help-43

[floral fjord]

@deft tangle

Let {x,y,z} be A

Let {1,2,x} be B

Let {x,2,4} be C

A intersection B = {x}

A intersection C = {x}

[kind viper]

missing {}

[warped oar]

Is {x,y,z} itself an element of A?

[floral fjord]

x y z are 3 elements

[warped oar]

Ok.

[deft tangle]

So the intersection will be in B and C

[warped oar]

?

[deft tangle]

A intersection B=A intersection C means the common elements will be B and C

A U B = A intersection B

Means A=B@kind viper

C

[compact pewter]

@deft tangle Has your question been resolved?

[deft tangle]

U agree

[sick shard]

A={1,2,3}
B={1,2,3}
C={1,2,3,4}

AuB={1,2,3}=AnC
AnB={1,2,3}=AnC

Clearly, B≠C, A≠C but A=B so option B or since it's not a proof, so option D can also be answer but options A and C can't be answers.

Bro did it here, real proof

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[fast comet]

I ended up in this limit trying to determine the convergence of the series with the ratio test. I ended up with two limits: the right one, I have absolutely no idea of how to solve. Any help? Or maybe there's an easier way to determine the convergence?

[compact pewter]

@fast comet Has your question been resolved?

[native shard]

it doesn’t look like you finished writing it

[fast comet]

Yes, it's because I don't know how to proceed once I have those two limits at the end of the second line

[native shard]

$\lim_{n \to \infty} \frac{4n^2 + 6n + 2}{n^2 + 2n + 1} \cdot \left(\frac{n}{n + 1}\right)^{2n}$

[boreal girder]

knief

[native shard]

so i guess you took ln for the right one?

[fast comet]

I tried to rewrite it as e^ln

But from there, I tried to do chain rule. I ended up with something much less intuitive than before

[native shard]

$L = \lim_{n \to \infty} \left(\frac{n}{n + 1}\right)^{2n}$

[boreal girder]

knief

[native shard]

$\ln L = \lim_{n \to \infty} 2n\ln \frac{n}{n+1}$

[boreal girder]

knief

[native shard]

now the ln bit goes to ln(1) = 0

and 2n —> inf so we can rewrite this and use lhopital

$\ln L = \lim_{x \to \infty} \frac{2\ln \left(\frac{x}{x+1}\right)}{\frac{1}{x}}$

[boreal girder]

knief

[native shard]

now we have 0/0

,w derivative x/(x + 1)

[boreal girder]

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[fast comet]

Shouldn't the argument be an exponent to "e"? I suppose you took down 2n as a coefficient by using e^ln

[native shard]

$\ln L = \lim_{x \to \infty} \frac{\frac{2}{(x + 1)^2}}{-\frac{1}{x^2}}$

[boreal girder]

knief

[native shard]

no

i took ln of both sides

we will do e^ after

then ln is continuous so we bring it inside the limit

[fast comet]

Alright, so you're doing this in a different equation from this one if I understand

[native shard]

i’m just evaluating the right limit

we know the limit of the rational function is 4 clearly

[native shard]

$\ln L = \lim_{x \to \infty} \frac{-2x^2}{(x + 1)^2}$

[boreal girder]

knief

[native shard]

the right side evaluates to -2

so ln L = -2 and so L = e^{-2}

[fast comet]

So 1/e^2

[native shard]

yea but then we also have the 4

so 4/e^2

e^2 ≈ 9

so 4/9 < 1

[fast comet]

So the series converge

[fast comet]

Did you apply substitution between this and the previous step?

[native shard]

,w \sum_{n = 1}^{\infty} \frac{(2n)!}{n^{2n}}

[boreal girder]

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[native shard]

i just made n —> x

because using lhopitals rule for an expression involving n is a bit cooked

it’s the same thing

[fast comet]

Does that property have a specific name? I still do not understand it

[native shard]

lhopitals rule?

[fast comet]

I mean, not that. I know L'hopital's rule is used if you have an indefinite limit

[native shard]

i didn’t change anything except writing x instead of n

it changes nothing

it’s just that n is for natural numbers and x is for reals usually in terms of notation

[fast comet]

I see, but what I'm confused about is that L'hopital's rule has to be applied to a fraction. And both 2n and ln are in the numerator here

[native shard]

because we need 0/0 or inf/inf

as it stands there we have inf * 0

so i rewrite 2n as 2/(1/n)

because then as n -> inf that goes to 0

so i can use 0/0

[fast comet]

Oooooookay, now I see it

[native shard]

nice

[fast comet]

I will try to do it in paper, and then take a picture

[fast comet]

Why so?

[native shard]

simple rational function limit

$\lim_{x \to \infty} \frac{-2x^2}{x^2 + 2x + 1}$

[boreal girder]

knief

[native shard]

divide coefficients

[fast comet]

Ah of course

The coefficients with the term of higher grade

Forgot to consider expanding (x+1)^2

[fast comet]

So, what we were doing this whole time was finding the limit of the exponent of e?

[native shard]

yea

i think this way is clearer than e^ln

[fast comet]

But the e is always there, right?

