help-33

[rapid geyser]

you get x = -1

insert x into xz+z=0

[proven timber]

Or y = 0

[rapid geyser]

yes 0,0,0 is one solution

but when you put x in

you get -1(z+z)=0

wait

hold up

[marble glacier]

Hmmm

[proven timber]

-1(z) + z

[rapid geyser]

but that elimated z

cause z-z

[marble glacier]

Yes it would be z(-1 +1) so It be 0

[rapid geyser]

but 0=0 is no z=0 right

not*

[proven timber]

It seems there are infinite points, yes

[marble glacier]

Yea you’ve just proved nothin lol

[rapid geyser]

ohhhh

i see

but that still is fine as an answer

because its find CP's not max or min

[proven timber]

I don't know how they want you to express it

[rapid geyser]

I will just write x=-1 amd z=5/y when y is not 0 for real numbers

and the other ones

let me try to solve all them

got it

thank you

bout to submit another question lol

.close

[marsh citrus]

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[eternal epoch]

can anyone help pls thanks

[eternal tundra]

do you know the right hand rule?

that is how you get the directio

[eternal epoch]

yea ik rh rule

a n b are perpendicular vectors

and here they are asking for the value of vector not direction so idts rh rule shud be used

<@&286206848099549185>

[eternal epoch]

chill out m8 been an active member here ik the rules
u mind answering the qn ?

[marsh citrus]

@eternal epoch Has your question been resolved?

[marsh citrus]

@eternal epoch Has your question been resolved?

[marsh citrus]

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[fast shoal]

I need help solving this using algebra

[fast shoal]

$$16.5639=-4.4407sin\pi/12(1-h)+12.2332$$

[elfin berry]

JollySaintQuinnichalos

[glass silo]

Are you sure you've typeset that correctly?

Do you mean $$16.5639=-\frac{4.4407\sin(\pi)}{12(1-h)}+12.2332$$, or something else?

[elfin berry]

chartbit

[glass silo]

Like $$16.5639=-4.4407 \sin \left( \frac{\pi}{12(1-h)} \right) +12.2332$$

[elfin berry]

chartbit

[fast shoal]

$$16.5639=-4.4407\sin\frac\pi/12(1-h)+12.2332$$

[elfin berry]

JollySaintQuinnichalos

[glass silo]

(for fractions, it's $\frac{top}{bottom}$)

[elfin berry]

chartbit

[glass silo]

(don't forget the curly braces)

[fast shoal]

$$16.5639=-4.4407\sin\frac{\pi}{12}(1-h)+12.2332$$

[elfin berry]

JollySaintQuinnichalos

[fast shoal]

This is exactly it

You can distribute the pi in but I was taught to format it this from my teacher

[glass silo]

Does the sin apply to the (1-h) as well, or nah?

[marsh citrus]

@fast shoal Has your question been resolved?

[fast shoal]

I’ve simplified it into $.8220=\sin\frac{\pi}{12}(2-h)$

[elfin berry]

JollySaintQuinnichalos

[fast shoal]

I changed my points to (2,15.8839)

<@&286206848099549185>

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[hot meadow]

Hey math peeps

[hot meadow]

I understand the answer, but I'm having trouble proving it

I tried proving it by induction, but doing so requires proving that every (valid) tree with order greater than 2 has a nonleaf connected to two leaves such that two leaves can be deleted to form a smaller tree

[marsh citrus]

@hot meadow Has your question been resolved?

[crystal lintel]

i think you can get that with pigeonhole principle

[hot meadow]

Oooooo

I'm listening 😛

[crystal lintel]

first show that if a tree only has degree 1/3 vertices, then any degree 1 vertex must be adjacent to a degree 3 vertex
then if the tree has n/2 + 1 degree 1 vertices and n/2 - 1 degree 3 vertices, there must be a degree 3 vertex with 2 degree 1 vertices adjacent to it

[hot meadow]

Hmmmmmmm

I got it

Gimme a sec...

[crystal lintel]

but your original idea is good if i'm guessing all the details of where it came from correctly lol

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

Please don't occupy multiple help channels.

[nova totem]

.close

[rigid light]

bruh

[marsh citrus]

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[nova totem]

#help-31

[marsh citrus]

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[agile abyss]

interior angle sum for a triangle

[sturdy ridge]

^

was gonna say that'

[keen raven]

wdym

[somber holly]

Add those 3 angles up and equate them to 180

[keen raven]

but i need to find x

[somber holly]

10x-1+125+26=180, find x

[sturdy ridge]

you know it has to equal 39

10x-1=39

i believe

havent done that for 5 years

[ionic field]

x is 4

[keen raven]

would it be 29

?

[blazing imp]

I think you add the two given angles and subtract it by 180

U get a -29 a remainder of 29

10(3)-1 = 29

I think X is equal to 3

[keen raven]

hey

howd u get 3

whaa

[agile abyss]

(10x-1)+26+125=180
(10x-1)+26=55
(10x-1)=29
10x=30
x=3

[blazing imp]

You gotta get something equal to 30

Subtract by 1 and you get 29

10(30) = 30

30 - 1 = 29

Therefore X = 3

The reason why we focus on 180 is because triangles always add up to 180 degrees

[keen raven]

bc 30/10?

