help-23

[blazing cloud]

good so that missing white angle is 120

[lean otter]

Subtract from 120 - 60

[blazing cloud]

that works....

[lean otter]

Yes but my teacher keeps disconnecting a lot as he is in school and I’m in another country

[lean otter]

60 and C

[blazing cloud]

here u go

i am mistaken

its not complentary

i used wrong worg

con gruent*

LOL

front C word

heres a nice picture

thoes are congruent angles

the given 60 degrtees and the b are congruent

[lean otter]

Basically like the 60 and The C because it’s the same place and line

Like this let me draw

One min

Ohh

They are in the same side

The smae

They have same measures

Because smae side

Oh I see

Verticals angle?

They sometimes could be the same angle

But not all the time

And they are the oppise of the same line

Wait its always smae angle?

Yes !

Oh ( the reaction was by mistake)

It shares same vertex?

Because like u can see it

They are same messure ?

It’s 60 !

It’s 60

They all 60

Now I understand

TYPE

OH

They both veritcle too

Ohhhh

So every verticals has same messure

Gonna put this in my notes

Yea we started exams last week and now I’m trying to learn them

Alr alr ic

For Z it’s smaller

[safe radish]

@lean otter Has your question been resolved?

[gaunt gull]

She belong to the streets

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[sharp crane]

What is the formula for the surface area of a cylinder? What is the formula for the surface area of a sphere?

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[kindred oriole]

What do they mean by find the flux away from the x-axis through S

[sharp crane]

@kindred oriole Hmm I'd guess they mean in the direction of the negative x-axis maybe? Or do they mean perpendicular to the x-axis?

A sketch might clear it up

[kindred oriole]

Here is a sketch of the region in question

So just the red surface in those bounds using its normal?

[kindred oriole]

Or some other vector perhaps?

<@&286206848099549185>

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@kindred oriole Has your question been resolved?

[sharp crane]

@kindred oriole Pretty sure "away from x-axis" means along the negative x-axis

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[timber atlas]

I want to find the volume of this cylinder

[timber atlas]

I used the formula pi r^2 h

And also did 100/360 x that

[lyric ingot]

Sector area * Height

You know how to find sector area?

Using proportions?

[timber atlas]

Yes

[lyric ingot]

Then do so and multiply by 14

[timber atlas]

What do you mean by that?

[lyric ingot]

Never mind

If you can do it then do it

[timber atlas]

Ok I will try

[lyric ingot]

I will work here and we can compare steps

[timber atlas]

Ok

So I did 100/360 x pi r^2 x 14

[lyric ingot]

Not 100...

Look at the figure

[timber atlas]

260?

[lyric ingot]

Yes

[timber atlas]

So we take the angle that’s inside right?

[lyric ingot]

Yes

[timber atlas]

Ok I get it

Ok thanks for help

.close

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[timber atlas]

I need help with 7 b.)

[timber atlas]

So I found the volume which is 182.83cm^3

Now I converted cents to dollars

Times it by 24 to get for 1 day

And then divided the volume by the answer

But it’s incorrect

So what am I doing wrong?

[novel edge]

i think u need to multiply by the volume and the amount u found per day

[timber atlas]

Oh

That’s weird but I’ll try

[timber atlas]

It works

But why ?

[novel edge]

when you are multiplying, you are finding the amount of heat per hour (or day in this case) in the whole greenhouse. however when u divide, you have found the amount of money it costs per m^3 . something like that. Sorry, its a little hard to write out.

[timber atlas]

It’s ok but I understand

So basically when multiplying we are finding the amount of heat and when dividing we are finding we find the amount of money per m cubed

[novel edge]

in this case, i believe so

[timber atlas]

Ok I see

Ok thanks

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[long oracle]

Hello, i was trying to wrap my head around how the inverse of polar coordinates was derived, and the same for spherical.

I figured i might first start easy but cant really seem to wrap my head around what the procedure to solve for a,b,c in this instance are. Wolframalpha doesn't give a step by step solution :(
Thanks.

[long oracle]

Also would there be a name for this sort of stuff? Tried googling but didnt find this type of coordinate sorts of transformations from cartesian to whatever it is and back again.

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[long oracle]

Nvm im just stupid lmao

.close

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[safe radish]

@desert radish Has your question been resolved?

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[azure turret]

the problem is Let one of the zeros of a function be x1. In terms of a, b, c, d, and x1, what are the other zeros of

the function h(x) = ax^3 + bx^2 + cx + d?

[mossy bramble]

Note:
If x1 is a zero of f, then (x-x1) is a factor of f.