And when you take the limit of a power, you take the limit of the base and leave the power as is

In this case, you took the limit of the power and left the base as is, as far as I understood

This is what I mean

[native shard]

i mean

eh

yes and no

[native shard]

i took ln of both sides

then after evaluating what ln L was i found L

like it’s the same thing

but i think my way is clearer

[fast comet]

Oh wait, I just saw it. After you got ln L = -2, you raised both sides to e

[native shard]

yea

[fast comet]

No, the other way around

But you get what I mean

Yeah, now I understand

[native shard]

nice

[fast comet]

I'll check everything briefly

[fast comet]

I did this derivative, and I'm getting a different result. Not sure what I'm doing wrong, I applied derivative of a product since I got x(x+1)^-1

I prefer to do that rather than derivative of a quotient

[native shard]

what happened here

[native shard]

quotient rule is the product rule after combining terms into one fraction

[fast comet]

Is the minus multiplying both x and (x+1) ?

[native shard]

$\frac{1}{x + 1} - \frac{x}{(x + 1)^2} = \frac{x + 1 - x}{(x + 1)^2}$

[boreal girder]

knief

[native shard]

and you get 1/(x + 1)^2

[fast comet]

Oh, damn

[native shard]

what you wrote there made zero sense

how did x get into the denomiantor

and what happened to the other term

and why is there (x + 1)^-1 in the numerator and (x + 1)^2 in the denomiantor

i have no clue how you got that

[fast comet]

Let me redo it

I rewrote it, it ended up being just a mishap from my part

[native shard]

no worries but you’re also missing an x here

it’s - x/(x + 1)^2

[fast comet]

Changed

[native shard]

why do you have an extra x + 1

everywhere

[native shard]

just go right here

$(x + 1)^{-1} + (-x(x + 1)^{-2}) = \frac{1}{x + 1} - \frac{x}{(x + 1)^2} = \frac{x + 1}{(x + 1)^2} - \frac{x}{(x + 1)^2}$

[boreal girder]

knief

[native shard]

then combine the fractions to get this

x cancels leaving you with 1/(x + 1)^2

[fast comet]

Wait a minute, you don't invert the -x too?

[native shard]

no

[fast comet]

Ahhh, right

It's two different factors, I saw

Damn

If it were a whole parenthesis it'd be another thing

[native shard]

don’t use product rule for this tbh

it makes it more confusing and more work

[fast comet]

I was trying to do this with the cross method, but I just got more confused

Then I just multiplied 1/x+1 by (x+1)/(x+1), and did it immediately

Good freaking damn...

[fast comet]

Alright, I finally understood everything

Damn, took a while

Mostly due to me getting confused over things 😂

You helped me a lot, thank you

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[tulip abyss]

is this correct i think it's missing a a in the in part

[grave field]

meant to type sin part

?

[tulip abyss]

yes

[dense dagger]

Russian textbook ass statement

Yes, you’re missing an a, but I think the +/- in sin should be -/+ no?

[grave field]

agree

i think this is from a trig handout for the usamo?

or usajmo

it had a lot of typos

so just double check everything

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[serene blade]

im confused

[stark marsh]

n[g(x)]^n-1
means you need to keep the inside function the same

[rugged glen]

U can't just multiply the 4 without expanding (x^2+3)^3

it stays outside

[serene blade]

oops

so 8x(x^2+3))^3

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[serene blade]

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[scarlet raft]

I can’t figure this out on how to make them have the same inside numbers, can anybody help? Step by step please.

[floral fjord]

How is cube root of 25=5?

@scarlet raft

Do you know prime factorisation?

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[scarlet raft]

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✅

[scarlet raft]

Wdym prime factorization

Like finding prime numbers

?

[scarlet raft]

5x5 is 25

I took notes earlier, and I just applied this to the question I’m having trouble with

<@&286206848099549185>

[compact pewter]

@scarlet raft Has your question been resolved?

[bold kayak]

your work there looks perfectly right

[scarlet raft]

can u tell me what I did wrong?

[bold kayak]

cube root of 25 is not 5, nor is cube root of 9 -> 3

those are square roots

[eternal pulsar]

Do you understand the concept of a cube root?

[scarlet raft]

yes I do

[scarlet raft]

what are the cubed roots then

[eternal pulsar]

Okay, so what are they?

[scarlet raft]

well cubed of 25?

I got no idea tbh

I might have to restart my work

But cubed roots are just number equal to number multiplied 3 times it self

Like 8 cubed is 2x2x2

or just 2

[bold kayak]

you seem to mix up "cubed" and "cube roots"

2 cubed is 8, 8 has the cube root 2

[scarlet raft]

what’s the difference

also it was cubed root of 250, so I think it would be 125 x square root of 2 with 3 or smth forgot what it was called

rather than 25x10 like I did

[bold kayak]

"square root of 2 with 3" is actually "cube root of 2"

"x cubed" and "cube root of x" are the inverses of each other

if x cubed is y, then cube root of y is x

it's like "twice x" and "half of x"

[bold kayak]

that's right

[scarlet raft]

I see

[scarlet raft]

How can I find the cubed root of 81

[bold kayak]

split it like you did with 250

find a different split than the 9 and 9 you did already (that one doesn't work)

[scarlet raft]

Does 27 and 3 work

They don’t got same radical thingy inside number

[bold kayak]

27 and 3 is perfect

[bold kayak]

you mean that $$\sqrt[3]{2}$$ from the first term mismatches $$\sqrt[3]{3}$$ of the second term?

[boreal girder]

vhj

[scarlet raft]

Is this good so far?

