help-28

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@charred escarp Has your question been resolved?

[charred escarp]

The question came up in a gcse paper and it doesn't specify

[somber sonnet]

drink water

stay hydrated

[charred escarp]

Oh

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@charred escarp Has your question been resolved?

[wild sleet]

expected number is not the same as most likely number

you want to keep the fraction

probably

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[coarse sinew]

How did they get those numbers for CD

[cursive dock]

as in, the length of CD? you can make a right angled triangle with CD as the hypotenuse

and you know the length of these 2 red sides

[coarse sinew]

Ok so what do they do vertically and horizontally h

How do they go

It goes 4,8

[cursive dock]

well that right angled point is at the same X as D

and the same Y as C

[coarse sinew]

Ok so

Does it go 4,8 ?

I mean I would do 8-4

But idk

[cursive dock]

well, what's the x coord of D

[coarse sinew]

4 ?

[cursive dock]

the length of that vertical side is (y coord of C - y coord of D)

the length of the horizontal one is (x coord of c - x coord of d)

[coarse sinew]

Is there a vid on this type of thing

I mean d is 1 at x

[grim skiff]

If you know distance formula, it's just applying that

If not, you can count the units of each side

[coarse sinew]

Ok got it

They used trigonometry here but how is there any sincostan here?

Heyy

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[wicked frost]

can I just integrate everything here directly?

like the u^3 becomes u^4/4 and the 1 becomes u and the u^2 becomes u^3/3

or do I need to use some sort of substitution or something

[minor crater]

no, why?

you have a fraction

[wicked frost]

how would I approach this problem then?

let x = u^3 - u^2 then find dx and substitute that in?

[left bone]

how are you gonna deal with the u^3 on top then?

[minor crater]

$\int_2^3 \frac{u^3+1}{u^2(u-1)}$

[glossy valve]

artemetra

[minor crater]

hm

ig you can do $\int_2^3 \frac{u^3-1+2}{u^2(u-1)}$

[glossy valve]

artemetra

[minor crater]

$=\int_2^3 \frac{(u-1)(u^2 + u +1)+2}{u^2(u-1)}$

[glossy valve]

artemetra

[minor crater]

by difference of cubes

[arctic wedge]

I personally would use partial fractions for this problem seems like it’d be slightly easier

[minor crater]

$=\int_2^3 \frac{(u^2 + u +1)}{u^2} + \frac{2}{u^2 (u-1)}$

[glossy valve]

artemetra

[minor crater]

but idk feels a bit forced

[wicked frost]

if I used partial fractions, would I have to long divide first?

[arctic wedge]

You can split up the fraction, and apply the linearity of the integral, you should get something easy to decompose, It will take me a while to type it up nicely on my phone

$\int_2^3 \frac{u^3+1}{u^2(u-1)}=\int_2^3\frac{u^3}{u^2(u-1)}+\int_2^3\frac{1}{u^2(u-1)}$

Then if you simplify the left integral and apply partial fractions to the right I’m pretty sure those should all be relatively easy integrals to do

But it’s 1am and I’m on my phone so I’m not doing the partial fractions myself to check, but I’m like 90% sure it should work out fairly nicely

[wicked frost]

alright i'll try it out thanks

[glossy valve]

Nope

[wicked frost]

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[desert musk]

how to find an affine function

[desert musk]

for that

[spice orchid]

what is your definition of affine function

[desert musk]

f(a)

• f'(a) * (x-a)

but here i have to variables

So do i take the gradient instead of f'(a)

[desert musk]

Thank you

that is basically what i've done

[desert musk]

but

f_x and f_y is just the gradient

cant i insert those without my a and b

just differentiate the function

[spice orchid]

not sure what you are asking, you need the a and b

[desert musk]

alright

[desert musk]

Ok

this is what i've done

The question was:
"Here g is the gravitational constant, m is the mass of the pendulum, and ` is the length of the pendulum. The angle θ ∈ R represents the swing of the pendulum (which we allow here to be more than one full revolution), and v ∈ R is the speed of the pendulum's weight. The pendulum rod is assumed to be rigid and massless, and the weight is assumed to have no extension. Energy is assumed to be conserved as the pendulum swings.

Therefore, a pendulum swing (θ(t), v(t)) as a function of time t, with v(t) = `θ0(t), will lie on a level curve for E.
(a) Determine the affine function that tangents the energy function E at a point (θ1, v1)."

