help-26

[dusk adder]

That is more fundamental

[ocean pewter]

Yeah, I've heard it's needed for eigen-stuff

[dusk adder]

It is also fundamental to making projection matrices

And just projections in general as you are doing right now

[urban grove]

wait, w1 = (1, 1, 0) and w2 = (0, 1, 1) though I thought?

v - c1 w1 - c2 w2 = (2 - c1, 4 - c1 - c2, 3 - c2)

[ocean pewter]

I thought w1 = <3, 3, 0> and w2 = <-2, 2, 4>

[urban grove]

No, those were the projections we calculated using the previous method

That method was wrong, my mistake

Sorry

This method should be faster and work correctly

By w1 and w2, I mean the two vectors that span S in the original problem

[ocean pewter]

Ohh, okay

[ocean pewter]

Now I'm getting c1=-11 and c2=8

[urban grove]

How'd you get that

[ocean pewter]

I used dot product to get the system

2*c1 + c2 = 6
c1 + 2*c2 = 7

[urban grove]

that's right, but your solution for c1 and c2 isn't right

(for example 2*-11 + 8 = -14, not 6)

[ocean pewter]

c2 = 8/3

c1 = 5/3 ?

Let me check with substitution

Yeah, I'm getting those solutions

[ocean pewter]

Are those the solutions you got?

[urban grove]

yup!

and now just plug that in to find c1 w1 + c2 w2

[ocean pewter]

👍

My final answer is <8/3, 13/3, 5/3>

Does that look correct?

[urban grove]

You mixed up c1 and c2, I think

[ocean pewter]

So c1 = 8/3?

[urban grove]

No I mean

c1 = 5/3 and c2 = 8/3

But it looks like you accidentally did c1 = 8/3 and c2 = 5/3 in your calculation for the final answer

[ocean pewter]

I just did the subscripts out of order

My bad

[urban grove]

yea

[ocean pewter]

So I have a visualization of what I think the porjection is, but I don't really know waht it means or does

https://www.desmos.com/3d/u6mo1qlxks

I notice that the projection bector lies on the span of <1,1,0> and <0,1,1> but it isn't orthogonal to <2,4,3>

[dusk adder]

Projection is the opposite of what you are thinking

It is the closest point to (2,4,3)

If (2,4,3) was in the span

[ocean pewter]

That makes sense now that I think about

[dusk adder]

The projection would have just been (2,4,3)

[ocean pewter]

So projection of a vector is the closest that a scalar multiple of the vector can be to another vector

and the projection of a subspace is the closest that a vector in the span of the of the subspace can be to another vector

[dusk adder]

Projection of a vector onto the subspace

[ocean pewter]

My bad

[dusk adder]

You can actually see this

Because the difference between vector and the it's projection is orthogonal to the span

[ocean pewter]

Let me try that

The orthogonal is supposed to be perpendicular, right?

[dusk adder]

It was already used here

[dusk adder]

It would be normal to the plane

[ocean pewter]

Like a normal line in calculus?

[dusk adder]

No like a normal line to a plane

[dusk adder]

If you are talking multivariable calculus where they taught you the definition of a plane yes

[ocean pewter]

I haven't taken calc iii yet

[dusk adder]

Just google

[ocean pewter]

I was kinda thinking this

[dusk adder]

Yes same idea

In more dimensions

[urban grove]

Ignore the red curve here, the blue line is normal to the green one

(normal just means perpendicular)

We're only working with linear stuff rn

[ocean pewter]

Okay

I just used the red curve to make the green and blue ones

I was going to ask about the least squares regression problem, but it's getting a little late

Thanks @dusk adder & @urban grove

.close

[dusk adder]

np

[topaz sinew]

Channel closed

Closed by @ocean pewter

Use .reopen if this was a mistake.

[urban grove]

you're welcome!

[ocean pewter]

Like this?

[dusk adder]

The blue line does look normal to that plane yes

[ocean pewter]

Cool, thanks again

[topaz sinew]

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[heady maple]

ok ques time

[heady maple]

for me

im kinda confused here

I wanted to practice some algebra

and I found this on one site

[boreal mulch]

gaussian elimination?

[heady maple]

hmm

lemme see

[grave karma]

what is 2+2

[heady maple]

im kinda stuck at the echelon form

what to do after this one?

@boreal mulch

[boreal mulch]

ig 2R1-R2->R2?

[heady maple]

oh right yess

i forgot

thanks dude

yes

thanks bro

.close

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[boreal mulch]

yw

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[queen cloud]

In a randomised controlled experiment are there no omitted variables

[astral blaze]

you supposed not to or can assume that the expectation of that is zero

There are omitted variables (I mean, randomization would not change the fact that I have a certain gene for example) but those should not influence the treatment (i.e., the distribution of people who have a certain gene should be randomized enough to not have effect on treatment effect)

[queen cloud]

So there are omitted variables but they don't possess an influence on the study

But then

Isn't the definition of an omitted variable one that DOES influence the study

[queen cloud]

And it has been wrongly excluded

So technically those variables are not "omitted variables"?

[astral blaze]

I mean. Yes. For example, suppose we have a new drug to reduce blood pressure. We let the participants consume it and let them go home and come back to check. What patients eat at home can influence their blood pressure, but does that count as an omitted variable? RCE ensures that those are evenly distributed (hopefully) and mitigate those biases.

Meal does influence blood pressure, but if we only care about the effect of the drug, RCE remove those effects

(hopefully)

[queen cloud]

So would you say that

"There are no omitted variables in a randomised controlled experiment" is True?

[astral blaze]

There should be no omitted variables bias in a randomised controlled experiment. The variables are still there (like meals) but the effects are even out from randomization.

[queen cloud]

Oh ok yeah

That makes more sense

[topaz sinew]

@queen cloud Has your question been resolved?

[topaz sinew]

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[dense cosmos]

idk what k is referring to

in part (a)

[glacial cliff]

k is just the normal variable people use for growth rate

[unreal kite]

$P(t)=60\cdot (2)^{(3t)} \to P'(t) = 60 \cdot 3 ln(2) \cdot 2^{(3t)}$\
or\
$P'(t)=3ln(2) \cdot P(t)$ \
giving by definition a relative growth rate of $3ln(2)$ (the amount $P(t)$ increases at a rate of $3ln(2)$ of its amount)

[thorny flame]

Crystopher

[dense cosmos]

so 3ln2 is correct?