[bold kayak]

yes, good

[scarlet raft]

yeah u see the cubed root, that’s the same but that inside number isn’t so we can’t combine at all

so we can’t do this

[bold kayak]

we can do it, it'll just be more than one term

nothing wrong with that

[scarlet raft]

also how do I find 125 square

so I can make it like 4(parenthis number) cubed root of 2

[bold kayak]

what you have there is the cube root of 125, not 125 squared

[scarlet raft]

cuz 125/2 ain’t doing anything

[scarlet raft]

ohh

[bold kayak]

look for a number such that x * x * x = 125

[scarlet raft]

Ohhhh

It’s 5 lol

Right?

[bold kayak]

yes!

[scarlet raft]

And for the other radical it’s 3

[bold kayak]

yes

[scarlet raft]

D?

[bold kayak]

yes, it's d

[scarlet raft]

Ohh Tysm

I understand it better now

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[ancient kiln]

how is 90+θ allied with θ?

[ancient kiln]

90+θ+θ = 90 + 2θ

if you let θ =21 then
90+2θ = 90+42
=132

which is not a multiple of 90

[kind crane]

132 is not relevant here

$\th = 21$ and $\th + 90$ are

[boreal girder]

riemann

[ancient kiln]

but I just disprobed their statment by giving a counter example

[kind crane]

find $\th + 90$ and compare it to 21

[boreal girder]

riemann

[kind crane]

your "counterexample" isn't one because you're not calculating the correct values in the definition

[ancient kiln]

wait I think I misunderstood the defination then. I thought "90 + θ" is allied to θ meant that I have to add θ to "90+θ"

[kind crane]

no

[ancient kiln]

so it meant that 90 + θ is the allied part with 90 and θ?

[kind crane]

the two angles in your question are $\th$ and $\th + 90$

[boreal girder]

riemann

[kind crane]

in here

and their difference is indeed a multiple of 90 degrees

[ancient kiln]

did you perhaps mean to say +θ and -θ here?

or no?

[ancient kiln]

ok. in that case the two angles 90+theta and theta are allied angles

and the difference of 90 + A and A is:

(90 + A) - A
= 90

which is a multiple of 90

but that's the differece not the sum, which is what I'm asking

(90+A) + A is NOT allied angle, correct?

[inland nacelle]

They said

Sum or difference

[ancient kiln]

alright. it said 90 ± A allied with theta so i assumed that ± here meant that it should be both plus and minus in case of 90+theta and theta

otherwise it would've said 90 - A allied with A which would've made more sense

do you agree?

well either way, I got it.

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[ancient kiln]

thanks everyone!

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[dusk isle]

how would you solve this in an easy and simple way

[whole flax]

You can say there's 4x ml of orange juice, 3x ml of pineapple guice and 2x ml of guava juice

then you can try and find x, and add everything up

[kind viper]

is that to say that you know how to solve it but your way is difficult and complicated?

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[dusk isle]

no

[kind viper]

ok

[kind viper]

click ❌ on bot's prompt @dusk isle

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[feral jewel]

ABCD is kite , |BC| = |DC| what is the value of x? I need help please

[compact pewter]

@feral jewel Has your question been resolved?

[bleak dock]

what have you tried?

[feral jewel]

bro i have tried to make a rectangle

like this

but it doesnt work

[bleak dock]

hmmm maybe try drawing BD, which must be perpendicular to AC?

I honestly don't know

[feral jewel]

ty

[bleak dock]

wow really?

yeah you do get similar triangles from that, but I was probably a step or two short from actually figuring it out myself

oh wait gah

[bleak dock]

np! if you're done type .close

[feral jewel]

okay

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[maiden plover]

I have a question to the monotonicity of graphs. Sorry if the terms are wrong I do maths in German. So for example x^3 has a “terrace” point so the derivative of x=0 is 0 but it continues up all the time apart from that so it is strongly monotonously increasing. But x^7 for example, its graph has a visible line from x is -0,5 until 0,5 so our teacher said it is not strongly monotonously increasing but only monotonously increasing. Although the f(x) is never really the same in those points, but has a tiny difference. Would x^7 be still strongly monotonously increasing anyways because the y amount is always bigger?

[eternal pulsar]

For x^n, n odd, as you increase the n the behavior around 0 becomes more and more "straight"

So I guess you could say that, but you have to define strength

[maiden plover]

Okay thanks

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[worldly wind]

Can anyone help me with --alpha beta gama

[quartz yoke]

Whats the question?

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@worldly wind Has your question been resolved?

[kind viper]

no question was ever posted

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[quartz yoke]

I don't understand why 0/0 is undefined where it means what number do we multiply 0 with TO get 0 again,Since any number we multiply 0 with..the result is always 0 so shouldn't 0/0 be ANY real number?Like 0 times 6 is 0 and 0 times 25 is 0 too.I'd appreciate if someone can clear it for me.

[lime juniper]

something can't be more than one number at the same time. you're right that it's undefined for a different reason from 1/0

0/0=0=1 implies 0=1 which is false

[quartz yoke]

Ah I see..SO since 0/0 equals multiple numbers..this implies that those numbers are equal as well which is wrong..I see it's clear to me why it's undefined.Thank you very much!

[eternal pulsar]

0/0 = quantum particle

[quartz yoke]

How is 0/0 a quantum particle?

[eternal pulsar]

It was a joke

[kind viper]

not the place for jokes

[eternal pulsar]

:(

[abstract bolt]

you mean ;(?

[quartz yoke]

that's fine I'll close it now.Thank you everybody

.close

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[forest token]

🤣🤣🤣

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[vale halo]

I thought that, P(U sub 2) would be the sum of all outcomes for U at week 2 eval.