[agile abyss]

just do this algebra

[keen raven]

yea ik

[blazing imp]

(10x - 1) = 29 x will be 3 to equal 30

(30 - 1 ) 29

[keen raven]

now yall gotta help me wit 6

[blazing imp]

💀

[agile abyss]

send

[blazing imp]

Send it den

[agile abyss]

its the exact same thing

[blazing imp]

Do the same thing

Try doing it yourself then come back

[keen raven]

alr

[blazing imp]

Same shit bro add up the given angles

119

180 - 119 = 61

Figure out a value to X to get 61

[keen raven]

u givin me the answer

u so nice

[agile abyss]

||(5x+1)+58+61=180
(5x+1)+58=119
(5x+1)=61
5x=60
x=12 ||

[keen raven]

howd u get 12

😭

[agile abyss]

mb fixed it

its still twelve tho

[blazing imp]

How did you not get 12

[keen raven]

i didnt finish

[agile abyss]

wdym how did I get 12

you didn't finish

[blazing imp]

5(12) + 60 + 1 = 61

X = 12

[keen raven]

is 5x

like 5^12

[blazing imp]

No

[keen raven]

bc is it js 1

[blazing imp]

5 to the power of 12 is a whole different answer

[keen raven]

yea

[agile abyss]

5x = 5 x 12

[blazing imp]

5x would be 5(12) in order to get 60 and it it to 1

To get 61

[keen raven]

thats an exponent right

[blazing imp]

It’s not

[agile abyss]

no

[keen raven]

oh

[agile abyss]

5 x 12 = 60

[keen raven]

what grade yall in

[agile abyss]

5^12 = 244140625

[keen raven]

oh

alright thank u

[agile abyss]

what grade are you in

[keen raven]

8th

[blazing imp]

Is that all the questions u had

[agile abyss]

bro 😭

[blazing imp]

I’m a college freshman I forgot most of this

[keen raven]

u smart

u like uhhh

elon

[blazing imp]

This math is actually very easy

[keen raven]

musk

[blazing imp]

I suck at math so

If I can do this so can you

[agile abyss]

algebra

is there any other questions

[blazing imp]

I’m doing statistics rn

[keen raven]

naur

i have a D in math

[blazing imp]

Close the channel 💯

[keen raven]

n the grade book clloses tmr

[agile abyss]

its fine you're in 8th

[blazing imp]

💀 rip man’s

Yea

[agile abyss]

it only matters in highschool

[blazing imp]

It’s not a big deal

[keen raven]

na cu

it matter to me

🙁

[blazing imp]

Close the channel and we could talk in #serious-discussion or something

[keen raven]

shazam

[bitter terrace]

Type .close

[keen raven]

.close

[marsh citrus]

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[wary dome]

help please

[still temple]

hi

[wary dome]

hi

[still temple]

y2 -y1 / x2 - x1 = slope

6-2 / 3 - 1 = 4/2 = 2

understand??

[wary dome]

yes

.close

[marsh citrus]

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[cold flare]

Does plugging in N = 65 and 75 into dN/dt just tell you the instantaneous rate of change of N(t) when N(t) = 65 and 75?

[lilac bane]

Yep

[cold flare]

Thank you 👍🏻

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[marsh citrus]

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[lilac bane]

[marsh citrus]

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[pale crag]

Hi
f'(x)= (1)/(x²+3)
h(x)=f(3x)
What is the derivative function h'(x)

[strange hazel]

lemme try it

well its simple

[still temple]

my instinct is just to tell you to integrate and compose, but i have a feeling this is one of those dumb questions where there is some sort of 'trick'

[strange hazel]

yup

you have to integrate it

[fleet dagger]

f'(3x) * 3 isn't it just chain rule?

[still temple]

oh yeah just ftc, i guess, maybe

yeah

dont integrate

[strange hazel]

hi doggo

[still temple]

apply FTC

[remote elbow]

You’ll need f(x)

The integration is pretty simple

[strange hazel]

yup

integration is good way

[fleet dagger]

I don't see why f(x) is needed ngl

isn't it just chain rule

[vestal forge]

you definitely dont need integrals here (neither is ftc needed)

[strange hazel]

integrate the first equtation and put 3x

[remote elbow]

Oh yeah integration isn’t needed

[marsh citrus]

@pale crag Has your question been resolved?

[pale crag]

How?

[pale crag]

I didn't know how 😢

[fleet dagger]

h(x) = f(3x)

h'(x) = f(g(x)) where g(x) = 3x

since it's two nested functions, the derivative is f'(g(x)) * g'(x)

[wintry wave]

To find the derivative of h(x), we need to use the chain rule. The derivative of h(x) is given by:

h'(x) = f'(3x) * 3

Substituting the given value of f'(x), we have:

h'(x) = (1)/(3x² + 3) * 3

Simplifying this expression, we get:

h'(x) = (1)/(x² + 1)

This is the derivative of h(x).

There.

[pale crag]

Oh thank

So with chain rule we get
h'(x)= f'(x)*3?

[wintry wave]

Yes.

[pale crag]

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[warm bear]

Find b if the abscissa of the top of the parabola y = 2x^2 +bx+4 is 3.