[spice orchid]

looks fine but i dont know what you mean by "we'll insert not" nor what the expression after that is supposed to be

got only one side of an equation there

[desert musk]

I meant

We will insert now

I insert this

@spice orchid

[spice orchid]

right but then what happened to this bit

[desert musk]

well im following the formula

[desert musk]

This is it right

[spice orchid]

thats fine yeah but i dont see the need to expand it all out fully

[desert musk]

what else should i do

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@desert musk Has your question been resolved?

[fringe sorrel]

2x2

[fringe sorrel]

9

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[fallen vector]

I understand how to do the problem but I don't actually understand what the problem actually represents, what would part B be?

[fathom saddle]

One way to think of it, if you're integrating in terms of time, you're multiplying the unit by a time

In other words, if r(t) is gallons per hour, then ∫ r(t) dt is just gallons

∫ r(t) dt is the total amount of liquid that has entered the barrel

[fallen vector]

oh

so would it just represent

the amount of gallons of watter flown out of the barrel from 9 to 6 hours after 9

would I also include the 1/6 as well

[fathom saddle]

So if ∫ r(t) dt is the total amount, what do we get when we take 1/6 of that? Reminder there's 6 hours here

[fallen vector]

just one sixth of the total amount right

one sixth of the total amount of water from 9 AM to 3PM

is that right

[normal gust]

won't it be the average value of r(t)

[fallen vector]

ohhhhh

wait that makes sense

so it would just be

average flow of water in gallons/hr from 9-3

[normal gust]

think so yeah

[fallen vector]

alrigh thanks

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[manic laurel]

Hi ive been having trouble with this word problem that you'll find below, I just have trouble putting it as a equation to solve

Jess walked for 45 minutes at 3 km/h and then ran for half and hour at x km/h. At the end of that time she was 6km from the starting point. Find the value of x

[fossil stump]

It may be helpful to split the problem up into two parts: walking trip, and running trip

[manic laurel]

should with that do i do it as speed x time = distance?

[fossil stump]

you can use distance = velocity times time

[manic laurel]

3 x 45 + 30 + x = 6

would it be something like that?

[fossil stump]

close

30 + x?

[manic laurel]

ohh

[fossil stump]

also be careful with units, maybe convert all the minutes to hours

[manic laurel]

3 x 45 + 30 x x = 6

3 * 3/4 + 1/2 * x = 6

soo that?

[fossil stump]

looks good to me!

[manic laurel]

alright ill solve it and see if it matches the answer

[fossil stump]

👍

[manic laurel]

so from that I got 1.5km/h

althought the answer is apparently 7.5km/h

ohhh

i know where ive went wrong

thank you!

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[granite shore]

@torpid siren are you comfortable breaking the original integral into three pieces?

[torpid siren]

I broke it into two, from -1 to 0 and 0 to 1

but it made it more complicated i think.

[buoyant pewter]

Every continuous function is integrable, but there are integrable functions which are not continuous. it is your case

[granite shore]

oh gotcha, yes I think youre right you dont need three pieces, just two

Ok let me write something down

$\int_{-1}^{1}f(x)dx = \int_{-1}^{0}f(x)dx + \int_{0}^{1}f(x)dx$

[glossy valve]

pencentre

[granite shore]

which "piece" of the piecewise function do we pick for the -1 to 0 part?

[torpid siren]

well -1 and x

[granite shore]

its actually the x^2 bit

$\int{-1}^{1}f(x)dx = \int_{-1}^{0} x^{2} dx + \int_{0}^{1}f(x)dx$

[glossy valve]

pencentre

[granite shore]

since f(x) is defined to be x^2 for values less than zero

let me know if that makes sense

[torpid siren]

ohh yea,

so would 0 to 1 be x

[granite shore]

Yes, exactly!
$\int_{-1}^{1}f(x)dx = \int_{-1}^{0} x^{2} dx + \int_{0}^{1} x dx$

[glossy valve]

pencentre

[granite shore]

Once you have this, then you can integrate each piece to find your final answer.

I think I get 5/6, but I just did that mentally

so double check me lol

[torpid siren]

Ok give me a sec

[torpid siren]

so are you integrating all of these?

[granite shore]

just the x^2 and x integrals

for the x^2, I got 1/3 and for the x integral I got 1/2

adding them together I think gives 5/6

[torpid siren]

Yea i got it

[torpid siren]

thank you so much!

[granite shore]

sweet, np

don't forget to close

[torpid siren]

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[mild star]

Where dies the 7 come from?