So I thought it should be
P(U sub 2) = 0.8 * 0.8 + 0.2 * 0.6
instead of
P(U sub 2) = 0.8 * 0.8 + 0.2 * 0.4
what am i missing

[kind crane]

are you asking specifically about the second equation

P(U2 | B1) = P(up to date week 2 | behind on week 1) = 0.6 and not 0.4 so you're right

[vale halo]

so the textbook is just wrong? @kind crane

[kind crane]

Yes typos exist. Google the errata for the book

[vale halo]

ok thank you

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[slim crystal]

am i right with A here?

[eternal pulsar]

Yes

[slim crystal]

thank you so much

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[warped oar]

???????

[compact pewter]

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[stone jackal]

\subsection*{41) $\begin{aligned}
&3x^2 + 2y^2 - z = 15 \
&(x, y, z) = (2, 2, 5)
\end{aligned}$}
$$f(x, y, z) = 3x^2 + 2y^2 - z - 15$$
$$\nabla f = \left<6x, 4y, -1\right> = \left<12, 8, -1\right>$$
$$\lm \nabla f \rm = \sqrt{12^2 + 8^2 + 1} = \sqrt{209}$$
$$\cos(\theta) = \left| \frac{-1}{\sqrt{209}} \right|$$
$$\theta \approx -1.6400230529$$

[boreal girder]

@stone jackal

[stone jackal]

I'm supposed to find the angle of inclination of a tangent plane to the surface at the given point. Can someone check my work?

[kind crane]

,calc acos(1/sqrt(209))

[boreal girder]

Result:

1.5015696006589

[kind crane]

not sure how you're getting a negative

,calc 12^2 + 8^2+1

[boreal girder]

Result:

209

[kind crane]

the rest looks right

[stone jackal]

I subtracted pi.

[kind crane]

why do you need to do that

[stone jackal]

It made sense to me because the surface’s z value reduces x increases, implying that the plane should be going down.

[stone jackal]

Is not needed? Should I just blindly apply the arccos function?

[kind crane]

i still don't see why you're subtracting pi

what does "surface's z value reduces x increases" mean

additionally, your angle doesn't even satisfy the equation before it

,calc cos(-1.6400230529) - 1/sqrt(209)

[boreal girder]

Result:

-0.13834289274244

[kind crane]

,calc cos(1.5015696006589) - 1/sqrt(209)

[boreal girder]

Result:

4.5602410736478e-14

[kind crane]

,tex .reflect trig

[boreal girder]

riemann

[kind crane]

maybe you were thinking about one of those?

,tex .shift trig

[boreal girder]

riemann

[kind crane]

or those?

[stone jackal]

No, that's not what I was thinking.

[stone jackal]

I'll draw it out.

This is the angle I would have found by simply applying arccos to both sides.

[stone jackal]

Instead, it made more sense to me that the plane was going in this direction:

[kind crane]

if that's the unit circle, the cosine of the black ray is negative while the cosine of the red ray is positive so i don't see how it can satisfy the cos(theta) >= 0 requirement

[stone jackal]

Why is cos(theta) >= 0 a requirement?

[stone jackal]

Also, haven't I satisfied the requirement by subtracting pi?

[kind crane]

absolute value bars means the right side is nonnegative

[stone jackal]

Oh, I see how I messed up now.

\subsection*{41) $\begin{aligned}
&3x^2 + 2y^2 - z = 15 \
&(x, y, z) = (2, 2, 5)
\end{aligned}$}
$$f(x, y, z) = 3x^2 + 2y^2 - z - 15$$
$$\nabla f = \left<6x, 4y, -1\right> = \left<12, 8, -1\right>$$
$$\lm \nabla f \rm = \sqrt{12^2 + 8^2 + 1} = \sqrt{209}$$
$$\cos(\theta) = \left| \frac{-1}{\sqrt{209}} \right|$$
$$\begin{aligned}
\theta &= \arccos(\frac{1}{\sqrt{209}}) \
&\approx 1.5015696007 \end{aligned}$$

[steel nest]

yo guys pppllllllssssssss dm me help me to understand powers must talk frensh but its ok if its english

[boreal girder]

@stone jackal

[stone jackal]

!occupied

[compact pewter]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[stone jackal]

If there's anything else I need to know, please ping me.

.close

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[slim crystal]

am i right with C here?

[slim crystal]

<@&286206848099549185>

[sage swallow]

Approximately, yes

[slim crystal]

tysm

[pine osprey]

!done

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[slim crystal]

am i right with D here? i got 10.945 and the closest was D

[slim crystal]

am i doing something wrong

<@&286206848099549185>

[sage swallow]

25 cos 47° t = 13
t=13/(25 cos 47°)

y=6.3 + 25 sin 47° 13/(25 cos 47°)- 16 13²/(25 cos 47°)²
≈ 10.939

[sage swallow]

You are getting away from the real answer because you've put in approximated values in y, to get the most accurate result always approximate at the last step

Error always adds

[slim crystal]

ohhh okay i see

yeah i messed up there

[slim crystal]

ill take note of that from now on

[sage swallow]

Just keep it in mind, let the calculator do the work

[slim crystal]

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[slim lodge]

to get tan i need to get dy/dx

am i right to say

$\ y\left = (1+\cos\theta)$ and y' would equal $(1+\cos\theta)\sin\theta - \sin\theta\cos\theta$

uhhh

idk how to use latex

[boreal girder]

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

[sage swallow]

Write it in pen and paper and share

[slim lodge]

uh

but its right there

[eternal pulsar]

Parentheses, possibly?

[slim lodge]

?