[unreal kite]

if we assume that your function P(t) is correct, then it should be 3ln(2)

[dense cosmos]

alr thanks

what about this?

[unreal kite]

how did you arrive at your results?

[dense cosmos]

i just used average rate of change in the given interval 💀

i have no idea what it wants

[unreal kite]

b) states basically that given two points $(1800,980)$ and $(1850,1260)$ find a function $P(t)=P_0e^{kt}$ with parameters $k$ and $P_0$ that goes through these points, then use it to find $P(1900)$.

[thorny flame]

Crystopher

[dense cosmos]

would this work

[unreal kite]

that should work

a bit different to what I had in mind but still seems to check out.

[topaz sinew]

@dense cosmos Has your question been resolved?

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[topaz sinew]

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[glass root]

.reopen

[craggy lily]

wait is the one solution 0?

I was confused for a second but I thinkk the only value that makes sense is 0

.close

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[tall quiver]

Can someone explain to me why my solution sets are wrong?

[tall quiver]

<@&286206848099549185>

life is hard when youre stuck on a problem for hours 😦

[topaz sinew]

Please don't repeatedly close and claim a new channel with the exact same question. This erases all previous progress made towards your problem and is confusing for helpers, making it more difficult to help you. Please be patient, even if your channel has not received much attention.

[muted relic]

you just fliped the sign

[tall quiver]

i was told to flip the sign to get rid of the -1/5

[muted relic]

that means all values where f'x <0 are values where fx > 0

so the answer is 5)

[tall quiver]

so should i have not flipped the sign

pretend im new to math

i have been stuck on this problem for hours

[muted relic]

5. and 1) is the correct answer

[verbal hound]

Moo

[tall quiver]

what is 5 and 1?

like

what

[muted relic]

correct answers

[tall quiver]

im not trying to find single points

im trying to find ranges where x > 0

[muted relic]

1. is fifth, 2) is first i mean

[tall quiver]

(−2, 3) ∪ (4, ∞)
(−∞, −2) ∪ (4, ∞)
(−2, 4)
(−2, 3)
(−∞, −2) ∪ (3, 4)

these are my options for the answers

LOL

your answer doesnt fit into any of these

[muted relic]

1. is (−∞, −2) ∪ (3, 4)

and 2) is (-2, 3) ∪ (4, +∞)

[tall quiver]

ok

can you circle on the problem where i did it wrong then?

so i can learn and know for next time

[muted relic]

if you don't divide both sides by -1/5 but 1/5, the answer is clear

[tall quiver]

ok can we take this step by step

so

[muted relic]

okay

[tall quiver]

f(x) = -1/5 (x+2) (x-3) (x-4)

original problem

I want to find f(x) > 0

so now its

(-1/5) (x+2) (x-3) (x-4) > 0

is that correct so far?

[muted relic]

this one is wrong

the fliped function isn't anymore f(x)

let's assume that is f'(x)

[tall quiver]

wait i dont want to confuse myself even more if i dont have to..

so should i just have not fliped the sign

my teacher didnt teach me inverses yet

i was just told to divide both sides by -1/5 to get rid of it

and because of that i flipped the sign

can i do it without flipping the sign

(-1/5) (x+2) (x-3) (x-4) > 0

divide both sides by 1/5 correcct?

[muted relic]

yes.

[tall quiver]

ok

so its now

(-1) (x+2) (x-3) (x-4) > 0

is that right?

[muted relic]

okay

if then the signs are like this

right?

[tall quiver]

ok one sec

im going to rewrite (-1) (x+2) (x-3) (x-4) > 0

[muted relic]

yeah,

[tall quiver]

do we have to do anything with the -1?

[muted relic]

of course

[tall quiver]

ok

so i now have the equation

(-1) (x+2) (x-3)(x-4) > 0

now i do the line with the points correct?

[muted relic]

yeah,

the crossing points are correct

[tall quiver]

ok

so i do it at

-2 , 3 4

correct?

[muted relic]

yeah

[tall quiver]

what about the -1

do we just ignore it for now? 🙂

[muted relic]

result is 20 i think

which is bigger that 0

[tall quiver]

?

what steps did u skip

im only at the number line rn

so now i have the number line

-2 , 3 , 4

do i do -1 too?

or ignore that

[muted relic]

if x=-1, (-1) (x+2) (x-3)(x-4) = 20

[tall quiver]

um

thats different from the method i did in the picture

😭

this is the part im confused on

and have been for the past 2 hours

bc no one explains it

this is what i have so far

[muted relic]

yes

[tall quiver]

ok so now i plug in all my test values

to check for + -

[muted relic]

sorry

if x=-1, (-1) (x+2) (x-3)(x-4) = -20

-1 * 1 * -4 * -5 = -20

[tall quiver]

so are we just not doing the number line?

[muted relic]

i made an error cuz i'm tired too

[tall quiver]

whats the calculation for?

im on the number line and i dont know what this calculation is for

[muted relic]

let me explain this

if x is bigger than 4,

x+2 >0, x-3 >0, x-4 >0

that means f(x)is smaller than 0

right?

[tall quiver]

wait

can i ask you why my method didnt work?

and why you chose this method instead

my teacher told me once I graphed the numbers on the number line

(-2 , 3 , 4)

we have 4 intervals

and then we choose a number within those intervals to test

so i chose -3 for the first interval

0 for the second interval

3.5 for the third interval

5 for the fourth interval

then

for each interval

i input it into x

so for first interval

[muted relic]

when x = -3

[tall quiver]

(-3 + 2) = -1

(-3 -3) = -6

(-3-4) = -7

so that section is negative

is it because i need to multiply eveyrthing by (-1)?

because i got - + - +

but since theres -1 do i just do the opposites?

[muted relic]

that's it

[tall quiver]

ok

thats literally the part ive been stuck on for 2 hours

i didnt know why it was opposite

[muted relic]

okay

let me try to explain

as the problem says fx = -1/5(x+2)(x-3)(x-4)

then if we introduce a new function g(x) which is equal to -1/5 * f(x)

g(x) = (x+2)(x-3)(x-4)

right?