[azure turret]

so what would the other zeroes be and how would I show my work?

[mossy bramble]

Since you have a cubic, you can write it out as (x-x1)(ix^2+jx+k)

Solve for i j and k in terms of a,b,c, d, and x1, and then you can solve the quadratic.

[azure turret]

i’m not sure how to solve that

[tame charm]

Ok think of it like this

ax^3 + bx^2 + cx + d = (x-x1)(a quadratic)

You’re trying to find “a quadratic” and then you can find its roots in terms of the other variables

How do you find the quadratic? You can divide (ax^3 + …) by (x-x1) using synthetic division or whatever method you prefer

@mossy bramble is this right

[mossy bramble]

I was thinking more like expand RHS and set up system of equations but I think that works too. It's just a matter of preference.

[tame charm]

Ah

[azure turret]

so what would be the end zeroes? would I need to solve for each variable

[tame charm]

Which route are you taking

Division or the ijk thing

[azure turret]

division sounds easier

[tame charm]

Ok

[azure turret]

either way I have never done this before so idk why my teacher gave it to me

I only know how to do it with a given x and equation

[tame charm]

You do know how to do it

You’re just using letters instead of numbers

[azure turret]

ohhh

[tame charm]

Do you know how to divide (ax^3 + bx^2 + cx + d) / (x - x1)

[azure turret]

I think so

also the x1 in the start of the equation would that count as x or x^1

[tame charm]

It’s $x_1$, just a fixed variable similar to writing $c$

[flat frigate]

abs_0

[tame charm]

It has basically nothing to do with the variable x

[azure turret]

ok

I meant x sry

[tame charm]

Oh

[azure turret]

it autocorrected to c

[tame charm]

Wait you’re talking about the “x1” in “x - x1” right? Or no

[azure turret]

from this

the given zero

[tame charm]

Ah, yeah x1 is the given zero

[azure turret]

would that count as x or x^1

[tame charm]

It’s not x or x^1

Think of it like a completely new symbol

[flat frigate]

abs_0

[tame charm]

(x - Δ)(quadratic)

[azure turret]

okay

that makes more sense

thanks abs_0 for ur help

[tame charm]

Np but

Did you divide the stuff?

[flat frigate]

abs_0

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@azure turret Has your question been resolved?

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[marsh walrus]

im struggling with a complex integral

[marsh walrus]

$\int _C (1 + \bar z) \dd z$ where $C$ is the line from 2 to $2i$

[flat frigate]

jan Niku (Shuri for Honorable)

[marsh walrus]

so parameterization is easy but

theres a weird part of where this function is not nicely behaved that were crossing right

do we have to do something special like split it around that?

or is it alright since were not crossing the real axis?

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[mystic wedge]

I need help figuring out what kinda derivatives these are

[tall bough]

constants

[mystic wedge]

Also this one

Power right?

Or quotient derivatives

[tall bough]

thats power rule

[arctic lagoon]

pls

[mystic wedge]

Oki thx

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[arctic lagoon]

i just want to know all the derivability rules

[mystic wedge]

Derivatives?

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[urban gale]

I subbed:

sqrt(4 tan(a))^2 +16/ 4 tan(a)

and got sqrt(4 sin(a)/cos(a))^2 +16/ 4 sin(a)/cos(a)

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@urban gale Has your question been resolved?

[urban gale]

<@&286206848099549185>

[charred flax]

you're pretty close, you just need to combine the fraction (try to view 16 as 16/1 and try to combine it with 16 (sin/cos)^2

[neon mason]

Hello I was wondering how I would solve this:

[urban gale]

this channel is already used

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[timber atlas]

How would I do this question?

[timber atlas]

I know polynomial division but then it gets complicated

[novel edge]

can you show what you have?

@timber atlas

[timber atlas]

Yes

So far I did this:

But then I’m getting confused

[novel edge]

first off, your 2x should be above -3x. the "x"s should always line up

[timber atlas]

What do you mean I don’t understand

[timber atlas]

Why should it be above the -3x I’m dividing by the first one

[novel edge]

when you place your two, put it above the -3x and not the 6x^2

the power of your x should match

[timber atlas]

Wait can you show me what you mean

I divided the 6x^2 by 3x which gave me 2x, hence I placed it above the 6x^2

BRB in about 10min

[novel edge]

I also just looked more into this, I don't actually think it is 100% vital, and it won't really impact your answer, but it will make it look more clear. (i was just taught to line everything up) but no worries if you have a different style since it won't change the final answer.

[safe radish]

@timber atlas Has your question been resolved?

[timber atlas]

Oh ok

[timber atlas]

So are you saying basically this?

