

[full forum]

@thorny spire Has your question been resolved?

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[willow mauve]

How to do 35?

[undone vector]

What are we looking at

[gritty rose]

What's the question

[full forum]

@willow mauve Has your question been resolved?

[willow mauve]

a circle has a radius 12m, sector area of 216cm, find length of sector arc

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[torn jolt]

why is a≠0, b≠0 denoted by a²+b²≠0?

[spiral vigil]

those aren't the same thing

no... a=0, b=1
a^2 + b^2 = 1

[torn jolt]

right okay

the textbook was wrong

.close

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[spiral vigil]

it does mean that a≠0 or b≠0 though for what it's worth

[torn jolt]

so it applies in some cases?

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[full forum]

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[zenith forge]

f(x) = x^3 - 2x^2 + 4x. Write the new equation when it undergoes Reflection across both X and Y axis, Horizontal dilation by 2 and horizontal translation by 1

[glacial pasture]

had a go?

[zenith forge]

-(-0.5(x-1))^3 - 2(-0.5(x-1))^2 + 4(-0.5(x-1)) Is this right?

I dont really understand the Hori dilation with horizontal shifting part

[glacial pasture]

lets see
(x-1)^3-2(x-1)^2+4(x-1)
(0.5(x-1))^3-2(0.5(x-1))^2+4(0.5(x-1))
(-0.5(x-1))^3-2(-0.5(x-1))^2+4(-0.5(x-1))
-[(-0.5(x-1))^3-2(-0.5(x-1))^2+4(-0.5(x-1))]

you just seem to be missing a bracket

[zenith forge]

Oh nice!

[glacial pasture]

or two

[zenith forge]

I just didnt know whether the -0.5 needed to be inside with the x-1

[glacial pasture]

you can distribute it if you want, but in the order we do it it has to act on the entire x-1

[zenith forge]

Isnt it always R then D then T

why did u translate first?

[glacial pasture]

it can get messy
say i have 2x+5
if i do my y reflection first
2(-x)+5
then i do a horizontal translation right by 2 units
2(-x-2)+5 is what i technically get

which wouldnt be what i wanted

[zenith forge]

my teacher taught me RDT method

[glacial pasture]

i tend to go:
T x
D x
R over y axis
R over x axis
D y
T y

but if RDT works for you, feel free

the only thing you had wrong was two missing brackets

your x axis reflection was only applied to the first term

[zenith forge]

Alright ty

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[heady idol]

I'm having some trouble with this question, if someone could go through it with me I'd be very grateful

[heady idol]

This is what I have so far

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[polar valve]

stop spampinging.

[full forum]

@heady idol Has your question been resolved?

[heady idol]

<@&286206848099549185>

[polar valve]

stop spampinging.

[heady idol]

• After 15 minutes, feel free to ping <@&286206848099549185>.

[polar valve]

4 pings in 30 minutes?

!15min

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[polar valve]

ping once!!!

[heady idol]

roger that. got time to look at the question?

(it was 4 in 45, but i agree that's excessive)

[heady idol]

.close

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[dim pumice]

The line L: (6,4,0) + t(1,2,-1) intersect the plane x+y+2z= 5 at the point P. Enter the line L1 that passes through P, lies in the plane and is perpendicular to L.

[dim pumice]

So n = (1,1,2)

Do I have to find the point P before I can find the line L1?

[full forum]

@dim pumice Has your question been resolved?

[dim pumice]

<@&286206848099549185>

I tried finding the point P and got the answer P = (1, -6, 5) but I don’t know if that is correct

How do I proceed now to find the line L1?

[full forum]

@dim pumice Has your question been resolved?

[worn linden]

So to make a sanity check if P correctly is on the plane, you could plug in P into the plane equation

$1 + -6 + 2*5=5$ which is correct

[glossy valve]

TRAMPELTIER

[dim pumice]

Okay great!