[tall quiver]

yea

[muted relic]

in this case, about a specific x if g(x) < 0,

what about f(x)?

f(x) = (-5) * g(x) > (-5) * 0 = 0

right?

[tall quiver]

where is the -5 from?

[muted relic]

we just assumed that g(x) = (-1/5) f(x)

that means f(x) = (-5) * g(x)

[tall quiver]

thats the part that confuses me

lol

[muted relic]

okay g(x) = -1/5 * f(x)

if we divide both sides with (-1/5)

[tall quiver]

i think if we were in vc it would make more sense to explain

[muted relic]

1/(-1/5) * g(x) = f(x)

here 1/(-1/5) = - 1 * 5

right?

[tall quiver]

1. fx = -1/5(x+2)(x-3)(x-4)
2. g(x) which is equal to -1/5 * f(x)

why are we multiplying f(x) by -1/5

is it to get a coefficient of 1?

[muted relic]

maybe. you can multiply by -1 instead of -1/5

the result is same.

[tall quiver]

ok

well thx for clearing it up

i have to go now but i appreciate it

.close

[topaz sinew]

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[topaz sinew]

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[ripe heron]

2. A wolf runs along a straight road that goes directly from her lair (W ) to Grandmother’s house
   (G). Some of this road is on flat ground and some is downhill or uphill. The wolf runs downhill
   at 20 km/h, on flat ground at 15 km/h, and uphill at 12 km/h. It takes the wolf 3 hours and
   27 minutes to run from W to G. It takes the wolf 2 hours and 33 minutes to run from G to W .
   Which of the following is closest to the distance between W and G?
   (A) 33 km (B) 36 km (C) 39 km (D) 42 km (E) 45 km

[sly pasture]

!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

[ripe heron]

hmm

1

[topaz sinew]

@ripe heron Has your question been resolved?

[sly pasture]

x=vt => t = x/v is a good start

[ripe heron]

huh

[sly pasture]

x is distance
v is speed
t is time

[sly pasture]

if you still don't understand I will send an ss that creates the equation

[ripe heron]

k thx

i still dont understand

[ripe heron]

so after everything is the answer 45

[sly pasture]

yes

[ripe heron]

thx again

wanna help me with another question

[sly pasture]

I can try

[ripe heron]

What is the sum of the rightmost two digits of 106^2024?

chatgpt gave me 15 😄

idk if its right or wrong tho

[sly pasture]

i'm sure about I have to find a pattern or sth like that but I have no idea how can I find one but gonna try

[ripe heron]

k thanks

[ripe heron]

i think i got the answer so imma close this chat , cya.

.close

[topaz sinew]

Channel closed

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[topaz sinew]

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[final hollow]

hey

[final hollow]

can someone explain this to me

i don't get why they dont equal

[shadow salmon]

Pick for example theta_1 = theta_2 = pi/2

[final hollow]

mhm

[shadow salmon]

what happens?

[final hollow]

uhhh, im not sure? i don't understand what you are asking.

[glass maple]

substitute some random angles that arent 0

and see what results u get

[final hollow]

ah ok

[shadow salmon]

Im asking you to simply try some values

[shadow salmon]

like these

[final hollow]

oh yeah, i know they dont equal i just want to see some sort of proof.

[shadow salmon]

well that is a proof

[leaden hatch]

proof by contradiction?

[glass maple]

if u want a proof for more general case, have u been taught compound angles yet?

[shadow salmon]

nope, its a perfectly valid proof by counter example

[final hollow]

because my rational is they should be equal sine x(a) + x(b) = x(a+b)

[shadow salmon]

why?

[keen raptor]

sin is not being multiplied by a or b here

it's not the distributive property because it's not multiplication

[final hollow]

oh yea, its a ratio...

[keen raptor]

it's a function application, which is different and just happens to use the same notation

[final hollow]

well. thanks yall! i just needed to know that because im confused about the cast law.

[glass maple]

ive never heard of cast law

[shadow salmon]

the thing you have to learn here is that linearity is a special property and not something that works for just about any function

[glass maple]

i feel like the term linearity wont mean much to them at the moment

[final hollow]

how could sin (180) =-0.801152635734, how can you have a triangle with negative length?

[shadow salmon]

im calling it by name, to make it more obvious it's a special property

[keen raptor]

well first you should probably switch to degree mode instead of radians

[final hollow]

oh.

yeah

i forgor desmos used raians

[glass maple]

yeah fair enough

[final hollow]

radians

[keen raptor]

You can switch to degrees in settings (top right)

[glass maple]

Out of curiosity, has your teacher shown you the unit circle or are you familiar with it

[shadow salmon]

desmos sin(x) is a generalized notion, than that of the definition from a triangle

[final hollow]

but my point being, how can some trig values be equal to negative if you can't have a negative distance? no im just studying on my own.

[keen raptor]

sine is defined a little differently outside of 0-90 since you can't use right triangles

[final hollow]

yeah

[keen raptor]

so the definition is generalized

[shadow salmon]

it makes sense once you encounter the unit circle

[glass maple]

there are videos on youtube that will teach u the unit circle concept better than we can probably word here on discord

https://www.youtube.com/watch?v=1m9p9iubMLU

diagrams will probably help more than texts for this topic so i would consider just watching these videos

[final hollow]

i tried to look but i couldn't find any, they all used ref angles and i don't get why ref angle has to be the one adjacent to the x axis and why using the unit circle to get the trig values could translate to triangles.

my first road block was sin (110), i thought that you could just get sin (90) then add sin (20) but i see thats wrong.

so i was confused why you would need a ref angle

[shadow salmon]

infact distances aren't now negative, here the triangle is instead represented in a coordiante system, so the coordinates may be negative, but we still treat distances between points as non-negative

https://www.desmos.com/calculator/qmzx2skkzy

play around with this

notice how the triangle is still there

[final hollow]

yeah I was confused about that, how can a negative coordinate translate to a triangle which can't have negative distance. i was looking for some sort of geometric proof too.