[novel edge]

yes

i do it because it makes it easier to see what the answer will be in terms of how many coefficients and such but honestly do however works for you

[timber atlas]

O ok

Ok yh so I was up to here

Now what do I do

Ok I got it actually

Thanks

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[heavy osprey]

anyone help

?

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@heavy osprey Has your question been resolved?

[heavy osprey]

<@&286206848099549185>

[royal kiln]

wtf does the first bullet mean? is that a typo? can you verbalize it?

oh...Expected value

a fool is I.

[heavy osprey]

haha

[royal kiln]

(I was just looking at a logic problem)

[heavy osprey]

LOL

fairs

Yea I am a bit confuddled

if the cov (u, L) I would do it in a breeze

but its cov (e , L)

dont know how to connect the dots

im afraid

[royal kiln]

what is the Law of Iterated Expectations? That sure looks important.

[heavy osprey]

its basically

1 random variable = 1 variable conditioned on another variable

right

i have a question

can i post my working?

and you can affirm whether its right or wrong?

oh god knows

at this point tbh

[royal kiln]

no, lol. I can't analyze this

stats isn't my strongest subject

there's some niche formulas I don't have memorized

including the definitions of variance, covariance, etc

<@&286206848099549185>

[heavy osprey]

sadgems

[heavy osprey]

helloooo

is it me your lookin foorrrr

i can see it in your eyess.

[safe radish]

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[nova elm]

Is my answer correct or wrong pls tell

[tall bough]

it's correct

[nova elm]

Ok thanks

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[lean otter]

hi

[lean otter]

can somebody help me what can I do here in this problem

the integral of secx(tanx^2 +1)dx

kinda stuck

[amber hamlet]

is it tan(x²) or tan²x

[lean otter]

yes

[amber hamlet]

which one lol

[lean otter]

the second one

[amber hamlet]

tan²x + 1 = sec²x

a trig pythagorean identity

[lean otter]

then it will become secx(sec^2(x))dx

??

[amber hamlet]

are you sure you have the problem written down correctly

ohh

nvm

[amber hamlet]

this is right

so it becomes sec³x

[lean otter]

i know something is wrong with this one tbh, im actually doing an integral by parts problem and I ended up in that equation

[feral linden]

It equal (1/cos^2)(tan^2+1)dsin=((sin^2/(1-sin^2)+1)/(1-sin^2) dsin

It’s a rational function

[lean otter]

hmmm thanks

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[quasi bison]

do you want an equation that defines the red region, an inequality that defines the red region, or do you just want to find its area?

@teal frigate

[teal frigate]

ah sorry, just the equation

not the inequality

[quasi bison]

that's strange because an inequality is more sensible to write down

but if you insist

[teal frigate]

yeah

nevermind, wrong question xd

.close

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[quasi bison]

okay i was going to do some malicious compliance and give you something full of max() and min()

but i could not get that to work

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[somber plaza]

Can I get some help with this question?

[safe radish]

@somber plaza Has your question been resolved?

[somber plaza]

<@&286206848099549185>

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@somber plaza Has your question been resolved?

[velvet quiver]

uk the differential for decay?

[somber plaza]

?

[velvet quiver]

there's a differential equation for radioactive decay

[somber plaza]

hm

[velvet quiver]

ok so $\frac{dN}{dt}=\lambda N$

[flat frigate]

1234567

[velvet quiver]

where N is the number of radioactive particles

and lambda is a constant called decay constant

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[somber plaza]

.close

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[visual swift]

do I do constructive counting?

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[safe radish]

@alpine ravine Has your question been resolved?

[alpine ravine]

<@&286206848099549185>

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[normal wing]

Quick Question about the Mean Value Theorem

[normal wing]

Does the theorem find the average slope from point a to point b?

interval [a,b]

Here is an example I was trying to understand

blue is from point a to point b

green = derivative of f(x)

[shy temple]

Yes. If f is continuous on [a,b], differentiable on (a,b), then there is a c in (a,b) with f'(c) = f(a)-f(b)/(a-b), which is the slope of the line segment from A to B, which is the "mean value" of the slope

[normal wing]

Thank you. So it is NOT the slope of the blue line?

blue line = point a to point b

it is Instead the average slope from a to b?

[shy temple]

the slope of the blue line is actually the average slope from a to b, P.S below assumes f' continuous, but this need not be the case

Mean value of f'(x) on [a,b] is $\dfrac{\int_a^b f'(x)}{b-a}$

[flat frigate]

1345631

[shy temple]

which is precisely what the MVT gives you

[normal wing]

oh, The slope of blue line = 1?

so the average slope is = to 1?