[cloud iron]

By top does the problem mean extrema?

Because that parabola has a minumum but not maximum

[warm bear]

I need b, but not the maximum and minimum

[crisp mantle]

this parabola doesnt have a top point

it only has a bottom-most point

ie a minimum and not a maximum

[warm bear]

Can you explain?

[crisp mantle]

the x^2 coefficient is positive

so this is how ur parabola looks

it has a bottom-most point

the top most point is at infinity

as it goes up forever

[warm bear]

Thanks a lot. can you help me with one more question?

[crisp mantle]

i can try

np

[crisp mantle]

what's the question?

[warm bear]

At what value of the variable does the function y = -3x^2 + 12x - 5 have the greatest value?

[crisp mantle]

do u know how to differentiate a function?

[warm bear]

No.. 🥲

[crisp mantle]

ah

okay basically u differentiate this and equate it to 0

[crisp mantle]

find the value of x

[crisp mantle]

now if u substitute that x here, u get the answer

[crisp mantle]

do u know how to complete the square?

[warm bear]

Yeah

[crisp mantle]

okay do that

express it as a sum of a constant and a square

or negative of a square

[warm bear]

I'll try

[crisp mantle]

great

lmk how it goes

[warm bear]

i have problem
"Complete the square" what is this?
||I don't know English well and google can't translate properly ||

[marsh citrus]

@warm bear Has your question been resolved?

[crisp mantle]

its a method to solve this problem

[marsh citrus]

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[warm bear]

Thanks

[marsh citrus]

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[crisp mantle]

.reopen

[crisp mantle]

-3x^2 + 12x -5
= -3(x^2 - 4x) -5
= -3(x^2 -4x +4 -4) -5
= -3 (x^2 -4x +4) -12 -5
=-3 (x-2)^2 - 17

[warm bear]

💀

[wintry wave]

what

[crisp mantle]

so y= -3 (x-2)^2 - 17

now minimum value of (x-2)^2 = 0

if u take any other no., then u get y= -(that no.) -17 which is always less than 7

so the max value is -17

[warm bear]

💀

[dry spindle]

step 4 when u take -4 out it should be +12

cause its multiplied by -3

[crisp mantle]

ahh thanks my bad

[dry spindle]

oh wow its still not there

haha

[crisp mantle]

-3x^2 + 12x -5
= -3(x^2 - 4x) -5
= -3(x^2 -4x +4 -4) -5
= -3 (x^2 -4x +4) +12 -5
=-3 (x-2)^2 +7

Im getting +7 even with differentiation

[dry spindle]

dude

wait no

wait yeah hold on

you technically multiplied it by -3 in the inequality

wait

no

huh

wtf am i sayin

[warm bear]

[dry spindle]

yeah it should be 7

[crisp mantle]

by differentiation:
y = -3x^2 + 12x - 5
y' = -6x + 12 = 0
=> x=2
max value = -3(4) + 12(2) -5
= -12 +24 -5
=7

[dry spindle]

maybe the function itself has a typo

do u mind sending thr actual question @warm bear ?

[warm bear]

it is in Russian)

[dry spindle]

the math itself should be readable

we only need the function

[warm bear]

give me a second

[warm bear]

this?

[dry spindle]

yeah no typo

hm

dunno man the question might be wrong

[warm bear]

Really?

[dry spindle]

yeah

[warm bear]

it's a pity

At what value of a does the parabola y=x^2+ax+9 touch the x-axis?
and in this case?

[marsh citrus]

@crisp mantle Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[flat junco]

can someone help with dynamic programming?

[flat junco]

i dont get it what is optimal play

[main idol]

cs server in #old-network

[flat junco]

oh, i dont know that it existed. thanks a lot

.close

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[crisp mantle]

Again you gotta use differentiation
This question is the same as the parabola having a minimum value at y=0
Differentiating,
y’ = 2x + a = 0
=> x= -a/2
y=0= (-a/2)^2 + a(-a/2) +9
=> (a^2/4) -(a^2/2) + 9 = 0
=> 9= a^2/4
=> a=6

[marsh citrus]

@crisp mantle Has your question been resolved?

[marsh citrus]

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[fallen stump]

%x = (-0 +/- sqrt(0^2 - 4 * 0 * 9)) / (2 * 1)
x = +/- sqrt(0) / 2
x = 0 / 2
x = 0%

[fallen stump]

.close

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[meager gull]

is my answer correct on the laptop?

[north sparrow]

,rcw

[meager gull]

thats the solution

but i did it in another way

[meager gull]

here

[glass silo]

$2^6 e^3 = (2^{2} \cdot e)^{3}$

[elfin berry]

chartbit

[meager gull]

yea but my answer is correct too right?

[glass silo]

Yea, they're the same, because of my quote

You can rewrite yours in terms of theirs, no?

[meager gull]

so answers may vary in integration and all be correct?