[worn linden]

Next step: If we know a plane and a point P on the plane, then intuitively that should be enough to idenitfy the perpendicular line L to that plane

Makes sense right

Now we just need to see how we can do that

[dim pumice]

Yes okay!

[worn linden]

So we have a plane, that is basically also the same as having 2 vectors that span the plane right?

[dim pumice]

Yes

[worn linden]

Now if we choose two vectors that are perpendicular to the other one this plane, then we only need to calculate the vector that is perpendicular to both of them, and this third vector will basically give us the line

[dim pumice]

Oh okay!

[worn linden]

Now I need to think for myself how to do that

So first step would be to select another point P1 on the plane. Then we can take $P1-P$ being one vector that "spans" the plane

[glossy valve]

TRAMPELTIER

[worn linden]

Since we know basically there is for each (x, y) pair in the space a point on the plane, we can just choose an arbitrary (x, y) pair, e.g. (1, 1).
Now we plug it into the plane equation to find z , which gives $z=(5-2)/2 = 1.5$. Then we'd have P1 = (1, 1, 1.5) and we can say our first "spanning vector" is $P1-P$ = (0, 7, -3.5).

[glossy valve]

TRAMPELTIER

[dim pumice]

Oh okay

[worn linden]

oh actually something is wrong with my calculation because the point is not on the plane if i plug it into the equation

[dim pumice]

Oh no

[worn linden]

oh wait, it's correct

I was just plugging in P1-P, which is not a point, but a vector on the plane

[dim pumice]

Okey good

[worn linden]

But if I plug in P1, then its correct

Okay. now we have P1 and P on the plane and P1-P is a vector that is parallel to the plane right

[dim pumice]

Right!

[worn linden]

Now we need another such parallel vector, but this vector should also be perpendicular to P1-P

Make sense right

[dim pumice]

Yes!

[worn linden]

To do that, we can make use of the dot product (do you know how dot (or scalar) product works?

[dim pumice]

Yes i do!

[worn linden]

alright, so if you have a vector A and a vector B, which has unit length, then the dot product between those will project A onto the vector B and give the length of that projection

[dim pumice]

Yes okay

[worn linden]

Actually not sure if we need that, hmm let me think

actually maybe this is not needed and all we need to do is find another point P2 on the plane, that is not P1. Then we'd have P1-P and P2-P parallel to the plane. And if we then calculate the perpendicular vector to both of those, we've got the line pretty much

[dim pumice]

Okay that makes sense

[worn linden]

Now, assuming we have chosen some P2 (same method as P1, just different x, y selected and found a different z; you can do that yourself later),
do you know how to create this third vector that is perpendicular to the P1-P and P2-P ?

Maybe you've learned about an operation between two vectors that can give us this

[dim pumice]

Hmm I might know but am not sure

Can I select any x and y?

[worn linden]

Yea any that is not same x, y as P1 or P

[dim pumice]

Okay

[worn linden]

Okay. So basically, the cross product gives us the perpendicular vector taht we want

did you learn about it?

[dim pumice]

Yes I did I got p2 = (2,2,0.5)

[worn linden]

okay

[dim pumice]

So now I do cross product on p1 and p2?

[worn linden]

no, but on P1-P and P2-P

[dim pumice]

Oh okay

[worn linden]

And then you have a vector V perpendicular to P1-P and P2-P.
Now, your line is basically given by first going to P and then move along multiples of V

so it's $P + t*V$

[dim pumice]

Okay I got (17.5, -3.5, -7) as the cross product

[glossy valve]

TRAMPELTIER

[dim pumice]

Okay

So v is (17.5, -3.5, -7)?

[worn linden]

yea

(or it can be any multiple of it, i.e. any scaling by some real number)

[dim pumice]

Okay so (1,-6,5) + t(17.5, -3.5, -7)

[worn linden]

yes

[dim pumice]

Okay so that is L1?

Can we check now if that is perpendicular to L?