[shy temple]

looks like it, yes

[normal wing]

Ok so when I find c = \frac{2\sqrt{3}}{2}

[shy temple]

Try to solve f'(x)=1. You will find c in (-2,2), so that verifies the theorem

[normal wing]

c =

and negative of that

then I plug c into f'(x) but that = 6

[shy temple]

should be 2sqrt(3)/3

[shy temple]

you made an error in the denominator

[normal wing]

oops

Thank you

Oh, ok now I understand.

@shy temple Thank you so much! Now I understand this concept

[shy temple]

Glad to help! You're welcome 🙂

[normal wing]

Have a great day!

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[lean chasm]

oh no! get an ambulance! it's dying!

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✅

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@lean otter Has your question been resolved?

[tame charm]

@lean otter I think they’re wrong

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[limpid kayak]

Textbook says I should be getting 2 - √3 as my final answer? But.. yeah-

[novel edge]

what is the question?

[limpid kayak]

The first line

Just this "solve"

((well okay that's a partial lie. It's just a small part in a much larger question butive already worked through that.))

[limpid kayak]

I just kinda need to know if this can be brought to (2 - √3) somehow

This is the preceding part of the question; I deem it irrelevant though . It was " find the value of tan π/12 " (using trig function)

[proper crypt]

Multiply the denominator and numerator by sqrt(3)

That should be the first step

[limpid kayak]

And then...? Rationalize it?

[rocky gazelle]

i would only do that to the inner fractions
don’t have to do it to the whole thing, might make it more complicated

[limpid kayak]

Yeah that's it actually

I got the answer!

I don't think it gets simpler than that

Thank you!

[rocky gazelle]

get rid of the 1/3 by finding common denominators first

[limpid kayak]

.close

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[lament fern]

can someone help me with this simple trig equation please? I need full explanations on how to solve this

[stoic dune]

Convert sin to cos
You have a quadratic in cos, sub u = cos(x) to see it
Solve as a quadratic

[lament fern]

ahh okay thank you

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[tall storm]

doing long division w equations, should that 19 at the bottom be -19x or -19x^2?

[queen parcel]

Neither

They're unlike terms

[tall storm]

shit
u,m im gonna try to figure out what to do next LOL

wtf is my remainder gonna be? or my quotient..?

or do i just scoot the -5x over to 7x?

or wait should i just put a 0x^2 below -14x^2??

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[steel lark]

Why can the power reduced from 1801 to 1?

[winter pivot]

let me use an example.
7 mod 9 = -2
7^2 mod 9 = (-2)^2

[steel lark]

So does this mean that if a mod b = c, then a^n mod b = c^n?

[winter pivot]

yes.

[steel lark]

does it works for any integers a,b and c?

[winter pivot]

yes

[steel lark]

But 765432^1801 should then equal to 74^1801 mod 209?

[stoic dune]

If a = c, then raise both to the power b to get a^b = c^b

Doesn't change even in mod 209

Though, yeah I don't personally see how the reduction worked here

[steel lark]

I'm not so sure that I get it

so if I have some integer say a mod n

where a is large

how do I get the result for x = a mod n

[stoic dune]

So you might be seeing mod here as the programmer mod, where mod n is a function that takes an integer and returns an integer

[steel lark]

like a is 33^7

[stoic dune]

In math it's far more common to see (mod n) as an algebraic structure

[steel lark]

I'm actually studying algebra

[stoic dune]

So for example, 7 = 10 (mod 3) is a true statement

[steel lark]

the explanation I have for mod is if a = b mod n, then a and b have the same remainder upon division by n

[stoic dune]

Yeah I kinda got on a tangent there, and the tangent was potentially unrelated

So one example is something like 11^1000 (mod 120)

11² is 121, so that's really just 1^500 = 1 (mod 120)

[steel lark]

hmm, how do you get the 1^500 thing

oo wait is it 1000/2

[stoic dune]

All in (mod 120):
11^1000
= (121)^500
= (1)^500
= 1

Something similar definitely happened with your question, but I can't quite put my finger on it?

[steel lark]

so you can calculate 121/120= 1*120+1 first

and then replace 121 by 1 the remainder there

[stoic dune]

Yeye. 121 is 1, so you can freely swap them

[winter pivot]

i feel like this has something to do with order

[steel lark]

I think I get it.

can you guys wait for a min longer to see is my understanding right?

[winter pivot]

if:
a mod b = c
then what is the smallest value of n such that:
a^n mod b = c?

oh right i should probably say that n>1

because the remainders of powers form a cycle don't they?

what do you call this

[steel lark]

so if I have 15^100 (mod 224), can I say that 15^100 = 225^50 = 1^50 =1?