[eternal pulsar]

Aaa

$y=r\sin(\theta)$

[boreal girder]

;(

[eternal pulsar]

“Message failed to send” fuck is discord talking about

[slim lodge]

oh yea

[eternal pulsar]

Looked a little unreadable but I get it now

[boreal girder]

jesko

[eternal pulsar]

Ain’t NOBODY memorizing that

[sage swallow]

True

I'll memories the first line

[slim lodge]

isnt y suppose to be y=(1+cos(theta))sin(theta)

i assume that dy part on top

is js the deirvative

[eternal pulsar]

Yep

[slim lodge]

i can get its derivative by product rule right

[sage swallow]

Yeah, but that is rather tedious, I'd just plug in r, theta in this

[slim lodge]

is it that tedious? i mean derivative of (1+costheta) isnt just -sin

[sage swallow]

You have sin theta multiplied too

[slim lodge]

so thats why this formula exists

[sage swallow]

If you can remember it great, but I personally can never

[slim lodge]

ill note it down

[sage swallow]

I'll simplify it for u w8

[slim lodge]

its ok

nw

i understand

[slim lodge]

what is dr in this case?

-sin?

[sage swallow]

Remember this one

[sage swallow]

To avoid that, I've sent...

[sage swallow]

This

[slim lodge]

oh

i see

[sage swallow]

You can find out dr/dtheta

[slim lodge]

thx

.close

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[river wasp]

Could someone jog my memory pls

[sage swallow]

SohCahToa

[river wasp]

OH YEAH

Some old hippies caught another hippie tripping on acid

[tight badger]

Or you can use the characteristics of a 30-60-90 triangle.

[river wasp]

SOHCAHTOA

[river wasp]

Isn't that like multiply 5 by 2

Making y = 10

@dense dagger it's a phrase to remember the acronym

[tight badger]

Wait no, I mixed up the long and short leg.

[dense dagger]

weirdest mnemonic ive ever heard

[river wasp]

Whar

Oh I see

Wait

Why is it multiplying by √3

[tight badger]

Never mind. Was right the first time. Hypotenuse is double the value of the side across from 30 degrees.

[dense dagger]

Notice that the outer triangle is equilateral, so you can get the side length ebtween the 30 deg and 60 deg angle. (you dont even need this actually) Afterwards, you can work backwards from the properties of a 30-60-90 triangle to get x

[river wasp]

So y is 10

[tight badger]

Yes.

[river wasp]

Wait so how would I get x with the same property

[tight badger]

Well, if you have the hypotenuse, you divide by 2. If you have the long leg (i.e., the side across from 60 degrees), you divide by (sqrt(3))

[river wasp]

I have both so ill do the hypotenuse and double check with the sqrt

10/2 = 5

What

How can x be 5

That's wrong

@tight badger

[dense dagger]

I think you're confusing the definition of x

[tight badger]

x is 5.

[river wasp]

I really doubt

[river wasp]

Huh

[dense dagger]

Do you mean the x in this

or your question?

[river wasp]

My question bruh

[tight badger]

The side across from 60 degrees is x(sqrt(3)).

[river wasp]

My question chat

Let's focus on my question

[dense dagger]

Yeah, @tight badger is referring to the x in the 30-60-90 triangle (like my photo), while youre referring to the triangle in your question

[river wasp]

So chat

How do I find x

If we've already determined y = 10

[dense dagger]

Draw your triangle, and my triangle side by side

the length of the side between the 90 deg and 60 deg angle is "Something"

Accoirding to the rule in my triangle, the hypotenuse must be 2*Something

and the other side must be sqt(3)*Something

[river wasp]

2 x 5 which is the adjacent

[dense dagger]

It's not the adjacent, it is the hypotenuse

side "y"

[river wasp]

5 is the adjacent I mean

[dense dagger]

adjacent to what

[river wasp]

It's just the adjacent

[tight badger]

In terms of 60, yes.

[river wasp]

All triangles are split into adjacent opposite and hypotenuse

[tight badger]

You can prove the value of x (in your diagram) with SOHCAHTOA if you want also

[dense dagger]

the term "hypotenuse" is universal, but the term "adjacent" and "opposite" is not, the latter two must be defined relative to an angle

5 is the side adjacent to the 60 degree angle, but x is the side adjacent to the 30 degree angle

[river wasp]

X is 8.7

I think

Is it

[tight badger]

Yes, if you were instructed to round to the nearest tenths place.

[river wasp]

Well seeing as it's infinite I'm gonna yeah

[compact pewter]

@river wasp Has your question been resolved?

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[urban glen]

why is B wrong?

[urban glen]

answer key said D

[proven wren]

wait

[proven wren]

my bad, D is correct

you're (and i'm) confusing with derivative

notice that D has two extrema, and the original graph intersects with x-axis at two corresponding points

[cloud granite]

f is derivative of antiderivative

The sign of f is the monotonicity of antiderivative of f.

Determine where f is positive and negative, then compare with the graphs.

[compact pewter]

@urban glen Has your question been resolved?

[urban glen]

oh i thought this was graph of derivative

i get it

thanks

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[wary mulch]

I dont know why I got Q5 wrong

[compact pewter]

Please don't occupy multiple help channels.

[wary mulch]

x = 9/2 mb but still confused why i got it wrong

[tight badger]

-6/3 is not -3.

[compact pewter]

@wary mulch Has your question been resolved?

[wary mulch]

yeah I fixed that but my answer is still incorrect

[tight badger]

There is a considerable change after fixing that error. What's your new graph look like?