[worn linden]

ehm Okay I misunderstood the task 😄

I mean we calculated L1 to be perpendicular to the plane, not to L

[dim pumice]

Oh no

[worn linden]

Oh I can't read 😄 also L1 need to lie in the plane

ehm

[dim pumice]

Oh yes sorry if I worded it weird

We don’t need to start completely over tho we still know that p is correct

[worn linden]

no problem not your fault

I need the task here again: "The line L: (6,4,0) + t(1,2,-1) intersect the plane x+y+2z= 5 at the point P. Enter the line L1 that passes through P, lies in the plane and is perpendicular to L."

[dim pumice]

Yes of course

It is totally fine if you don’t want to redo the entire question I can just try again tomorrow

[worn linden]

I'm having an idea but need a few minutes

[dim pumice]

okay thats fine!

[worn linden]

Okay forget everything from previous misunderstanding. delete all variables in the head

[dim pumice]

okay done!

[worn linden]

except P

[dim pumice]

yes okay

[worn linden]

P is correct. now we need to find $L_1 = P + t \cdot V$, that means we need ot find $V$. And V should be on the plane and perpendicular to the vector (1, 2, -1)

[glossy valve]

TRAMPELTIER

[dim pumice]

okay!

[worn linden]

Now we have 2 constraints for V:

1. be on the plane
2. perpendicular to (1, 2, -1)

is that clear why?

[dim pumice]

yes it is!

[worn linden]

okay

If two vectors are perpendicular, their dot product must be 0. With that, the 2nd constraints tells us $<(1, 2, -1), V> = 0$ which means $1\cdot V_x + 2 \cdot V_y -1 \dot V_z = 0$

[glossy valve]

TRAMPELTIER

[dim pumice]

yes!

[worn linden]

Now to the first constraint: For this, we plug in $x = P_x + V_x, y = P_y + V_y, z = P_z + V_z$ into the plane equation

[glossy valve]

TRAMPELTIER

[dim pumice]

okay

[worn linden]

Now the problem is we have 2 constraints but 3 unknowns.

[dim pumice]

yes we do

[worn linden]

But! It's also okay to scale V by any arbitrary scalar and use that V instead, i.e. the scale of V doesn't matter

Intuitively, that's reducing one dimension

[dim pumice]

oh okay

[worn linden]

Now, with this, we can (like in the misunderstood derivation), choose an X-coordinate for our V

an arbitrary one. But since it's not a plane, but a line, we can't choose also a Y coordinate. We can only choose an X coordinate.

[dim pumice]

okay!

[worn linden]

So, basically, we can choose our $x = P_x + V_x$ arbirarily. Let's say $x=1$

[glossy valve]

TRAMPELTIER

[dim pumice]

yes okay

[worn linden]

Now, frmo x=1, we can get $V_x = 1 - P_x$.

[glossy valve]

TRAMPELTIER

[worn linden]

We can plug that into our 2 constraints. Then, we only have two unknowns in the two constraints: $V_y$ and $V_z$

[dim pumice]

yes!

[glossy valve]

TRAMPELTIER

[worn linden]

So basically, then we only need to solve for the unknowns doing standard linear system solving

[dim pumice]

yes okay but is vx 0 then?

[worn linden]

yea it happens to be that

[dim pumice]

okay!

[worn linden]

Now after solving that system, we got V

then the line should be $L_1 = P + t \cdot V$

[glossy valve]

TRAMPELTIER

[worn linden]

(or any scalar multiple of V in place of that. To create a unique solution (which might be asked in the course you're taking), maybe we should normalize $V$ such that the length of $V$ is 1)

[glossy valve]

TRAMPELTIER

[dim pumice]

yes okay!

but vy and vz could be any values?

like vy = 4 and vz = 2?

[worn linden]

I mean , they're basically fixed by the constraints once you chose v_x

[dim pumice]

okay but is (0,4,2) an answer for v?