[shadow salmon]

it's just a rather convenient generalization

if you tried to keep the coordinates positive, youd have ugly jumps

and again, the distances arent negative its the coordinates

we've changed represenation here

[glass maple]

The negative sign should be thought more of as direction than anything else

[final hollow]

so, what i am confused on, is why we would use a ref angle, and if there was any geometric proof to show that a triangle with a sin value would correspond to a coordinate that on the unit circle.

i can see that the unit circle is already a geometric proof of most trig functions, but i want to see if there was anyways to translate transformation in the unit circle into math steps, like how would you represent a turn of the radius to some angle theta on an equation.

well, thats all i guess. thanks for the clarifications.

[shadow salmon]

im not sure what you mean by ref angles, but for the latter imagine you're on the 1st quadrant and have a unit circle. Place a right angle triangle such that the hypotenuse is the radius of the unit circle, so it must be a distance of 1. (refer to the desmos link)

[final hollow]

yup

[shadow salmon]

now (again refer to the desmos link) given some angle a, which is between the adjacent side, origin and hypotenuse

[final hollow]

ok

so a complement angle?

[shadow salmon]

so like this

[final hollow]

oh ya

[shadow salmon]

again we know the hyp is 1, could you express the other sides in terms of sin and cos?

[final hollow]

yes you could repersent the adjacent side as cos and the opposite as sin

[shadow salmon]

cos of what?

and sin of what?

[final hollow]

cos (a) and sin(a)

[shadow salmon]

yeah and why?

[final hollow]

because both those functions equal to some number x over the hypotenuse which is 1

[shadow salmon]

Sorry, i cant quite follow; let me ask you the defintions of cos and sin instead

[jaunty badge]

What are we doing here?

[final hollow]

oh its opp/hyp for sin and adj/hyp for cos, we are doing trig.

[jaunty badge]

Question?

[shadow salmon]

yup!

[jaunty badge]

What is the Question?

[shadow salmon]

okay cool, now since we're in a coordiante system

[shadow salmon]

im clearly already helping them rn?

just scroll up idk

[final hollow]

i just didn't understand how the sin values could be negative since that implys a negative ratio which means a triangle with negative distance. and how the unit circle is relevant to a triangle.

[jaunty badge]

K but still

[shadow salmon]

now, since we're in a coordiante system

[final hollow]

but yes he is helping me.

ya

[shadow salmon]

notice how from the origin traversing a distane cos(a) to the right

we land on the coordiante (cos(a), 0)

[jaunty badge]

Ok, but if you guys still need help ping @jaunty badge

[final hollow]

yeah.

thanks.

[shadow salmon]

so if we imagine a line segment from (0,0) to (cos(a),0) what would that represent?

[final hollow]

a value of x

[shadow salmon]

Sorry, whats x?

[final hollow]

the distance from the origin to the x axis which is also equal to the cos(a) in this context.

[shadow salmon]

oh, you mean the x-coodinate or?

im more trying to relate the triangle right now

[final hollow]

yes the x- coodinate

[shadow salmon]

like if you had a line segment from (0,0) to (cos(a),0) what would that look like?

[final hollow]

a straight line across the x-axis

a triangle?

[shadow salmon]

yeah sure, but not all the way right?

yeah so think in relation to the triangle

[final hollow]

yes

[shadow salmon]

so what im trying to get at is that this line segment is just the adjacent side on our triangle

do you see that?

[final hollow]

ya

[shadow salmon]

and how the point (cos(a),0) would then be the right endpoint of that line or in other words the bottom right corner of the triangle

[final hollow]

yes

[shadow salmon]

so now imagine we're on the point (cos(a),0)

[final hollow]

mhm

[shadow salmon]

let's traverse upwards by sin(a)

[final hollow]

yes

[shadow salmon]

where would we land you reckon?

[final hollow]

cos(z),sin(a)

cos(a)

[shadow salmon]

yeah!

and in relation to the triangle, where is this?

[final hollow]

at the end point of the hypotenuse

[shadow salmon]

yup!

so if our right triangle lives in the 1st quadrant, we're able to use the definitions from geometry (since our coordinates will be positive) and the points on this triangle can be expressed with cos(a) and sin(a), and so we havent really done anything different yet apart from representing points of the triangle as points in a coodiante system

[final hollow]

yeah

[shadow salmon]

now notice, once we move the right triangle say to the 2nd quadrant, our triangle will be flipped horizontally if it must obey the same conditions as we said earlier, i.e hyp is the radius of the cirlce etc

but now our x coordinate for example will be negative if we talk about the corners of the points we talked about eariler

we'd like to keep using say the point (cos(a), 0) to describe that corner we mentioned before

or even (cos(a), sin(a))

[final hollow]

yes but, when we do that, a becomes the exterior angle though.

[shadow salmon]

the natual thing to do here is if our angle a is greather than 180, then the cos(a) should be negative, to account for this change in the coordiate sytem

Yeah sure

[final hollow]

and if that happens then the values of the trig functions would be respective the exterior angle not the interior one, and the exterior angle does not form a right angled triangle but an oblique triangle

[shadow salmon]

Yeah so that's perfectly valid if you want to keep the coordinates positive, but what we want here is be able to represent the same points the same way we moved it, i.e the x-coordinate of our corner will be negative. Also notice that the point (cos(a),sin(a)) will act as a rotation about the origin

if we let cos be negative, once in the 2nd quadrant, we can start rotating things rather nicley aswell, as you would in real life

[final hollow]

yeah but the interior angle is not equal to the exterior angle so how could you use the interior angle to represent the exterior angle.

[shadow salmon]

So youll have that the interior angle + exterior angle = 180 (if we treat the x-axis as our floor in this case)

just solve for whatever youre keen on knowing

[final hollow]

wait...

nvm, i am still confused, thanks for the help though - you cleared up most of the confusions i had

can i ask you a different algebra question?

[shadow salmon]

im glad i could help, sure!

[final hollow]

it one that i have been having trouble with

ok so...

can you solve this system of equation step by step, i have been having trouble with it. i think i do the steps correctly but the results never make sense

[shadow salmon]

Hm let's try

I should start by asking what progress you made, so i could focus on something in particular

[final hollow]

so, i can solve for x in the first equation, then i plug it into the third equation and solve for y

then i plug x again into the third equation

this yealds me some sort of quadratic equation with two variables instead

then i plug in y from the previous equation and plug it into the third equation

this only leaves z then i solve the quadratic and the results dont make sense

when i input the value of z in desmos, the graphs don't get a point where they all intersect

[shadow salmon]

Alright, let's try something different

[final hollow]

oks

[shadow salmon]

so let me just label these, so we have the system

$$\begin{cases}
x+2y-z=2&\quad (1)\
3x-y+2z=7&\quad (2)\
1/x+y-z=0&\quad (3)
\end{cases}$$

[final hollow]

ok

[thorny flame]

Aslan

[shadow salmon]

notice that if we multiply equation (1) with 2, we get 2x + 4y -2z = 4, right?