[glass silo]

Yeah, they may look different at first glance but usually there's a way that you can rearrange one of them to get the other

[meager gull]

how do ik?

like alot of times i get different answers

but in the solution manual it’s different

[glass silo]

E.g., for your one:
$\frac{(2^6 e^3)^x}{\ln(2^6 e^3)} = \frac{((2^{2} \cdot e)^{3})^x}{\ln((2^{2} \cdot e)^{3})} = \frac{(4e)^{3x}}{3\ln(4e)}$

[elfin berry]

chartbit

[meager gull]

yes

[glass silo]

(Using rules of logs/powers)

Generally there should be a way you should be able to spot how to convert one solution into another using stuff like that, or, for example, if you're dealing with trig functions, the trig identities should help

One common example I say is that depending on how you go about it, $\int 2\sin(x)\cos(x) dx$ looks to have multiple different answers, but you can rearrange those answers into each other, if that makes sense?

[elfin berry]

chartbit

[meager gull]

yes i see

thank you

[glass silo]

No problem!

[meager gull]

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[meager gull]

is ln(1/x) = -ln(x)?

[meager gull]

if so how?

and why?

[viral lily]

a*ln(x)=ln(x^a)

[meager gull]

oh yes i just realize

d

[viral lily]

It's a property of logarithms

[meager gull]

-1 . ln (x)

alright thank you

.close

[marsh citrus]

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[marsh citrus]

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[edgy jungle]

Isn't the denominator 1 anyways

Since l2+m2+n2=1

[marsh citrus]

@edgy jungle Has your question been resolved?

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[manic timber]

please help

[cloud dawn]

You cannot find the square root of a negative number (because a negative number multiplied by itself once will not yield a negative number)

So you need to set all of these expressions greater than or equal to zero

[manic timber]

yes

ik

but answer?

[cloud dawn]

So what are you having trouble with?

[manic timber]

Getting an asnwer

for them

[cloud dawn]

Show me your work

[manic timber]

no i dont have any

i tried in my head alos i dont undrstand

it

I was supposed to go to a class but, I was sick so i missed the class

then they give US hoemworok

homwork*

[cloud dawn]

Gotcha

So go ahead and try it on paper

[manic timber]

I dont understand tho like what tod o

omg

im so sorry

i ntg

can i dm u

[cloud dawn]

Take what’s inside the square root, so x+5, and set that greater than or equal to zero, so $$x+5>=0$$

[manic timber]

and u help later?

[elfin berry]

Derpy

[cloud dawn]

Kek Idk how to do the > or = to sign

But then you just solve it algebraically

You can come back and tag me here I usually miss dms with this new dm system

[nova totem]

\geq

$a \geq b$

[elfin berry]

Umbraleviathan

[marsh citrus]

@manic timber Has your question been resolved?

[cloud dawn]

Appreciate it 👍

[marsh citrus]

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[wicked moon]

The question was asking for the x values of f(x) that makes the tangent line horizontal. Where did I go wrong?

[still temple]

looks fine

what are you concerned with?

[marsh citrus]

@wicked moon Has your question been resolved?

[wicked moon]

0 and 2pi don't have horizontal tangent lines

[still temple]

how did you get 0 and 2pi?

[marsh citrus]

@wicked moon Has your question been resolved?

[arctic vector]

do you know what is a horizontal tagent line to a graph

[wicked moon]

sin(x)=0

on the interval [0,2pi]

[still temple]

ah i see

you divided by sine here

so we need sin(x) to be nonzero

[wicked moon]

ohh

ok

[marsh citrus]

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[idle jetty]

i don't really get this one, how would I know which ratio it is , i know the first would be adj but idk whether the second is hyp or opposite

[idle jetty]

it's df by cd but i dont see why

[dry prawn]

what are "first" and "second"

[idle jetty]

since i need to find whether i should use sin/cos/tan

wants to know DE so i know it's adj by something else which i dont know

[dry prawn]

adj and opp are all relative to some angle

[idle jetty]

don't get it

[dry prawn]

what don't you get

[idle jetty]

i don't know why it's tan

[dry prawn]

DF is the side opposite the angle that measures 54 degrees

and since you know CD = 9m and CD is adjacent to the 54 degree angle, you use tangent to find DF

[idle jetty]

i see

.close

[marsh citrus]

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[dusty acorn]

just looking for clarity

[dusty acorn]

exhibit 1

exhibit 2

exhibit 3

can anyone see how exhibit 1 derived from 2 and 3

[marsh citrus]

@dusty acorn Has your question been resolved?

[marsh citrus]

@dusty acorn Has your question been resolved?

[marsh citrus]

@dusty acorn Has your question been resolved?

[marsh citrus]

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[hot mantle]

I made some math equations for this game

[hot mantle]

I’m trying to figure out whether or not 15% more critical hit chance does more long term damage than 15% more power gained per hit

The current crit chance is 9.25% which is also 37/400

[upper briar]

mmm

[hot mantle]

And the crit chance multiplier is times 5 (it does 5 times more damage than a normal hit)

I developed these equations

Is this right?

The first one’s for 15% more crit chance

The second is 15% more power

I wrote the equations assuming there were ten total hits

[upper briar]

um

so

power is just a universal multiplier?

then its just better

[hot mantle]

Well these are just my equations

I feel like there may be a flaw

[upper briar]

ye

games give u more bonuses

ltr on

so just

search a guide

[marsh citrus]

@hot mantle Has your question been resolved?