[worn linden]

If oyu plug it into both connstraints, they should be satisfied

I think the perpendicularity constraint is not satisfied with this v

[dim pumice]

oh okay then that is wrong

i mean my answer

okay hold on

okay i meant -4 and 2

vy = -4 and vz = 2

then it is satisfied

[worn linden]

$1 * 0 + 2*(-4) + (-1) * 2) = -10 \neq 0$

[glossy valve]

TRAMPELTIER

[worn linden]

Something's wrong

[dim pumice]

oh i thought it was supposed to equal 5

[worn linden]

that's the other constraint

[dim pumice]

oh okay

i dont really know what it supposed to be

[worn linden]

I'm currently trying myself

[dim pumice]

what about (0,2,-1)

or does it have to equal 0 exactly?

[worn linden]

4 +1 =5 != 0

[dim pumice]

I’m so sorry but I really have to go now but I will continue tomorrow, thank you so much for the help

.close

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[worn linden]

Alright. Here I finished my solution with all the steps we took @dim pumice

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[atomic barn]

ive crossed multiplied 10/38 by 26/x

and ive tried 10/36 = 16/x

but neither of those were correct

i dont know what to do

[harsh pilot]

,calc (26/10)^2*38

[glossy valve]

Result:

256.88

[harsh pilot]

Since triangle LSP and triangle LRN are similar, you know that the ratio between corresponding sides will be the same number. You are already given that

LN/LP = 26/10, so you know also that RN/SP = 26/10 and RL/SL = 26/10.

You are also given the area of the small triangle LSP, which is 38. This is half base times height, so

0.5*LS*SP = 38

The area of the large triangle LRN is

0.5*LR*RN,

and this is what you want to find.

[atomic barn]

ooooooh

[harsh pilot]

But you know that LR = 26/10 * LS, and you know that RN = 26/10 * SP. You can plug these expressions into the area formula for the large triangle. You will get

0.5*26/10 * LS*26/10 * SP = 26/10 * 26/10 * 0.5*LS*SP

[atomic barn]

okay that makes os much more sense

thank you !

[harsh pilot]

Nice

np

[atomic barn]

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[hasty kraken]

May someone help validate (or maybe invalidate) my answer

[worn linden]

can't read it's too small

[hasty kraken]

You can zoom in?

I personally find it readable when you zoom in

[worn linden]

how does one zoom into picture in discord

[full forum]

@hasty kraken Has your question been resolved?

[hasty kraken]

Just press on the picture if you're on mobile. If not maybe download it

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[heavy jetty]

Is there a formula used to find x?

[spare nacelle]

x = log(7.776)/log(6)
x*log(6) = log(7.776)
log(6^x) = log(7.776)
6^x = 7776 = 6^5
x = 5

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[heavy jetty]

.close

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[ruby wolf]

10.55

[ruby wolf]

What does it mean by finding the roots

If the equation

[gritty rose]

equation=0

[ruby wolf]

Oh wut?

[gritty rose]

Solve for x

[ruby wolf]

Finding all the roots just means when the equation is 0?

[gritty rose]

Yes

[ruby wolf]

Uhh alr then

Ic

[ruby wolf]

I think I did something bad

[gritty rose]

?

[ruby wolf]

Oh wait no I figured out what I did lol I just need to hopefully solve it

[gritty rose]

Why'd you ping me just to tell me thay

[ruby wolf]

Uhh

I usually respond

And it pings

It's a habit srry

[ruby wolf]

Alright I solved it hopefully this is a useful ping 😭

[gritty rose]

,w solve (2x-1)(8x+4) + (2x+1)(9x+1)+(9x+1)(17x-3)=0

[glossy valve]

Results provided by WolframAlpha

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

Requested by rie.mann

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[glad whale]

Hi

[glad whale]

I got D here

I just solved to be honest

Q-4 times q-4

and nthen just went along

nice ok

.close

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[winged brook]

given ABC is a straight line and tan x is 3/4. i need to find AC in cm without using solving triangle formula

[hot herald]

wdym by "solving triangle formula"

[winged brook]

this one

[hot herald]

nothing is being solved there

that's just conventional labelling/naming angles/sides of a triangle

[urban condor]

You mean not finding via angles or what?