[final hollow]

yeah

[shadow salmon]

so by adding (1) and (2) we get 5x + 3y = 11, right?

[final hollow]

yup

[shadow salmon]

alright lets remember this one and instead notice something else

[final hollow]

ok ok

[shadow salmon]

if we say subtract (1) from (3), what do we get?

[final hollow]

we get, 1/x - y -x = 0

[shadow salmon]

almost!

[final hollow]

oh

wait , 1/x - x +3y -2z=0

[shadow salmon]

oh and i already made a typo lol, it should say 2y and not 3y

or no, nvm ignore that

[final hollow]

hm. i really cant fact check you rn so im just comming along until it clicks :d

[shadow salmon]

Oh, how come?

i want you to feel like youre following along

[final hollow]

because i have been focused on that equation for the past week and every where i search its always just a quadratic system with only two variables and rn im just desperate for some sort of mental ease, don't worry, im following along.

[shadow salmon]

alright!

[final hollow]

🙂

[shadow salmon]

so going back to this

[final hollow]

ok

[shadow salmon]

this was a bit more closer

you just forget the -2 on the RHS

do you see how?

[final hollow]

oh wait. wouldnt it be 0 though? because you substracted a negative by a negative so the negative becomes additon.

[shadow salmon]

so when you subtract or add equation, formally what we really mean is that we are subtracting or adding each side separately, if that was unclear

so on the RHS, we are doing 0 - (+2)

[final hollow]

oh so like x+y+z - ( x+Y+z) not each term substracted seperatly?

[shadow salmon]

Let me show it more explicitly maybe

[final hollow]

ah yeah i got it dont worry

[shadow salmon]

you sure?

Well, just to make sure we're on the same page. Say we have the equations w = 4 and k = 2, could you add these two for me?

[final hollow]

wait, is it 1/x + y -z - x + 2y-z or 1/x + y -z - ( x + 2y -z)

6

w + k

[shadow salmon]

yeah! so that would be the RHS and LHS resp. !

I.e we have that w+k = 6

makes sense?

[final hollow]

so is it the one with the brackets or the one without.

[shadow salmon]

it would be the one with brackets, but this is only the LHS, we also have a RHS of the equation

[shadow salmon]

so going back to this example

[final hollow]

yeah?

[shadow salmon]

could you subtract say w=4 from the eq. k=2?

[final hollow]

w - k = 2

[shadow salmon]

Thats correct, if i told you to subtract the eq. k=2 from the eq. w=4; but it seems youre on the same page atleast

[shadow salmon]

alright, so do you now see why we're missing a -2 on the RHS here?

[final hollow]

yeah, so for the other equation it would equal 1/x -x -y = -2 oh yeah my bad i forgor about the rhs

[shadow salmon]

no worries!

good

alright so now, let's go back that that other equation we got

[final hollow]

yeah

[shadow salmon]

5x + 3y = 11

[final hollow]

mhm

[shadow salmon]

Let's solve for y here

[final hollow]

yeah

[shadow salmon]

What do we get?

[final hollow]

we get y = (11-5x)/3

[shadow salmon]

yup!

lets substitute this in the eq. 1/x -x -y = -2

[final hollow]

k

[shadow salmon]

we should now have an equation in one variable

[final hollow]

yeah

[shadow salmon]

could you write it for me?

[final hollow]

ah sure wait

$\frac{1}{x}-x\ -\left(\frac{11-5x}{3}\right)$

[thorny flame]

element

[shadow salmon]

missing a RHS here!

[final hollow]

noice it worked

oh ait

$\frac{1}{x}-x\ -\left(\frac{11-5x}{3}\right)=-2$

[thorny flame]

element

[final hollow]

: - )

[shadow salmon]

neat!

after a bit of algebra we have the equivalent equation (3+6x)/x = 11 - 2x

would you agree?

[final hollow]

wait

yes

it makes sense

[shadow salmon]

alright cool, you might see where this is going, now multiply the nonzero x to both sides to get a quadratic equation.

Namley 3+6x = 11x - 2x^2

[final hollow]

ya

[shadow salmon]

I suppose you can solve this one on your own?

[final hollow]

ah yes let me see

this is what i got

$x=\frac{5+7}{4},\ x=\frac{5-7}{4}$

[thorny flame]

element

[shadow salmon]

I can provide atleast one solution, x=1

Check this by plugging it in

There’s one more however

[final hollow]

yeah i already knew the values of those varables, so this works. and yes there is the second value

but my one problem is

[shadow salmon]

But uh, this isn’t right?

[final hollow]

this worked but why didn't it worked when i didn't do elimination and dis substitution through out the entire system.

i mean why didnt it work when i did substitution through out

the entire system

[shadow salmon]

In theory it should of worked

[final hollow]

this time you used elemination and then substitution but when i used substitution through out the entire system it didn't work.

i don't know why that is, do you? or it was just a miscalculation on my part.

[shadow salmon]

It could be just due to a mistake, but in principle it should work too

I haven’t seen the explicit work for me to judge though

[final hollow]

because when you don't use elimination you get massive numbers like $54x^{2}-132x+46=0$

[thorny flame]

element

[final hollow]

and when you solve it, it gets bigger

but anyways, thanks for your help, atleast now i know how to solve em.

i guess ill ask a teacher when i get back to school.

[shadow salmon]

If you’re getting different solutions to that quadratic, then you’ve made a mistake

[final hollow]

yeah, its when i solve the quadratic i get a different answer.

but its not the quadratic that i miscalculate, its at the steps before. because when i check my quadratic equation answers in a calcualtor they make sense.

[shadow salmon]

You did however also not provide the correct solutions to this, so it could very well be a mistake on your part

Like when solving the quadratic

[final hollow]

wait what was my answer?