[hot mantle]

thats conditional

not applicable to this game ig

the percentage doesnt really matter

<@&286206848099549185>

[marsh citrus]

@hot mantle Has your question been resolved?

[hot mantle]

<@&286206848099549185>

[marsh citrus]

@hot mantle Has your question been resolved?

[whole sleet]

You care about the expected value. That is, the sum of (damage of a type of hit)×(probability of that type of hit)

@hot mantle

What's 10? Why's there a 1.15 on the critical hit?

[hot mantle]

10 is irrelevant

I just wanted to find the total damage after 10 total hits

[flat raft]

critical hit does 5x more damage, and 15% power just means 15% extra damage on all hits?

is that what you're asking?

[hot mantle]

Well when you put it like that

[flat raft]

well I don't know if it's true or ont

I'm trying to clarify how the game works

[hot mantle]

Yeah

[flat raft]

so 15% power means that you'll do 15% extra damage on any hit you make, critical or not?

[hot mantle]

I should’ve realized that

Seems so obvious now

Thx

.close

[marsh citrus]

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[azure kiln]

A model of the Moon has a radius of 12.5 cm. Find the volume of the globe (sphere).

[sturdy schooner]

Look up the formula for the volume of a sphere

[sturdy schooner]

Now apply it

[azure kiln]

apply the what?

[brisk scarab]

do they mean earth by globe

[bright jay]

You got help for this before

[sturdy schooner]

The formula

[bright jay]

Plug in the value given into the equation

[azure kiln]

i didnt get it, sorry

[azure kiln]

12.5 x3 ???

[bright jay]

No

$$V = \frac43 \pi \red{r}^3$$
and you are given
$$\red{r} = 12.5\text{cm}$$

[elfin berry]

dldh06
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[bright jay]

Plug in the value for r

[sturdy schooner]

,calc

4/3 * pi * 17^3

[elfin berry]

Result:

20579.526276116

[sturdy schooner]

There's an example assuming the radius is 17

[azure kiln]

4/3pi12.5^3

[sturdy schooner]

Yes

[azure kiln]

8181.231???

am i right

?

[sturdy schooner]

,calc
4/3 * pi * 12.5^3

[elfin berry]

Result:

8181.2308687234

[bright jay]

If you plugged in the value and used a calculator correctly

[azure kiln]

8181.23086872

[sturdy schooner]

Thats the answer

In centimetres

[azure kiln]

8181.23086872 cm???

[sturdy schooner]

Yes since you use cm as your dimension when you plugged it in

[bright jay]

Don't troll

[sturdy schooner]

Go Google it

[azure kiln]

ohh, okay thank you

[bright jay]

<@&268886789983436800> Got a troll

[sturdy schooner]

@azure kiln if you're done please type .close

[bright jay]

You shouldn't be doing this for everyone fyi, actually let the people do it. I know that the OP did do it, but if they honestly used a calculator and plugged in the number correctly then they should result in a correct answer

[azure kiln]

.close

[marsh citrus]

Channel closed

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[bright jay]

If they get some number not close to the answer, you can ask what they typed in and assist from there

[sturdy schooner]

Mhm

[marsh citrus]

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[dense crater]

Is ${v_1,v_2,v_3}={1,x,x^2}$ an orthogonal basis for $P_2(\mathbb{R})$? If yes, is the norm of $v_3$ equals to $||x^2||=\sqrt{\langle x^2,x^2\rangle}=x^2$?

[elfin berry]

Trenton

[dense hill]

What have you done so far?

[dense crater]

Let $V=P_2(\mathbb{R})$ be the inner product space of polynomials with degree at most $2$ equipped with the inner product $\langle\cdot ,\cdot\rangle$ defined as $\langle f,g\rangle=\int_0^1f(t)g(t)dt$ for any $f,g\in V$. Let $T:V\to\mathbb{R}$ be a linear functional given by

$$T(f)=f(0)+f'(1)$$

for all $f\in V$.

Find a vector $g\in V$ such that

$$T(f)=\langle f,g\rangle$$

for all $f\in V$.

So now I let $B={v_1,v_2,v_3}$ be an orthogonal basis for $P_2(\mathbb{R})$.

if we call $h(f(x))=\langle f(x),g(x)\rangle$, then we can say
$$h(f(x))=\sum_{i=1}^3 a_ih(v_i)$$

Since $B$ is orthogonal, we know $a_i=\frac{\langle f(x),v_i\rangle}{||v_i||^2}$.

Then
$$h(f(x))=\sum_{i=1}^3 \frac{\langle f(x),v_i\rangle}{||v_i||^2}h(vi)= \left\langle f(x),\sum_{i=1}^3\frac{h(v_i)}{||v_i||^2}vi\right\rangle$$
Therefore, $$g(x)=\sum_{i=1}^3\frac{h(v_i)}{||v_i||^2}v_i$$

[glass silo]

One thing is that the norm of an element should be a number

[elfin berry]

Trenton

[dense crater]

I see. So the norm of $x^2$ is just $1$?

[elfin berry]

Trenton

[glass silo]

Erm, let me check, you should have that $(x^2, x^2) = \int_{0}^{1} x^4 dx$?