[wary mulch]

Theres not much changes because 0 is not smaller than -2 as I used the test point (0,0)

[tight badger]

,rccw

[tight badger]

The x and y-int need to be altered for the new equation of y = (2/3)x - 2, as they are not the same values when the equation is y = (2/3)x - 3.

@wary mulch

[compact pewter]

@wary mulch Has your question been resolved?

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[carmine garden]

An equivalence relation on $S$ is determined by the subset $R$ is determined by the subset $R$ on the set $S \cross S$ consisting of those pairs $(a,b)$, Such that $a \sim b$. write the axioms for an equivalence relation in terms of the subset $R$
\
\begin{enumerate}
\item reflexivity : The Relation is reflexive if $a \sim a \implies (a,a) \in R \forall a \in S$
\item symmetry : The relation is said to be symmetric if $a \sim b \implies (a,b) \in R \implies (b,a) \in R$. In short if $(a,b) \in R$, then $(b,a)$ must be in $R$
\item transitivity : if (a,b) and (b,c) in R, then (a,c) must be in R
\end{enumerate}

[boreal girder]

What a wonderful world !

[compact pewter]

@carmine garden Has your question been resolved?

[kind viper]

you definitely wrote that wrong

also where is your question

[carmine garden]

the top

[carmine garden]

An equivalence relation on $S$ is determined by the subset $R$ is determined by the subset $R$ on the set $S \cross S$ consisting of those pairs $(a,b)$, Such that $a~b$. write the axioms for an equivalence relation in terms of the subset $R$

[boreal girder]

What a wonderful world !

[kind viper]

\sim for a tilde symbol btw

~ produces a non-breaking space

[carmine garden]

An equivalence relation on $S$ is determined by the subset $R$ is determined by the subset $R$ on the set $S \cross S$ consisting of those pairs $(a,b)$, Such that $a \sim b$. write the axioms for an equivalence relation in terms of the subset $R$
\
\begin{enumerate}
\item reflexivity : The Relation is reflexive if $a \sim a \implies (a,a) \in R \forall a \in S$
\item symmetry : The relation is said to be symmetric if $a \sim b \implies (a,b) \in R \implies (b,a) \in R$. In short if $(a,b) \in R$, then $(b,a)$ must be in $R$
\item transitivity : if (a,b) and (b,c) in R, then (a,c) must be in R
\end{enumerate}

[boreal girder]

What a wonderful world !

[carmine garden]

<@&286206848099549185>

.close

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[slender ferry]

the denominators should be 9 and 36 why is it 3 and 6

[bleak dock]

$\sqrt{\frac{1}{3^2}} = \frac{1}{3}$

[boreal girder]

south

[slender ferry]

oh my god im so fucking dumb

.close

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[misty jewel]

Plz help

I got no idea how to start

Like the lowest point is 0.5 but the starting is 1

The highest point is 4

I just don’t know how to graph it

[bleak dock]

a few mistakes, so yes 4 m and 0.5 m are correct

the midline is (4 + 0.5)/2 = 2.25

also her lowest point is at 1.5 s, so that means you should stretch your curve out more

[misty jewel]

So I should start my point at 2.25

?

Something like this?

[bleak dock]

,rotate

[bleak dock]

yep that's correct!

[misty jewel]

Is that how you set the equation?

[bleak dock]

just be careful of this starting bit

[bleak dock]

wait you didn't follow instructions: that's only 1.5 periods

highest to lowest = 1 second = 1/2 a period

so 4 seconds, say 0 to 4 s, would be 2 periods

[bleak dock]

you've mixed up a and d in a cos(b(x - c)) + d

[misty jewel]

Emmm?

[bleak dock]

should be 1.75 cos(b(x - c)) + 1.25

[misty jewel]

That’s my friend answer

Did he do it right?

Because I don’t know how the 2.25 come from

[bleak dock]

the amplitude is 1/2 * (max value - minimum value)

,calc 1/2 * (4 - 0.5)

[boreal girder]

Result:

1.75

[bleak dock]

wait yeah I messed up the mental maths

[misty jewel]

But why +2.25

[bleak dock]

yes, that's correct

the period is 2 so b = 2pi/period = pi

[misty jewel]

I don’t get it shouldn’t it add 1?

[bleak dock]

2.25 + 1.75 = 4
2.25 - 1.75 = 0.5

your task is to find d, a such that d + a = 4 and d - a = 0.5

[misty jewel]

Ohhhhh

I c

[bleak dock]

d is the midline, a is the amplitude

so adding both equations, 2d = 4.5

[misty jewel]

So if cos =1 is 1.75

[bleak dock]

subtracting both equations, a - (-a) = 3.5

[bleak dock]

yes, 1 is the max value of cos and -1 is the min value

same is true for sin

[misty jewel]

4.5?

[bleak dock]

yes

2d = 4.5

[misty jewel]

4-0.5/2?