[shadow salmon]

This

[final hollow]

isn't that the solution?

[shadow salmon]

As I said, one of the solutions is x = 1, but neither of them is that

[final hollow]

i used the completing the square method this time and it made sense through out, lemme do that again

[topaz sinew]

@final hollow Has your question been resolved?

[shadow salmon]

@final hollow how’s the progress?

[final hollow]

ah i see, i messed up because i divided by -2 instead of 2

i just got that...

wait lemme send you wnat i got

$x=\frac{5-1}{4},x=\frac{5+1}{4}$

[thorny flame]

element

[final hollow]

well thats all. thanks 👍

[shadow salmon]

Great! Ur welcome

[final hollow]

ima close dis now, adios

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[gaunt quartz]

x²+3x

6x²+24x

can i cancel out the x² numerator on top with the 6x² on the bottom to make it 5x²? and can i also just divide 24 and 3 individually?

[deft holly]

if you want to cancel things on fractions, it has to apply to every term. you cant do anything to individual terms

[gaunt quartz]

so how would i simplify this fraction?

[deft holly]

factorise so that the numerator and denominator has common factors and then cancel

[gaunt quartz]

yessirrr okay thanks thanks

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[lyric cobalt]

Need help with this stats question 😃

[astral blaze]

what is percentile?

if you understand the definition, this should be more obvious to you

[cinder sequoia]

quite handy that you have exactly 10 observations ._.

ahem in increasing order

[astral blaze]

I will say that there are a lot of ways to answer this question, however. Many of them result in a slightly different answer.

but that depends on what is being taught in your class

[lyric cobalt]

I’ve put the answer 6.8 it didn’t work and I can’t understand any other way of finding the answer

[cinder sequoia]

i believe 6.8 would be the 30th percentile?

[astral blaze]

I mean, you are not entirely wrong. 6.8 is lower bound for the 40th percentile

but I think you can do better

Hint:|| Think about the position. What does that decimal point tell you?||

[lyric cobalt]

Are you telling me I’m supposed to arrange them by the lowest to highest decimal point?

[astral blaze]

no

when you find the percentile, you need to find the position for that, correct?

what is that position?

[lyric cobalt]

Well I mean it would be the 4th box no? 🙂‍↕️

[astral blaze]

no

well, it is at least at that point

but not exactly there

[lyric cobalt]

Okay

So how would I be able to find the actual position

[astral blaze]

use definition of percentile...

$\frac{k}{100} (N+1)$

[thorny flame]

nobody

[lyric cobalt]

I mean I got a number

1.4

[astral blaze]

are you sure?

Anyway... if it is like .4, you know that it should be far off from 4 but not that far that it becomes 5.

you know that your 4th position is 6.8 and 5th position is 16.1

so your percentile should lie inbetween that

[lyric cobalt]

Ahhh Okay

So I would always use the position above the percentile?

[astral blaze]

not really

one of the way to approximate that is to think that if the length between 6.8 to 16.1 is one unit, what is the length of 0.4 of that unit?

like... if it is 4.5, you know that it should be in the middle between 6.8 and 16.1

[lyric cobalt]

Honestly dude I’m actually fried I don’t get this last part at all

[lyric cobalt]

Alr Alr I UNDERSTAND THAT

[astral blaze]

so you approximate that

[lyric cobalt]

Soo I would think 9.3 or 10.3

[astral blaze]

well... you know that it is 0.4 position more than 4th position. so if I know that one position around there is 16.1 - 6.8, then 0.4 of that is...

(like, you need 16.1 - 6.8 much to turn 6.8 [4th position] into 16.1 [5th position])

[lyric cobalt]

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[brazen inlet]

A ball weighs 0.3125 pounds. How many foot-pounds does it take to throw a ball 90mph.

[brazen inlet]

@cinder sequoia

Follow up question to the kinetic energy proof

[ivory sorrel]

foot-pounds?

what's that in SI

[brazen inlet]

Newton meters

[ivory sorrel]

ah

[brazen inlet]

Just imperial

Work

[ivory sorrel]

so work

[brazen inlet]

Yeah

[ivory sorrel]

ok, maybe I can help

[brazen inlet]

Seeet

Sweet

[ivory sorrel]

Power is Fv

or wait

[brazen inlet]

uh

[ivory sorrel]

find the change in KE

0-> KE at 90mph

,w 90mph to m/s

[thorny flame]

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[brazen inlet]

1/2(mv_2^2)-1/2(mv_1^2)

[ivory sorrel]

yup

but v_1=0

no?

[brazen inlet]

Yh

And v_2 is 90

But I get it wrong

The answer is somehow 85😭

[ivory sorrel]

what units is the answer in

[brazen inlet]

Foot pounds

[ivory sorrel]

,w 0.3125 pounds to kg

[thorny flame]

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[ivory sorrel]

,w 0.5(0.14175)(40.23)^2

[thorny flame]

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[ivory sorrel]

,w 114.707 Newton meters to foot pounds

[thorny flame]

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[brazen inlet]

Howew

Bro what 😭

[ivory sorrel]

must have made a calculation mistake

[brazen inlet]

How

[ivory sorrel]

maybe work in SI and then covert to imperial ?

[brazen inlet]

90^2x0.3125x0.5

[ivory sorrel]

90^2?

why

mph

[brazen inlet]

Because that’s the units we using

[ivory sorrel]

not feet per second

[brazen inlet]

Why not feet per hour

🤯

[ivory sorrel]

no, I mean convert the speed to feet per second

[brazen inlet]

Why not feet per hour though

Why seconds

[ivory sorrel]

I'm not familiar with calculations in imperial, so I can only guess

but because the standard unit of time in imperial is second

[brazen inlet]

Idk

I have transitioned to metric 😎

I can’t do imperial stuf

[ivory sorrel]

stick to metric, calculations are much easier

[brazen inlet]

Yeah I don’t need to know there’s 37393746.27384 feet in a mile

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[neon iron]

question

[neon iron]

why is f(x)=2x^3-3 an even number?