[elfin berry]

chartbit

[glass silo]

Taking your inner product as per here

[dense crater]

oh yes

oh I see

I forget to consider that

that's why it looks weird

[glass silo]

For orthogonality, you'd want to see whether the inner product of any two distinct elements is zero?

Which I believe it isn't?

[dense crater]

Ok I don't think it is orthogonal too

But how can we construct an orthogonal basis for $P_2(\mathbb{R})$?

[elfin berry]

Trenton

[glass silo]

You can do better and get an orthonormal one, via Gram-Schmidt

You know that ${ 1, x, x^2 }$ is a basis by being linearly independent

[elfin berry]

chartbit

[dense crater]

yes

it is the standard ordered basis

[glass silo]

Hmm, I guess it is the "standard" basis, but then I haven't ever heard of "standard" basis in the context of polynomial or function spaces tbh 😂

[dense crater]

hmmm I am not sure lol. But this is one of the well-known example to get a basis for $P_2(R)$

[elfin berry]

Trenton

[dense crater]

Alright let me try to apply G-S process

[marsh citrus]

@dense crater Has your question been resolved?

[dense crater]

Alright so I have got ${1,x-1/2,x^2-x+1/6}$ after applying GS

[elfin berry]

Trenton

[glass silo]

Ah, cool cool, I'm taking your word for that as I haven't tried it 😂

[glass silo]

For that, you can just use it from there as per this?

Actually hang on, my notes have this!

[dense crater]

wow

noice

[glass silo]

You need to normalise the last element I think 😂

[dense crater]

oh yes

but seems in this question it doesn't matter

since any orthogonal basis will do

[glass silo]

Yea was about to say that, as long as it's orthogonal (which it is), then it's fine

That said, if you normalise, it makes your life easier as you know that this "collapses"

As the norm of each element becomes 1, so you just have to work out the $h(v_i)$'s

[elfin berry]

chartbit

[dense crater]

oh yes

[dense crater]

oh nice I got the answer to be $210x^2-204x+33$

[elfin berry]

Trenton

[dense crater]

Thanks a lot!!!

.close

[marsh citrus]

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[umbral kraken]

hello, can you help me?

[umbral kraken]

i am having a bit trouble but this is my work
11. (120 + 80)1/2 = 100
12. (70+50)1/2 = 60

i am not sure on what theorem should i use

<@&286206848099549185>

.close

[marsh citrus]

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[umbral kraken]

.reopen

[marsh citrus]

✅

[marsh citrus]

@umbral kraken Has your question been resolved?

[umbral kraken]

.close

[marsh citrus]

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[raw solar]

Slope of a Line question is this correct? i dont know what to do when im in this part now..

[raw solar]

By the way (-4, 3) and (1, -3) are the given

I suppose to find the slope of a line

[eternal epoch]

(-3-3)/(1+4)

-6/5

is sllope

[raw solar]

oh wait its suppose to be 5?

oof

the given is 1 - 4 though aren't i suppose to subtract

[eternal epoch]

1-(-4)

becomes 1+4

[raw solar]

Ahhh right

ok then!

So -6/5 is the final answer or is there something else i should do after?

[verbal trail]

actually

yeah your right

that’s it

[raw solar]

ok cool!

[verbal trail]

-6/5 is your slope

[raw solar]

Thank you both ='w'=

This is done now, have a nice day u 2

.close

[marsh citrus]

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[marsh citrus]

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[spare spruce]

How do I do Angle of elevation word problems?

[marsh peak]

It's helpful to draw a diagram when reading such problems

[spare spruce]

well yah

I drew one and it says I have to find how far is he standing from the base of the lighthouse

and im confused on how I solve this diagram

[marsh peak]

Can you show the problem and what you've drawn?

[spare spruce]

Casey sights the top of an 84 foot tall lighthouse at an angle of elevation of 58 (degrees). If Casey is 6 feet tall, how far is she standing from the base of the lighthouse?

[late geode]

the diagram is off

[spare spruce]

in number placement?

or like how

idk how to do any of this

[late geode]

do you think lighthouses float mid-air?

you should have something that represents the ground

[spare spruce]

thats how my teacher drew it

[marsh peak]

Also 58 degrees represents the angle measure, not a side length

[spare spruce]

so this?

[spare spruce]

so it goes inside the triangle?

[marsh peak]

Yes, at the angle on the left

[spare spruce]

like this?

[marsh peak]

No, the other acute angle

Wait let me draw the diagram

[late geode]

no

start with a horizontal line to represent the ground

[late geode]

no

try to follow my guided instructions

[spare spruce]

horizontal is straight isnt it?

[late geode]

well lines are straight...

horizontal as in flat

like this symbol -

[spare spruce]

Yah

[late geode]

start with a horizontal line to represent the ground
then towards the left side, draw a vertical line coming up from the ground to represent 6ft tall Casey (the top would represent Casey's eyes/head)
towards the right side, draw a vertical line coming up from the ground to represent the 84ft high lighthouse

[spare spruce]

vertical

[late geode]

vertical lines look like |

[late geode]

then draw a horizontal line from Casey's head until it reaches the lighthouse
and draw a line connecting Casey's head to the top of the lighthouse

[spare spruce]

?

like this?