[bleak dock]

that's this other equation

[misty jewel]

Should the highest minus the lowest

Ohh

My bad

So a is the like where mid line add a is the highest or lowest

[bleak dock]

d is the midline

a is the amplitude

[misty jewel]

And in this question 2 sec is one period

So we need pi

Because 2 pi /2

[bleak dock]

yes

[misty jewel]

And we move x-0.5

Because 0.5 is the highest point

We start from X=0

[bleak dock]

actually your friend's answer is also correct

both give the exact same graph

it's a direct sine graph

but it's also a cos graph moved 0.5 units to the right

[misty jewel]

So in all the case if we wanna shift( just making sure), we start no matter what x=0

To shift the graph

[bleak dock]

well, it depends if you start with sin first or cos

[misty jewel]

Why does that matter? Both start at 0 right, the matter is that how much it shift

The shift is different

[bleak dock]

no

cos(0) = 1 but sin(0) = 0

[misty jewel]

X=0

What I mean is

Not y

[bleak dock]

I know

you have to start with a base sin or cos graph

so the shifts for the sin graph and the cos graph will be different

[misty jewel]

Phase shift

[bleak dock]

yes

[misty jewel]

Yeah, but like both start at x=0 no matter what right

[bleak dock]

no

I mean it depends

what I'm talking about is

[misty jewel]

But this both start at x=0

[bleak dock]

I know

[misty jewel]

I think what u mean is the phase shift will be difference because cos start at 1 and sin start at 0

[bleak dock]

yes, the phase shift for sin and cos is different

[misty jewel]

What I try to said is if like I wanna shift a graph is either sin or cos, I always start with x=0 from that point I start phase shift

Like x=0 not y

[bleak dock]

which graph are you using then

sin or cos

[misty jewel]

Ok is getting kinda confusing

I wanna phase shift

[bleak dock]

instead of doing this

[misty jewel]

But where should I start my phase shift?

[bleak dock]

look at the maximum of the graph

and then look at the maximum of the cos function, let's choose cos then

[misty jewel]

Should be at x=0 and then I see my highest or lowest point to phase shift to that point

[bleak dock]

yes, the maximum of cos is at x = 0

[misty jewel]

Ohh I see what you try to say

[bleak dock]

and (one of) the maximum of the graph is at x = 0.5

so it's a shift right by 0.5 units

so there's an (x - 0.5)

specifically, pi * (x - 0.5)

[misty jewel]

Do you think we can use sin in that problem?

[bleak dock]

yes

the phase shift is 0 actually

for sin

[misty jewel]

This ?

[bleak dock]

yes

[misty jewel]

Okok

Last double check

D is the middle line

And a is the distance between the middle line and the highest or lowest line

Is there other thing I need to knows

?

[bleak dock]

you already know that b = 2pi/period, which is nice

working in radians of course which are the natural units for all trig functions

the other thing you need to remember is to pick the maximum on the graph, and know the maximum of the original sin or cos graph, then compare for the phase shift

you could also compare minimum with minimum

[misty jewel]

Okok, tysm

❤️❤️❤️

[bleak dock]

no worries!!

[misty jewel]

.close

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[deft tangle]

how can I solve 789135mod1001

[deft tangle]

,w 789297 mod 1001

[boreal girder]

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[deft tangle]

,w 789135+162

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[inland nacelle]

@deft tangle you done?

[deft tangle]

not yet

actually i am solving this one

so 1000000mod 1001 is 1

so it will be 789135+162 right?

,w 789135000000mod1001

[trim plume]

yes

[deft tangle]

so answer should be 509

[trim plume]

seems right to me, yes

[deft tangle]

i see

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[boreal girder]

polar

[turbid folio]

idk where to go from here

[boreal girder]

polar

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[weak warren]

im a little confused on the base case

how can i make sure k isnt greater then one?

because it just breaks if k is

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@weak warren Has your question been resolved?

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@weak warren Has your question been resolved?

[weak warren]

got it lol

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[solemn karma]

help im so confused

[solemn karma]

what do i do with the 4

[latent ridge]

do you know the formula for area of trapezium?

[solemn karma]

No, whats that?

[eternal pulsar]

$\frac{a+b}{2}\cdot h$ where $a$ and $b$ are the bases and $h$ is the height of the trapezoid.

[boreal girder]

;(

[slate sand]

Double the trapezium, flip it upside down now you have a parallelogram

Cut the side up so it’s straight then move it to the other side

[solemn karma]

ohh

[slate sand]

You find that the side length is base 1 + base 2

It’s a rectangle so multiply by the height

But there’s of them so divide by 2 at the end

[solemn karma]

so would my base be 4???

and i just do 4+4/2 x height

[slate sand]

There are 2 base numbers

[solemn karma]

OHH

okay yes tysm i get it now

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[bleak canopy]

I have no clue where to begin

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[serene blade]

woiuld this be an acceptable way to solve this type of question?

[kind viper]

the way you write your sevens is wonky

and... well, i guess it's okay, but it isn't free of risk.

[native shard]

$\rangle$

[boreal girder]

knief

[serene blade]

hehe

[kind viper]

for more complicated graph shapes it could well happen that the tangent line intersects your curve at other points than the point of contact

better to find where dy/dx = the slope of your line

and then check all such points for whether they lie on your line

[serene blade]

so your saying that wont always work out as nicely?

if i were to do it that way dy/dx=-7?

then wym check all points

@kind viper

[kind viper]

yes

[kind viper]

find all x at which dy/dx = -7

then for each of those values of x check whether the y value of the curve and the line are the same

[serene blade]

What would I need to do to find all x at which dy/dx = -7

[kind viper]

work out dy/dx first

[serene blade]

On the 63/x-4?

[kind viper]

yes

take the derivative of that

[compact pewter]

@serene blade Has your question been resolved?

[serene blade]

for this i just plug in numbers for x for the equations to be equal so in this case -7

[kind viper]

you... haven't done what i wanted you to

you found y' = -63/(x-4)^2, now solve the equation -63/(x-4)^2 = -7 for x

[serene blade]

oh i just set them equal now

so just algebra from there

from there i plug in my x values

so either answer will work?