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[weak whale]

Anyone can help me here?

[novel edge]

have u tried anything yet? any ideas?

@weak whale

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[spark yoke]

Find the minimum value of $$(x_1-x_2)^2+\left (\frac {x_1^2} {20}-\sqrt{(17-x_2)(x_2-13)}\right)^2$$ where $x_1$ is in $R^+$, $x_2$ is in $(13,17)$.

[flat frigate]

ron143

[safe radish]

@spark yoke Has your question been resolved?

[spark yoke]

<@&286206848099549185> ?

Yo all helper dead or what?

[worthy hemlock]

Probably, a good amount of people are from America so it's sometime between 9pm to 12am

[spark yoke]

Hmmm

[safe radish]

@spark yoke Has your question been resolved?

[safe radish]

@spark yoke Has your question been resolved?

[tall bough]

am dead

[woven kiln]

y = that
partial derivative
dy/dx1 = 2(x1-x2)
when that is zero
x1=x2
so min value=
when x1=x2 and the only term left is (x2^2/20)
so x2 = 13

(169/20)^2 is your answer

@spark yoke

I could be wrong

feel free to tell me if I am

oh wair

I missed the second term

the big one

NOOOO

I’m stupid as hell

[spark yoke]

But Answer key give n is $54- 20 \sqrt(2)$

[flat frigate]

ron143

[spark yoke]

Umm

[woven kiln]

yeah no, I MISSED THE CHUNKE TERM

[woven kiln]

that

[spark yoke]

But x_1 = x_2 is what we got right?

I don’t get what you missed

[woven kiln]

that’s, if the big bracket doesn’t exist,

that has x1 too

so you’ll get 2(x1 - sqrt(term)

too

[spark yoke]

Ohh umm 🤔

[woven kiln]

yeah, sorry man. I should have noticed. Instant realisation there

[spark yoke]

I don’t get it

[woven kiln]

which part?

[spark yoke]

.

[woven kiln]

partial derivatives?

[spark yoke]

Ya i guess

Can you systematically write the steps

Sry for such trouble

[woven kiln]

chain rule, oh wait 4x(x^2/20 -sqrt term)

yeah, I probs should

I’m making way too many mistakes

[woven kiln]

which topic is this again?

just to see if i’m at least on the right lines

[spark yoke]

Its one from application of derivatives

And before this we used tangent normals something…. Which was right before this problem

So maybe it has something to do with it

[woven kiln]

Can I see that question and answer please

[spark yoke]

Is copied was from board

In class

HW

And please don’t ask how I know answer, its funny.

My friend peeked into sir’s notes

[woven kiln]

cheeky😂

[spark yoke]

I think it maybe has something to do we tangent, normals and all that

[woven kiln]

do y’all do normal differentials or partial differentials?

I’m genuinely curious

For these

That’s why I wanted to see your question

as in previous question

[spark yoke]

General differentiation

Idk partial diff

[woven kiln]

ohhh

Can I please see your last question and answer ?

[safe radish]

@spark yoke Has your question been resolved?

[safe radish]

@spark yoke Has your question been resolved?

[rancid wagon]

this channel is occupied , use another available channel

[plucky elk]

.close

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[jade bloom]

How do you simplify these type of exponents so I have (x-5)^2k+5 over (x-5)^2k+4

[jade bloom]

wait i think i figured it out nvm

just split the exponents into parts and cancel [(x-5)^2k * (x-5)^5] / ](x-50^2k * (x-5)^4)] = x-5

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[regal ivy]

I never rly get how to do these type of questions

[ancient escarp]

have you tried anything

[regal ivy]

yes

I differentiated it and I set y=41

but that’s probs wrong and I don’t even know what I am doing

[ancient escarp]

do you know how to find a tangent line?

[regal ivy]

like the equation of it?

yes

[ancient escarp]

that would be a solid start

[regal ivy]

find the gradient and plug it into the y=mx+c equation

[ancient escarp]

not quite

well i guess you could

but it's easier to plug into point slope form

whatever works for you

[regal ivy]

okok

so the change of y/change of x?

[ancient escarp]

that's not how you're gonna wanna get slope here

you can differentiate

you want a tangent line at x=3

[regal ivy]

wait so can I set x=3 and solve for y?

[ancient escarp]

in your original equation, yes

[regal ivy]

So 7=7

Y*

then now my equation would be y=-7x-45

[woven kiln]

ok, give me equation of dy/dx?