[dusk adder]

That's a function and it is also odd for all integer x

[neon iron]

oh f I'm stupid 💀💀💀...

i just realised

sorry for wasting your time idk what got into me

[dusk adder]

No worries

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[glass maple]

It is more of a general question in techniques of evaluating power series

[glass maple]

context is highschool math and power series isnt a stand alone topic

Overall I find it quite hard to evaluate power series and idrk where to proceed with them

i haven't been taught any techniques or general guidance in how to evaluate these series so I was wondering if anyone here have readings or general suggestions in solving these

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[ocean prairie]

There is not really a standard way of evaluating power series, except it’s like a known series

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[vast vector]

Hi! I'm making a tower defense game and I want to calculate where the tower should shoot the bullet to hit the monster and the equation would look like this where everything is constant except for x,y :
x = monster.rect.centerx + (monster.direction[0]*sqrt((x^2) + (y^2))) - self.rect.centerx
y = monster.rect.centery + (monster.direction[1]*sqrt((x^2) + (y^2))) - self.rect.centery

[vast vector]

I want to somehow solve these equations to:
x = something
y = some other thing

[odd pagoda]

well are the monsters only moving in the same direction always?

otherwise your turret will definitely miss if they turn a corner

[vast vector]

yeah rn I don't care about that this will be good enough

please someone help me <@&286206848099549185>

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@vast vector Has your question been resolved?

[topaz sinew]

@vast vector Has your question been resolved?

[topaz sinew]

@vast vector Has your question been resolved?

[old cloak]

Dist = sqrt((monster_pos_x + monster_vel_x * t - turret_pos_x) ^ 2 + (monster_pos_y + monster_vel_y * t - turret_pos_y) ^ 2) = t * proj_vel
solve for t

then
monster_pos_x' = monster_pos_x + monster_vel_x * t
aim turret at (monster_pos_x', monster_pos_y')

@vast vector

[vast vector]

hmm

[old cloak]

solving for t should just be a quadratic

[vast vector]

t = (-monster_pos_xmonster_vel_x - monster_pos_ymonster_vel_y + monster_vel_xturret_pos_x + monster_vel_yturret_pos_y - sqrt(-monster_pos_x**2monster_vel_y*2 + monster_pos_x**2proj_vel*2 + 2monster_pos_xmonster_pos_ymonster_vel_xmonster_vel_y - 2monster_pos_xmonster_vel_xmonster_vel_yturret_pos_y + 2monster_pos_xmonster_vel_y**2turret_pos_x - 2monster_pos_xproj_vel*2turret_pos_x - monster_pos_y*2monster_vel_x*2 + monster_pos_y**2proj_vel*2 + 2monster_pos_ymonster_vel_x**2turret_pos_y - 2monster_pos_ymonster_vel_xmonster_vel_yturret_pos_x - 2monster_pos_yproj_vel*2turret_pos_y - monster_vel_x*2turret_pos_y*2 + 2monster_vel_xmonster_vel_yturret_pos_xturret_pos_y - monster_vel_y**2turret_pos_x*2 + proj_vel**2turret_pos_x*2 + proj_vel**2turret_pos_y*2))/(monster_vel_x**2 + monster_vel_y**2 - proj_vel**2)
,
t = (-monster_pos_xmonster_vel_x - monster_pos_ymonster_vel_y + monster_vel_xturret_pos_x + monster_vel_yturret_pos_y + sqrt(-monster_pos_x**2monster_vel_y*2 + monster_pos_x**2proj_vel*2 + 2monster_pos_xmonster_pos_ymonster_vel_xmonster_vel_y - 2monster_pos_xmonster_vel_xmonster_vel_yturret_pos_y + 2monster_pos_xmonster_vel_y**2turret_pos_x - 2monster_pos_xproj_vel*2turret_pos_x - monster_pos_y*2monster_vel_x*2 + monster_pos_y**2proj_vel*2 + 2monster_pos_ymonster_vel_x**2turret_pos_y - 2monster_pos_ymonster_vel_xmonster_vel_yturret_pos_x - 2monster_pos_yproj_vel*2turret_pos_y - monster_vel_x*2turret_pos_y*2 + 2monster_vel_xmonster_vel_yturret_pos_xturret_pos_y - monster_vel_y**2turret_pos_x*2 + proj_vel**2turret_pos_x*2 + proj_vel**2turret_pos_y*2))/(monster_vel_x**2 + monster_vel_y**2 - proj_vel**2)

@old cloak how do I know which one to use for t?

[old cloak]

should just be the positive one

no solution if both are negative

t = max(t1, t2);
if (t < 0) {
return Null;
}

well no if both are positive then you should pick the smaller one

not sure if possible

[vast vector]

proj vel is the projectile's x or y?

[old cloak]

it is both combined

[vast vector]

in what way?

[old cloak]

pythagoras

[vast vector]

ooh

[old cloak]

velocity as a vector

but also you should go from proj_vel to proj_vel_x and y and not the other way around

it's a constant that you should decide

[vast vector]

okay I'll get back when I implemented this in my code

and what is the monster_pos_y'?

[old cloak]

' is used for changed / updated value, in this case it means future position

[vast vector]

oh I subsitute everything into a solved version for y of this?
y = monster.rect.centery + (monster.direction[1]sqrt((x^2) + (y^2))) - self.rect.centery

[old cloak]

what I wrote should do everything that you want, not sure what you were doing

[old cloak]

.

[vast vector]

you never said what monster_pos_y' is did u?

[old cloak]

.

[vast vector]

but what is the future value of the monster's y position?

[old cloak]

you aim to where the moster will be given its current velocities and position

[old cloak]

same for y

mb

[vast vector]

thx

I multiply both with the same t?

[old cloak]

yes, t is the time it takes for the projectile to go the distance to the future position of the monster

[vast vector]

hmm

for some reason it says no solution every time

there are math domain errors (negative numbers inside sqrt) in the sqrt

[old cloak]

should check for the discriminant first yes

[vast vector]

like both t1 and t2 fail always for some unholy reason

[old cloak]

the error tells you everything

discriminant = b ^ 2 - 4 * a * c
if (discriminant < 0) then error

a * t ^ 2 + b * t + c = 0

[vast vector]

?