@kind viper

[kind viper]

no

you need to check whether (1, -21) and (7, 21) lie on the line y = -7x + 70

and you'll find that (1, -21) does not

[serene blade]

oh

thanks

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[bleak canopy]

Is this right?

[warped oar]

everything is basically given

[bleak canopy]

Nah i calculated everything based on these

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[strange pendant]

can I get some hints?

[compact pewter]

@strange pendant Has your question been resolved?

[devout latch]

what are you thinking?

imo i think you will have to compute H_1 perp with the k to verify when it will be contained in H_2 and find k there

[strange pendant]

H1 perp = <(1,0,k,2)>

1 - 0 -k + 2 = 0

k = 3

[strange pendant]

I am trying to figure out the dimension of T

T is dim 2

[devout latch]

yeah because dim(H_2) = 3 and dim(H_1 perp) = 1

[strange pendant]

yeah and also bec they are in oplus

H1 perp oplus T

lets find S n H2

(a+b,2a+b,-a-2b,b)∈ S

H2 = {x1-2x2-x3+x4=0}

(a+b)-2(2a+b)-(-a-2b)+b=0

a(1-4+1)+b(1-2+2+1)=0

-4a+2b=0

-2a+b=0

b=2a

(a+2a,2a+2a,-a-2(2a),2a)∈ S n H2

SnH2=<(3,4,-5,2)>

,w nullspace{{1,2,-1,0},{1,1,-2,1}}

[boreal girder]

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[strange pendant]

Sperp = <(3,-1,1,0),(-2,1,0,1)>

S = {3x1 - x2 + x3 = 0, -2x1 + x2 + x4 = 0}

,w nullspace {{3,-1,1,0},{-2,1,0,1},{1,-2,-1,1}}

[boreal girder]

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[lime juniper]

hi!

[strange pendant]

i made a mistake somewhere

[lime juniper]

what were you looking for with Sperp, whats the purpose

[strange pendant]

I was checking if my intersection is correct

but its not

[lime juniper]

but Sperp isnt used in the problem at all

[strange pendant]

is hard to explain but

I was trying to get the cartesian equation of S

[lime juniper]

why?

[strange pendant]

if you are given S = <(1,2,-1,0),(1,1,-2,1)>

[strange pendant]

because I wanted to check if I found S n H2 correctly

[lime juniper]

ah

why

[strange pendant]

because i think i made a mistake when finding SnH2

[lime juniper]

im not sure why S cap H2 is something you want to know anything about

[strange pendant]

is very simple because

dim(SnT) >= 1

[lime juniper]

yes

[strange pendant]

And

T ⊂ H2

[lime juniper]

yes

[strange pendant]

so if T is entirely contained in H2

and SnT is non trivial

particularly SnH2 is nontrivial aswell

what i mean is, maybe we can use the vector of the intersection SnH2 for the basis of T

[lime juniper]

aah

ah, that's smart!

good shortcut

I didn't see that shortcut

https://tenor.com/view/high-five-gif-16062267483909963763

and what did you find?

[strange pendant]

but i messed up the intersection

[lime juniper]

how do you know?

[strange pendant]

well I think so

[lime juniper]

what did you get for the intersection?

[strange pendant]

.

[lime juniper]

okay so lets go through this

S= span{(1,2,-1,0),(1,1,-2,1)}

H2 = span, uh, did you find a basis?

[strange pendant]

we didnt but for finding the intersection we dont need to

🥺

like what I usually do Is write S as a generic point in R4

[lime juniper]

right, sure

[strange pendant]

and make it satisfy H2 eq

[lime juniper]

so let's do that

except

for the basis of S

[devout latch]

i tried to redo your computation here and i got this

but i’m prone to errors be aware

[lime juniper]

rather

for a generic linear combination of the basis vectors

[strange pendant]

right, there is the mistake its -2a +2b = 0

[devout latch]

yeah i think

[strange pendant]

SnH2=<(2,3,-3,1)>

[strange pendant]

like in here

[lime juniper]

this is correct I thjink

a=b is also correct

[devout latch]

then it’s vector picking time with the conditions given

[lime juniper]

so your shortcut actually doesn't conclude anything! but it was an interesting idea

[strange pendant]

T = <(2,3,-3,1), t2>

t2 = (1,0,1,0)

[lime juniper]

hold up, do you know the dimension of T ?

[devout latch]

not in H_1 perp but in H_2

yeah

[devout latch]

yeah it’s 2 imo

[lime juniper]

how do you figure?

[strange pendant]

because the

grassman formula

[lime juniper]

H2=H1(+)T

dim H2?

[strange pendant]

H2 consists of 1 equation and 4 unknowns

4-1=3

[lime juniper]

good

dim H1?

[strange pendant]

well H1perp is dim 1

[lime juniper]

I meant Hperp

[strange pendant]

because

[lime juniper]

wait we still have
H2=H1(+)T

[strange pendant]

the normal of the plane is orthogonal to all the vectors contained in the plane

[devout latch]

H_1 perp (+) T = H_2 ?

[strange pendant]

yeah but we can see that H1perp is the coefficients of the cartesian equation of the plane H

[lime juniper]

I forgot to write down the perp ugh

thats why I was confused

ty

dim(H1)=3 so dim(H1 perp) = 1

[strange pendant]

yes because they are complementary subspaces

T = <(2,3,-3,1), (1,0,1,0)>?