[late geode]

you should put some braces/brackets to clearly mark which segment is 84
can you also try marking in the elevation (this will be the angle between Casey's line of sight to the top of the lighthouse and the horizontal)

[late geode]

not quite what i meant

[spare spruce]

😔 My bad

hold up

[spare spruce]

ohh

[late geode]

something like that to indicate that whole thing is 84 (ft)

can you also try marking in the elevation (this will be the angle between Casey's line of sight to the top of the lighthouse and the horizontal)

[spare spruce]

yah I get it

cool

Casey sights the top of an 84 foot tall lighthouse at an angle of elevation of 58 (degrees). If Casey is 6 feet tall, how far is she standing from the base of the lighthouse?

so where do I put 58?

[late geode]

yah I get it
exactly as i described and you said you got it

[spare spruce]

nah I mean I understand it thx

[late geode]

you replied to the comment about angle of elevation implying you understood the definition

you should be able to apply that if you did, unless this wasn't the case

[spare spruce]

i mean lemme see

Thats where it goes?

[late geode]

no

as is, it looks like you're marking distance from the tops of casey and the lighthouse as 58

[spare spruce]

ohh

[late geode]

no

now you're marking the horizontal distance as 58

[spare spruce]

fascinating

[late geode]

don't even put any numbers in

[spare spruce]

"lighthouse at an angle of elevation of 58 "

[late geode]

use a small arc, to mark the angle
between between Casey's line of sight to the top of the lighthouse
and the horizontal line of sight from Casey to the lighthouse

[spare spruce]

what does elevation mean in this?

[late geode]

i've typed it out multiple times

[spare spruce]

The distance between her head and the light house?

[late geode]

this will be an angle between two lines

angle of elevation

angle

you shouldn't be thinking about lengths here

[spare spruce]

so in degrees

[late geode]

angles are measured in degrees yes...

but forget about even that for now

can you highlight

Casey's line of sight to the top of the lighthouse

[spare spruce]

the top?

[late geode]

highlight (and only highlight, don't add in any addition numbers or anything else)
what you think i'm referring to

[spare spruce]

the thing in blue

[late geode]

what's the 58 still doing there

[spare spruce]

oh lemme remove it

[late geode]

but that thing in blue was indeed the line of sight to the top

now highlight

the horizontal line of sight from Casey to the lighthouse

[late geode]

yes

the angle of elevation from Casey eyes/head to the top of the lighthouse
would be the angle between those two lines you just highlighted

[spare spruce]

sooo

basically

this?

[late geode]

yes

[spare spruce]

so thats where 58 goes

[late geode]

58°, yes

[spare spruce]

idk how to do the degrees thing

[late geode]

alt 0176

(easily written)

[late geode]

yes

now you have a much more reasonable diagram representing the situation
(not quite to scale but doesn't matter that much as key features are there)

[spare spruce]

I can redo it

make it look neater

💀 Artist and im struggling to draw triangles

Now thats its setup, how do I solve it?

[late geode]

ideally you'd also mark in right angles
you have a right triangle
try to determine the lengths of its sides

[spare spruce]

so

uhh

hmmm

[late geode]

note that one of its sides could be determine from (segment) addition/subtraction and properties of rectangles

[spare spruce]

😶 so

the sides are

I can set it up like this?

[late geode]

x isn't the final goal

you also erased important info from your initial diagram

[spare spruce]

its still there

so whats the final goal?

[late geode]

its mentioned in the question

how far is she standing from the base of the lighthouse?

[spare spruce]

well

if thats the case

she looking up 58 degrees and shes 6 feet tall and the lighthouse is 84 feet tall, and I need to find out how far away she is from the lighthouse

😶 Now how am I supposed to figure that out

[late geode]

note that one of its sides could be determine from (segment) addition/subtraction and properties of rectangles

consider what i've said and try determining the length of that green segment

[spare spruce]

alright

[spare spruce]

idk what that means ngl

like ik what a segment is

but what properties?

[late geode]

segment is just a line of fixed length

do you know the properties of a rectangle

i mean you've also made some markings on your diagram implying the property needed here

[spare spruce]

no

[late geode]

do you know what a rectangle is

[spare spruce]

You mean a regular rectangle?

[late geode]

yes

[spare spruce]

yes

an elongated sqaure

pretty much

[late geode]

so currently in your diagram you've marked the left edge as 6ft

what does that tell you about other parts of the rectangle

[spare spruce]

that the other side is also 6 feet?

[late geode]

which side do you mean by other

there are 3 "other" sides

[spare spruce]

oh um

The line the indicates the height of the light house

84 feet

this?

[late geode]

the red part is 6ft yes,

[spare spruce]

yah

cuz the other part is 6 feet right/

[spare spruce]

yahhh

[late geode]

use that to determinet he length of the green

[spare spruce]

so do i do 6-84?

[late geode]

no

other way around

84 - 6

green + 6 =84
green = 84 - 6

[spare spruce]

oh yah

why 64-6?

[late geode]

typo

[spare spruce]

oh

[late geode]

fixed

[spare spruce]

84-6 is 78

thanks for the help

[marsh citrus]

@spare spruce Has your question been resolved?