[winged brook]

i need to find the length of AC

[hot herald]

we are aware of the end goal

the issue is its not entirely clear to us what you mean by

without using solving triangle formula
so we don't know what we're allowed to use

[urban condor]

Wait

do you mean the sine and cosine rules?

@winged brook You there?

[winged brook]

yeah

[winged brook]

yes sorry i called it solving triangle 😭

[winged brook]

i need to find AC but i cant use this formula

[spiral vigil]

those wouldn't even really help you here. what can you do with the information that tan x = 3/4?

,tex .sohcahtoa

[glossy valve]

hayley!

[winged brook]

yeah and i went and calculate it and i get 36'52'

the degree for x

[hot herald]

do not make any attempt to determine x

[winged brook]

oh?

[spiral vigil]

yeah it doesn't tell you anything useful. tangent is a ratio between two side lengths. Which side lengths?

[hot herald]

you "can", but there's not much point

[winged brook]

ohh

so...

12.2422cm

kinda logic

[hot herald]

how are you getting that?

[winged brook]

tan-1 (3/4) = 8/x? since we dont know the adjacent

[hot herald]

no

[winged brook]

and then i went and calculate the cos

[urban condor]

x is AC not BC

[winged brook]

yeah so like opposite/adjacent = 8/x right?

[hot herald]

no

still no

[winged brook]

oh

[urban condor]

nope

[hot herald]

tan(angle) = opp/adj

[winged brook]

omg i mix my pre calculus knowledge with basic math gurl😭

[hot herald]

what is the side opposite x in your diagram

[winged brook]

they dont give it

[hot herald]

whats an expression for the side opposite x in your diagram

[urban condor]

assume it to be some variable

[hot herald]

you can name the side using the given points

[urban condor]

👍

[winged brook]

ABC is a straight line and tan 3/4 is x. thats what they give and from what i can conclude is AD = hypotenuses, AC is adjacent and CD is opposite

[urban condor]

nah

[hot herald]

focus only on this triangle

[winged brook]

wait shit i just realized i could seperate it

[winged brook]

yeah omg

wait i think i know hold

err 5.976cm?

[hot herald]

how are you getting that

[winged brook]

tan 3/4 = x/8 since i want to find BC

[hot herald]

no

[winged brook]

i used a calculator

[hot herald]

you shouldn't be using tan on the ratio

tan(angle) = opp/adj

your angle here is x

[winged brook]

so i have to change it to degree and minutes?

[hot herald]

no

do not make any attempt to find the angle

apply ONLY this defintion of tan and nothing else

tan(angle) = opp/adj

what is the side opposite x (i don't care about the numerical value, just give me the name/expression for the side)

[winged brook]

straight line?

[hot herald]

not the answer i'm looking for

[winged brook]

90 degree?

[hot herald]

no

you've already demonstrated that you're able to name/describe sides

just do that for what i'm asking for

[winged brook]

ohh is it the triangle that has the same length?

[hot herald]

do you know what opposite , adjacent means

[winged brook]

yeah

[hot herald]

don't overthink what i'm asking

what's the side opposite the angle x in the diagram

[winged brook]

lemme contemplate hold on

[hot herald]

there are three sides,
(relative to the angle x)
one side is opposite
another is adjacent
and the other is the hypotenuse

are you able to identify which is which

[winged brook]

yeah

[hot herald]

so answer

what's the side opposite the angle x in the diagram

[winged brook]

uhh CD

?

[hot herald]

no

what's your definition of opposite in this context?