[woven kiln]

yep

[regal ivy]

6x^2-18x+12

then i can plug in 3 for x

[woven kiln]

when x = -1

[regal ivy]

oh

[woven kiln]

oh

nvm

you’re right

[regal ivy]

i thought that was the x in mx+c

so my gradient would be 12?

[woven kiln]

it’s x=3

[regal ivy]

okok

[woven kiln]

what’s y when x =3

[regal ivy]

so now -41=12(-1)+c

[regal ivy]

12

[woven kiln]

no, that’s something you have to show my friend

[woven kiln]

you gotta use x =3 and this one

[regal ivy]

oh

[woven kiln]

I hope you got the right values

[regal ivy]

emmm

i will try again

6(3)^2 - 18(3) +12

so that would be 6 * 9 - 18*3 +12

54-54+12

wait or did i get a step from

wrong*

[woven kiln]

that’s dy/dx

[regal ivy]

.....

i forgot

lol

[woven kiln]

and yeah, your dy/dx is right

[regal ivy]

would my y be -7?

[woven kiln]

idk man, just sub it in and tell me

[regal ivy]

ok

2(3)*3 - 9(3)^2 +12x -2

so 2 * 27 = 54

-81

+36

-2

[woven kiln]

54-81+36-2

[regal ivy]

e

wait why is it 27?

[woven kiln]

oh no

[regal ivy]

7

[woven kiln]

y = 7?

[regal ivy]

yea

[woven kiln]

looks like it

[regal ivy]

i put in calc

[woven kiln]

now do the eqn and find c

[woven kiln]

sweet

[regal ivy]

yep its 7

ok

so now i can do -41=7(-1)+c

-34=c

ok so now my straight line equation would be y=7x-34

but i do i show that the tangent to the curve when x = 3 passes through the point (-1,-41)

do i need to work out my curve using 1st derivative test?

[safe radish]

@regal ivy Has your question been resolved?

[regal ivy]

<@&286206848099549185>

[full zephyr]

You’ve made an error

Okay so let’s go back

You have your derivative

And you’ve set x = 3 to find the gradient of the tangent at that point

Now what?

Hint: you don’t know yet that the tangent passes through (-1, -41)

So what point do you know

Remember we want to find the equation for the tangent

[regal ivy]

I know my x and y

[full zephyr]

Which are?

[regal ivy]

actually I know both x1 y1 and x2 y2

right?

so I should use slope equation instead of mx+c

[full zephyr]

Give me the numbers

[regal ivy]

3- - 1/7- - 41

which is 4/48

wait what am I doing lol

[full zephyr]

Give me the two points

That you think you have

No calculation

[regal ivy]

-1, -41?

[full zephyr]

You said you had two?

[regal ivy]

and

3, 7

[full zephyr]

Alright

So think about the question

We want the tangent

But the question wants us to show (-1, -41) lies on the tangent

[regal ivy]

OH

[full zephyr]

Can we really assume (-1, -41) lies on the tangent, and then find the equation of the tangent that way?

[regal ivy]

-1 -41 is normal

line

meaning that it’s perpendicular?

[full zephyr]

(-1, -41) is nothing right now

[regal ivy]

ok

[full zephyr]

It’s just a point we know nothing about

We don’t know it lies on the tangent yet, because we don’t know the tangent

The question doesn’t tell us it does, we’re supposed to tell the question it does

Don’t assume what you’re trying to show basicallt

basically*

[regal ivy]

ok

hm

[full zephyr]

So what do we know, we do know that (3, 7) lies on the tangent

Think about ways of finding the eqn of a line

[regal ivy]

what is eqn

[full zephyr]

equation

[regal ivy]

m1 * m2 = -1?

[full zephyr]

Why?

[regal ivy]

since we have gradient of one surely we can find out the other one and work out from there

[full zephyr]

‘The gradient of one’

The gradient of what?

[regal ivy]

wait

I mean

-1, -41

which the gradient being 7

[full zephyr]

(-1, -41) has no gradient it’s a point

[regal ivy]

…………

[full zephyr]

Do you mean the line connecting (-1, -41) and (3, 7)?

[regal ivy]

yea

[full zephyr]

But remember we don’t know anything about (-1, -41) yet

Don’t use it

[regal ivy]

can I use ax+by+c?

[full zephyr]

So the two standard methods of finding the equation of a line

1. Two poinys

points*

2. Gradient and a point

We only know one point

So try the second method

How can we find the gradient of the tangent at x = 3?