[late geode]

now that you have an angle and a side in your triangle,
you can use trig to determine what you want

[marsh citrus]

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[marsh citrus]

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[quaint chasm]

What does this task want me to do?

[marsh citrus]

@quaint chasm Has your question been resolved?

[marsh citrus]

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[karmic dawn]

h

[elfin berry]

dain

[karmic dawn]

but how

<@&286206848099549185>

[marsh citrus]

@karmic dawn Has your question been resolved?

[marsh citrus]

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[dense crater]

I am doing question 6

[dense crater]

But I am not sure why the argument circled in blue makes sense.

It is clear that $x$ is nonzero since it is not in the subspace $W$, which assumed to contain zero vector.

[elfin berry]

Trenton

[dense crater]

and we cannot assume $\langle x,y\rangle\not= 0$ yet.

[elfin berry]

Trenton

[dense crater]

oh wait actually can we?

[stoic saddle]

if y were 0 then x would be equal to w

[dense crater]

oh yes which contradicts the assumption that $x$ is not in $W$

[elfin berry]

Trenton

[dense crater]

Thanks a lot!

.close

[marsh citrus]

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[dense crater]

.reopen

[marsh citrus]

✅

[dense crater]

.close

[marsh citrus]

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[delicate wadi]

man why the hell did my friend made me join this

[stoic saddle]

🚪

[marsh citrus]

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[runic otter]

can someone help me with this

i need to understand

<@&286206848099549185>

hello

hello

yo

where

hello

<@&286206848099549185>

hello

[oak pollen]

explain what you have understood so far

[marsh citrus]

@runic otter Has your question been resolved?

[still temple]

If I reach 8’10, how high would I need to jump to dunk.

[marsh citrus]

@runic otter Has your question been resolved?

[runic otter]

<@&286206848099549185>

[spark berry]

For angle 6 use this

Then as you can see, point q seems to be the center of the circle

So all intercepted arc = central angle

[spark berry]

Use this for 2

For angle 4, we see that point q is the center and line af intersects it, making ef the diameter of the circle. So that means that ARC AF = 180
Find Arc AG and subtract it to ARC AF = ARC GF
And as we know earlier, all intercepted arc = central angle so Arc GF = Angle 1

From previous answers (if you answered them), we already know the value of arc EF and arc FG. So create line EQ and you will have an isosceles triangle. Since we know the arc of EF and FQ, add them. You will get angle EQG. Isosceles triangle have 1 angle and 2 angles that are the same and equal. So angle EQG + 2x = 180. Solve for x and you will get angle QGE and angle QEG. Use either of them to solve for the missing angle 5 and 4. Create triangle QG4, and we know all angles insides all triangle always have 180 degrees. So angle GQF + angle QG4 + Angle D4G (which is our target) =180

Equate it, and we know that angle D4G = 5 via vertical angle postulate.

And we know that angle 5+4 = straight line or 180 degrees.
Equate for 4 using the value you got at 5

[runic otter]

damn thanks

.close

[marsh citrus]

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[ionic owl]

To convert 600 inches per minute into an angle in radians. Would you divide 600 by the radius of the pulleys? 600/2 = 300 revolutions per min and 600/5 = 120 revolutions per minute. One revolution is 2pi, so 2pi * 300 = 600pi and 120 * 2pi = 240pi? Does this sound right?

[strange hazel]

dude

wait

see this

60 min = 1 degree

3600 sec = 60 degrees

so the radius

of larger pulley is 5 inches

now you can solvethis

[ionic owl]

Good point, I forgot about that

[strange hazel]

no problem

[ionic owl]

[\omega = \frac{\theta}{t}]
[\omega = \frac{60}{5}]
[\omega = 12]

@strange hazel For the larger pulley?

Okay idk why latex isn't doing it

[strange hazel]

ohh

u know the formula for velocity

i mean anglar velocity

[ionic owl]

Yes

[strange hazel]

you have to find with the belt rate that how faster is the angle is changing or wheel is rotating

so it travels 600 inches per minute

thats the rate of angle

is changing

in the wheel

try to figure out yourself

if the wheel

is moving with rate 600 inches per minute

then

how you can solve this

for angle

[ionic owl]

I would assume convert the 600 inches per minute into revolutions

[ionic owl]

Thats what I was trying to do here

[strange hazel]

its very simple(use the radius here)

[strange hazel]

nice approch

convert

the 600 inches perminute into

inches per degree

and one point

to be noted aswell

that the radius is givcen that means

[ionic owl]

Using this, would it be 1/60 of a degree per minute?

[strange hazel]

circumference

[strange hazel]

yes

well

try to figure out what the circumference of both wheels will do

basically circumference is the distance the belt travels in each wheel in particular time thats the hint

[ionic owl]

I don't think latex is working so I can't type it out

But like omega = 1/60 / 1 minute ?

[strange hazel]

you r rght

[ionic owl]

Sure but, that doesn't change for each pulley

[strange hazel]

yup

[ionic owl]

Don't I need to do something with the radius of the wheel?

Since it's asking for angular speed of both the smaller and larger one

[strange hazel]

thats where you hAVE TO use circumference

to determine the

anglar vel

[ionic owl]

The circumference?

I thought you used the radius?