[winged brook]

so in the small triangle the side opposite of x is BC

[hot herald]

yes

[winged brook]

but the big one is CD

ohh i thought ur asking like the opposite like literally

[hot herald]

i was...

doesn't matter

the answer will be the same

[winged brook]

yeah

so opp/adj?

[hot herald]

english definition, mathematical definition, whatever reasonalbe definition

the side opposite x here is BC

[winged brook]

yeah

[hot herald]

and what's the side adjacne to x?

[winged brook]

CD

[hot herald]

note that x is an acute angle in the smaller triangle i cropped
so you don't really care about the bigger triangle when considering tan(x)

yes

[winged brook]

okay

[hot herald]

now go back to
tan(x) =** 3/4 = opp/adj**

you've identified expressions for the opp and adjacent
and you are given one of the lenghts

use those in your equation, focussing on the bolded part

[winged brook]

so... 3/4 = x/8 ?

[hot herald]

no

what's the opp

[winged brook]

x

[hot herald]

no

[winged brook]

bc

[hot herald]

capitals but yes

[winged brook]

oh sorry i meant the x is bc so like 3/4 = BC/8 ?

[hot herald]

you shouldn't be using already extablished variables

x is already representing your angle

you shouldn't be using x again to represent something else

[winged brook]

sorry i cant help😭 cuz when i want to find something i let it to be called x

[hot herald]

well don't get hung up on that

[winged brook]

okay ill change

[hot herald]

x isn't the only thing that can be used as a variable

[winged brook]

yeah

[hot herald]

it wasn't clear to be, so every time you used x like that before,
i was interpretting that as you trying to divide an angle by side length

which yielded an immediate no

3/4 = BC/8
anyway, solve that to determine BC

[winged brook]

oh no no i just let the the value called x no wonder we're stucked bcz of the expression😭

6

[hot herald]

yes, do you think you can do the rest from here?

[winged brook]

yeah

bro thanks my god really appreciate it

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[sharp helm]

really?

[sharp thorn]

wdym by solving it

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[eager holly]

I have to take the derivative of the mean to prove r= mean = 1/n * sum(yi)

My only question here is where does this "2*" go to from the 2 red marked spots

[eager holly]

that does not seem like its even the same equation but this is the solution to the assignment we got

actually nvm i think i understand but i need clarification

if i have a summation (the big E thing) and i divide the entire equation by 2, does it affect whats inside the summation or no?

[gritty rose]

If 2 * something= 0 then something=0

They divided both sides by 2

[eager holly]

Yeah my question is kinda like this give me a second

or is that effectively the same thing? i am bad at math apologies

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[lilac sun]

I need help

[lilac sun]

would it be wrong if I asked for help again?

[brave crater]

no it won't be wrong

shoot

[devout valley]

It would be wrong if you opened a new help channel without posting your question in the first post, considering the very first bullet point here

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@lilac sun Has your question been resolved?

[gritty rose]

Come back when you have a question

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@polar bone Has your question been resolved?

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[lofty heath]

could someone help me solve for s in question 7?

i don’t understand how to solve for it

[nimble crane]

Have you found the slope of PR?

[lofty heath]

no

[nimble crane]

yeah do that first

[lofty heath]

okk

[nimble crane]

Then use the fact that PR and PQ are perpendicular

[lofty heath]

it’s 1/4

[nimble crane]

Are you sure?

[lofty heath]

well rise is 1 and run is 4

[nimble crane]

rise is 1?

look carefully again

[lofty heath]

i see 1

[nimble crane]

R is below P on the y-axis

[lofty heath]

so its -1?

[nimble crane]

rise is -1 yeah

[lofty heath]

-1/4

[nimble crane]

So the slope of PR is -1/4

Yeah

Do you know what the relationship is between perpendicular slopes?