[regal ivy]

substitute the 3 into the derivative and work it out

[full zephyr]

Okay so do that

[regal ivy]

6(3)^2 - 18(3) +12

which is 12

[full zephyr]

Alright

So now find the equation of the tangent

[regal ivy]

so now I can put the gradient into mx+c=y

[full zephyr]

Yeah

[regal ivy]

so x=-1 and y = -41

[full zephyr]

Or, what is probably better to learn, $y - y_1 = m(x - x_1)$

[flat frigate]

dk.dkn

[full zephyr]

Why?

Where are you getting (-1, -41)?

We want the eqn of the tangent

[regal ivy]

I think this is where I am confused

I don’t know what to use

[full zephyr]

The gradient is 12

[regal ivy]

yea

[full zephyr]

And (3, 7) lies on the line

[regal ivy]

OH

we have 2 x and y

so

the change of y / change of x =12

[full zephyr]

No

We have 1 x and y

We know nothing about (-1, -41) until we show it

We have 1 x and y and the gradient

No other points

[regal ivy]

What

I am very lost

then how can I use the eqn?

[full zephyr]

Have you learnt $y - y_1 = m(x - x_1)$?

[flat frigate]

dk.dkn

[regal ivy]

I’ve learnt y2-y1/x2-x1

=m

[full zephyr]

What I’m showing you there is the equation for a line if you know the gradient and one point on it

It works for the reason you just showed

$\frac{y - y_1}{x - x_1} = m \Rightarrow y - y_1 = m(x - x_1)$

[flat frigate]

dk.dkn

[full zephyr]

We know a point, (3, 7)

And the gradient, 12

Also note $(x_1, y_1)$ is the point, m is the gradient

[flat frigate]

dk.dkn

[full zephyr]

y and x are general y and x

[regal ivy]

So I can do for now y-7=12(x-3)

[full zephyr]

Yes

You can simplify it if you want

[regal ivy]

12x-36=6-7

y*

y-7

[full zephyr]

So y = 12x - 29

[regal ivy]

Yea

[full zephyr]

Um

Much better

Alright so NOW how can we show (-1, -41) lies on the line, given we know its equation?

[regal ivy]

plug x and y into the eqn

[full zephyr]

Yep

[regal ivy]

-41= -12-29

thx

.close

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[iron bolt]

just to confirm is the greater than or less than it's asking me to find equivalent to : (r, -theta) ?

[stable bobcat]

Can you maybe rephrase the question

[iron bolt]

hmm

[stable bobcat]

Like, what it's simply saying is that theta is negative and r is positive

[iron bolt]

"where r>0 and theta<0" is equivalent to (r, -theta)?

ahh okay

[dapper forum]

((theta, r)

[iron bolt]

would it be (2 , -5pi/3)?

[stable bobcat]

whats the question?

[tender island]

can someone help me in #help-13

[iron bolt]

oh my bad i thought i included the question

[stable bobcat]

dont do that

[dapper forum]

I was just about to say that

[stable bobcat]

yeah

[tender island]

sorry

[iron bolt]

https://cdn.discordapp.com/attachments/903481106819612714/959179833324220436/Screen_Shot_2022-03-31_at_12.58.10_PM.png
https://media.discordapp.net/attachments/903481106819612714/959178009091407892/Screen_Shot_2022-03-31_at_12.50.20_PM.png

i got -5pi/3 by identifying -2pi/3 which is 4pi/3 but then since it's a -2 radius it means its flipped over? so it would become pi/3 which in a negative degree it's -5pi/3?

[safe radish]

@iron bolt Has your question been resolved?

[safe radish]

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[gray estuary]

What is the length of a language containing only the empty string?

[gray estuary]

.close

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[sharp stone]

how do i find the image F[U]?

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@sharp stone Has your question been resolved?

[sharp stone]

<@&286206848099549185>

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.close

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[spark yoke]

Find the minimum value of $$(x_1-x_2)^2+\left (\frac {x_1^2} {20}-\sqrt{(17-x_2)(x_2-13)}\right)^2$$ where $x_1$ is in $R^+$, $x_2$ is in $(13,17)$.

[flat frigate]

ron143

[plucky elk]

did you take a derivative?

[spark yoke]

Umm how to take derivative

Like with respect to what

[tame charm]

Partial derivative @spark yoke

Although that looks miserably ugly

But it would work

[plucky elk]

I guess a change of variables could be helpful since you have a sum of squares

s=x1-x2 and t=(...) Other stuff being squared

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[viral current]

anyone know what im doing wrong

[granite compass]

I think you plugged x(0) = -1 instead of x(0) = 6.

[viral current]

sry u mind screenshotting where

oh nvm

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[wild thorn]

How come the domain here is x is Greater than or equal to 0

[wild thorn]

And not x cannot equal 1

I mean. To get rid of the square root, u would square both sides right

You’ll get x = 1

[drowsy moss]

what not?