[old cloak]

the quadratic you get from squaring both sides of this gives you the form
a * t ^ 2 + b * t + c = 0
the quadratic formula tells us that the discriminant (the part that is under the square root)
b ^ 2 - 4 * a * c
must be positive in order for solutions to exist

[vast vector]

yeah but this happens every time

[old cloak]

vary the starting conditions, check your variables

a = monster_vel_x ^ 2 + monster_vel_y ^ 2 - proj_vel ^ 2
b = 2 * (monster_vel_x * (monster_pos_x - turret_pos_x) + monster_vel_y * (monster_pos_y - turret_pos_y))
c = (monster_pos_x - turret_pos_x) ^ 2 + (monster_pos_y - turret_pos_y) ^ 2

simple case: it is impossible to hit if the monster is moving away at a speed faster than the projectile

[vast vector]

frick it I'll go with projectiles that follow the enemy until they hit it

I tried to do this in various ways, this one, an iterative one, one where I check the coords where the monster will be and if I shoot in that direction will I hit it

[old cloak]

this is just physics, it works if you implement it right

but up to u

[vast vector]

thanks anyways

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[cedar wagon]

Hey

[stray wave]

I was wanting some clarification on the first 2 and I was wondering if someone could explain the last one

So for 1. Code 5
2. Code 3

[stray wave]

Hi

And for question 3 I'm stuck at the beginning I have no idea how to start

[steep nebula]

its not code 3

[arctic igloo]

find gradient of 5x-8y+12=0

[stray wave]

What is a gradient I haven't learned that yet

[rose sierra]

#❓how-to-get-help

[arctic igloo]

in any linear equation y=mx+c m is the gradient

[stray wave]

And why isn't it code 3

[steep nebula]

solve for y

what does it give you?

[arctic igloo]

so arrange the equation in the form y=mx+c where y and x are variables whereas m and c are constants

[stray wave]

Oh so code 4

[stray wave]

Ok

[stray wave]

Is it code 4

[steep nebula]

yes it is

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[wicked burrow]

for the first matrix in the span

i get the 0101

and they get 1010

is it the same?

basically the answer i get has the columns switched

[opal vault]

so x_4 is supposed to be fixed and equal to 1

[wicked burrow]

yes

[opal vault]

so I doubt the x_4 coefficient would be in the span

[wicked burrow]

makes sense

then ill try and figure out what i did wrong

[opal vault]

!show

[topaz sinew]

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

[wicked burrow]

its a bit messy but

x1 = a and x2 = c

but no matter how i calculate it, i get a = 0

when i chose b as 1 and c as 0

so basically x2 and x3 are free

so when calcualting the span, i chose x2 as 1 and x3 as 0

and i always get x1 = 0

but in the answers x1 = 1

nvm i goofed with the values

.close

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[cedar wagon]

Factor lhs by x_1

And divide to isole it

p_1x_1 + p_22x_1 = m
x_1 ( p_1 + 2p_2) = m
x_1 = m/( p_1 + 2*p_2)

[hollow drum]

You're suppose to factor out x1 first before dividing

It'll make it easier

Because if you divided by p1, you're dividing all the terms by that

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[hollow drum]

It'll be like this

If that makes sense

So if you divided by 2p2 then you'll have x1/p2 + x1/p1 on the left

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[trim trench]

is there something wrong with this process?

[restive inlet]

just a lot of bad notatiion

[trim trench]

whereabouts

[icy sky]

youre missing dx in your integrals

you have nice handwriting

[restive inlet]

and ambiguous fraction

[trim trench]

first one is my friends soln and this one is mine

integral calculator says his is right though

are they the same? or am i wrong somewhere

[restive inlet]

they're the same

[icy sky]

you can use a change of base

,w -(ln(2))^3/(2(ln(x))^2)=-(ln(2))/(2(log_2(x)^2))

crap

anyway

you can do
$-\frac{\ln^3(2)}{2\ln^2(x)}=-\frac{\ln(2)}{2}\left(\frac{\ln(2)}{\ln(x)}\right)^2$

[thorny flame]

Underfull \hbox (badness 10000)

[icy sky]

then you can use change of base on the fraction on the right

[trim trench]

oh so they are the same

that's weird

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[trim trench]

.close

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[meager anchor]

I've been having trouble with a question here

[rigid cloak]

Good

Imma help u

<@&286206848099549185>

[rigid ivy]

!15m

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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.

[rigid cloak]

Bro answer his question

I am busy in my question

I k I am helper but seeking help myself

[rigid ivy]

Don't spam helper ping

[rigid cloak]

Just pinged one time bro

He is new bro does not know how help helpers work

[rigid ivy]

Don't ping until 15 minutes have passed

[rigid cloak]

Okok

[rigid ivy]

!noadvert

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Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, so advertising can quickly turn into spam.

[rigid cloak]

Okok

Sry

By mistake

[rigid ivy]

Your question is cut off a lot

[split cloak]

Evaluate fraction?

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[rancid forum]

What are the steps to solve this trig identity?
I tried it and the answer was wrong

[rancid forum]

I began to solve the left side by turning the sin^2theta into 1-cos^2theta

[rigid ivy]

Let u=cos and maybe factoring will work

[rancid forum]

wdym by u=cos
Would i now subtract the 1 and the 3 like normal from the denominator
so that it becomes
-cos^2theta+3costheta-2

[rigid ivy]

I'm saying instead of writing cos everywhere, write u

[torpid sparrow]

Factor after that numerator and denominator

Simplify and multiply by the conjugate

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@rancid forum Has your question been resolved?

[rancid forum]

ohh

i think i got it

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[rancid forum]

thanks

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[pearl patio]

solve this Differential equation

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[patent beacon]

Hey

Can someone please explain how to multiply / divide exponents that have coefficients?
I couldn't understand it
Something like $(5x^3)(7x^4)$ or $\frac{8x^2}{4x^3}$

Thanks

[thorny flame]

𝓣𝓪𝓼𝓴

[reef fjord]

note that $(bx)^n \ne bx^n$

[thorny flame]

🫎 Moosey 🫎

[patent beacon]

Yeah

[reef fjord]

but otherwise you can multiply just like ordinary numbers, recall your expontent rules as well

,tex .exp rules

[thorny flame]

🫎 Moosey 🫎

[reef fjord]

basically, as long as you don't have any addition attached, you can freely divide/multiply numbers just as you normally would

so you can just do 5*7 then apply product rule, or do product rule first, then mutliply the numbers

essentially you treat them separately

[patent beacon]

Ok I'll try

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