[lofty heath]

uhhh no

[nimble crane]

negative reciprocals? Have you seen that phrase

[lofty heath]

I know what reciprocal is

[lofty heath]

But I haven’t heard that phrase

[nimble crane]

yeah so

If you have a slope of a line, then the slope of perpendicular line to that line is the negative reciprocal

for example, if you have a slope of 1/2, then the negative reciprocal of that will be -2/1 = -2

[lofty heath]

ohh

do I have to find the slope of PQ?

[nimble crane]

Yes, but now you know what it is already

[nimble crane]

.

[lofty heath]

so it’s 1/4

[nimble crane]

No

[lofty heath]

wait

4

[nimble crane]

That’s the negative

[nimble crane]

There you go

So now you have a slope and a point

You are actually given the y-intercept for PQ, which is quite nice

[lofty heath]

oh it’s 14

[nimble crane]

Using this information, you can find what the y value is when x=2

[lofty heath]

oh its not 14

ok hold on

no it si 14 tho

but 14 isn’t an option

[nimble crane]

Show how you got that

[lofty heath]

4(2)+6

8+6

is 14

[nimble crane]

+6?

Why 6?

Is 6 the y-intercept of PQ?

[lofty heath]

yea

[nimble crane]

look again at your diagram

P is the y-intercept

[lofty heath]

ohh

so it’s 4(2)+2 is 10

Tysm

.close

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[thorny prawn]

woudln't this be 0.752

0.572 sorry*

would it be like 0.12 * 30 / 2 pi

[onyx forum]

yes

(use dimensional analysis if you arent already)

[thorny prawn]

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[sinful bloom]

Need help understanding this problem. Thanks

[sinful bloom]

It’s asking for the volume above h

Am I right that it would rotate about the x axis and I would use the shell method?

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@sinful bloom Has your question been resolved?

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@sinful bloom Has your question been resolved?

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[limpid junco]

Make n boxes

From people that have no friends to people that have n - 1 friends

Prove that at least 2 go in same box

[odd rose]

whats a starting point for this

[limpid junco]

What do you mean by starting point

[odd rose]

for the proof

[limpid junco]

How formal proof?

[odd rose]

can i say that for the person that has n - 1 is a contradiction because another person has 0 friends

[limpid junco]

No but you can say that at least one of those boxes must be empty

There doesnt have to be a person that has 0 friends and there doesnt have to be a person that has n - 1 friends. But they both surely cant exist. Because they contradict each other.

There cant be a group of n people where 1 has 0 friends and 1 has n - 1 friends that would imply that the person with n - 1 friends is friend with person with 0 friends. A contradiction

So there are n boxes but really there are only n - 1 that can have person in it

[odd rose]

hmm, but im not sure how i would start proving this

[odd rose]

so by showing this contradiction, it will prove that there must be at least two people who have the same number of friends.

[limpid junco]

Yes

It will prove that at least 1 out of n boxes are empty

Leaving n people into n - 1 boxes

[odd rose]

that makes sense, thanks

[limpid junco]

It seems that it works even with 28 games , 22 wins and everything else same

Maybe that will be easier to prove

[odd rose]

i dont follow

[limpid junco]

In their last 28 games, the pens won 22 games, lost 4, and tied 2. Prove that they must have won at least 4 games in a row at some point during their last 28 games.

Also valid

[odd rose]

i get that, im not sure why thats necessary

[limpid junco]

Its not but maybe its easier like this

[odd rose]

ok, how would i start this

[limpid junco]

Why not just look at first 4 games , then second 4 games etc

[odd rose]

is there some sort of principle id have to follow for this?

[limpid junco]

If there are n boxes and n - 1 pigeons there is at least 1 empty box

I mean its trivial but you will need it

[odd rose]

maybe another contradiction? say instead of winning 4 in a row, we assume its 3?

[limpid junco]

That wouldnt be a really good contradiction

You are assuming something that isnt proved

Just make 7 boxes first 4 , second 4 , ... , seventh 4

You have 4 loses and 2 ties so what do you get from here

[odd rose]

they won 4 in a row at some point

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