[rocky shadow]

Why would u wanna square both sides

Even if u rationalize by mutltiplying the function by $\dfrac {\sqrt{x}-1} {\sqrt{x}-1}$ the domain would remain the same

[flat frigate]

Glory

[rocky shadow]

Bc u can't plug in negative numbers into a square root

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[waxen chasm]

The last 2 bits of this proof are confusing me

[waxen chasm]

I tried to figure it out but can’t

So my textbook says Ax = lambda x

Why does P^-1 Ax become lamba P^-1 x

[feral linden]

Your second last sentence implies your last sentence……

[waxen chasm]

Thanks cogwheels

But how

I’m just not seeing it

Sorry I’m dumb

To clarify I meant why does lambda move around? Is this possible

[plucky elk]

lambda is a constant. constants can "move around"

[waxen chasm]

Thank you

And because Ax=lambda x, we have P^-1 Ax = P^-1 lambda x?

[plucky elk]

yes, $Ax = \lambda x$ is the definition of eigenvector with eigenvalue $\lambda$

[flat frigate]

riemann

[waxen chasm]

Thank you so much!

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[waxen chasm]

Now that I think I understand the proof, is this right? It’s practice problem (a)

Actually this doesn’t seem right at all

Especially 1

I just proved the same thing for 1 and 2 so that takes me nowhere

[plucky elk]

what's the solution and what's your work?

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[waxen chasm]

.reopen

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✅

[waxen chasm]

I got the same answer for both and it kind of makes sense but at the same time I’m worried I just ran in a circle

But there’s no other way

Or at least I don’t think there is

[plucky elk]

i dont' even know what work is yours

[waxen chasm]

What do you mean

The first one is similar to an example in my notes i just went through more steps

[plucky elk]

"i got the same answer for both"

which one is even your work

[waxen chasm]

Both

I meant that I ended up at the same place for both matrices

I know they’re similar so that might make sense

But I don’t really understand if I’ve even answered the question

I think I proved every eigenvalue of A is an eigenvalue of B

I think that’s impossible though

So this proof is probably wrong

[plucky elk]

that's exactly what the problem tells you to show

[waxen chasm]

Wow I need coffee

Thank you 😅😅😅😅😅

I just realized

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[lament robin]

<@&286206848099549185>

[drowsy moss]

what have you tried?

also • After 15 minutes, feel free to ping @Helpers.

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

[woven kiln]

bro, do you see that the difference is a constant?

so $A_n = a + (n-1)d$

or $A_n = mn +c $

both of them work just as fine

[flat frigate]

Rylo

[woven kiln]

n is the term number

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[ancient summit]

Can someone explain how the answer is derived

[empty tulip]

are you familiar with rules of logarithms?

[ancient summit]

Yeah kinda

[empty tulip]

run me through what you know as of now

[ancient summit]

loga-logb = loga/b

loga+logb=logab

logab=c, a^c = b

logab = logcb/logca

bringing down the power of the log

think thats about it

[empty tulip]

okay looks good. So what part of the problem is tripping you up?

[ancient summit]

when i subject y i get a different answer to the real answer

how does loga/log5 = log base 5 of a

[stray socket]

That's a logarithm rule

[ancient summit]

oh

ok

.close

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[solemn vault]

Hi

[solemn vault]

How do I solve this

The

C number

One

Question

The question with (4, 5) coordinate

[sharp tiger]

can you write 2 points form of equation?

[solemn vault]

Ye

You mean

The

[sharp tiger]

assume x intercept to be some value 'a' then your 2 points for the line will be (a,0) and (0,2a)

[solemn vault]

Double intercept

One

??

[sharp tiger]

then you can write equation with these 2 points

[sharp tiger]

yea

[solemn vault]

Ye

So

Its

$$\frac{x}{a} + \frac{y}{b} = 1$$

[flat frigate]

BASU

[sharp tiger]

yes

[solemn vault]

Ok

[sharp tiger]

then the line has to pass through 4,5 so you can substitute that point and get value of a

[solemn vault]

Ahh ok

Ig

$$\frac{4}{a} + \frac{5}{b} = 1$$

[flat frigate]

BASU

[sharp tiger]

you know that b = 2a

[solemn vault]

Oh

[sharp tiger]

so write there 2a instead of b

[solemn vault]

Ye

Ye

[sharp tiger]

ez

[solemn vault]

Ok

[sharp tiger]

[solemn vault]

Got it

Thx

Bhai

.close

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