[mild star]

Does*

Is it from the 2/7 + 11/7

[desert pike]

“We notice that we can multiply both sides by 7…”

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[civic nebula]

so i was doing a job application question assessment thing and its math based but im so confused even tho i do a maths degree

[civic nebula]

i thought it was simply 19/28 and the rest of the info was just there to try trick you and add info for u to process

but part b makes me doubt my answer to part a (above) and if i need to use the other info

or maybe im just overthinking

[glacial pasture]

in a the graph isnt the number of people who get alerts
its just the average number of alerts a person gets per month
so it wouldnt be 19/28

youll have to use the from the 1000000 and the overall average, you can see there are around 15000000 alerts overall per month

youll have to use the other two averages to work out how many people of the 1000000 are on each method

roughly anyway

just what fraction of the 1000000 is what you need ig

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[shell crane]

Ferris Wheel Problem

As you ride the Ferris wheel, your distance from the ground varies sinusoidally with time. Let t be the number of seconds that have elapsed since the Ferris wheel started. You find that it takes you 3 seconds to reach the top, which is 13 meters above the ground, and that the wheel makes a revolution once every 8 seconds. The diameter of the wheel is 12 meters.

a) Write the function equation of this sinusoid.

b) What is the lowest height you go as the Ferris wheel turns?

[shell crane]

<@&286206848099549185>

[eager crag]

y = 7 - 6cos(x)

i suppose the function you need is the height

[shell crane]

thanks but Question wants sinusoid function

[eager crag]

then y = 7-6sin(90-x)

[full forum]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[shell crane]

when I gave x=3 it gives 12 . but should it give 13? and function will have period like 2π/8 because of the it makes revolution every 8 second. How can I write again function to use period?

<@&286206848099549185>

@queen crater are u available to help me

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@shell crane Has your question been resolved?

[shell crane]

<@&286206848099549185>

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@shell crane Has your question been resolved?

[mild star]

But cant you multiplie it with any number? How does one come up with 7?

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@shell crane Has your question been resolved?

[desert pike]

after combining like terms

u get (13/7)x

[shell crane]

<@&286206848099549185>

Ferris Wheel Problem

As you ride the Ferris wheel, your distance from the ground varies sinusoidally with time. Let t be the number of seconds that have elapsed since the Ferris wheel started. You find that it takes you 3 seconds to reach the top, which is 13 meters above the ground, and that the wheel makes a revolution once every 8 seconds. The diameter of the wheel is 12 meters.

a) Write the function equation of this sinusoid. ( example: f(t) = Asin(B(t-C)) + D )

A is the amplitude (half the diameter of the Ferris wheel, which is 6 meters),

B is the frequency (related to the period T by B= 2π/T)

C is the phase shift (time delay before the motion starts),

D is the vertical shift (the average height).

b) What is the lowest height you go as the Ferris wheel turns?

[shell crane]

thanks for helping

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[heavy nimbus]

How, in my head, should I conceptualize the expectation of a predicted relationship, given an input matrix X and a vector of estimated coefficients Beta-hat?

[heavy nimbus]

Am I supposed to conceive of the input matrix as a random variable?

The professor for this course has stated that X*Beta (without the hat, and not the expected value, just the multiplication) is a constant, though. So the input matrix X isn't a random variable?

Oh, wait, is it because Beta-hat is found by doing a least-squares regression of Y on X, and since it's an estimate there are always errors (i.e. noise) that mean(s) that Y doesn't fully correspond to X, and the error term is a random variable, so because our estimate of Beta-hat depends on whatever errors get thrown at us, Beta-hat is actually a random variable?

(Holy run-on sentence, Batman)

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[forest widget]

more simply put
Y is a rv
your estimator β^ depends on Y
so β^ is also a rv

[heavy nimbus]

Oh because Y is a function of the X inputs, which are known, but also an error term, which is a random variable?

[forest widget]

yes

[heavy nimbus]

Thanks.

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[limber flicker]

since it is oriented clockwise, we have to flip the result, is there some place to do this typically or would I just flip the end result?

also im getting $\int\int_D 1 dA$ so would it just be the area of the circle?

[glossy valve]

Clarkie

[sacred zodiac]

Yeah you can just flip it at the end if you'd like

[spice orchid]

i would probably just flip the bounds personally

[sacred zodiac]

It's really up to you as long as you know what's going on

[limber flicker]

alright nice,

oh

im getting -3 inside, ;-;

can I take it out and just multiply the area of the circle in that case?

trying to save time

[sacred zodiac]

Not sure what this means but you can always take constants out of your integral

[limber flicker]

so when I swap through green's theorem

I get (3x^2-3(x^2+1))

which just results in (-3)

[sacred zodiac]

ah yes

yeah its just -3 ∫∫dA

[limber flicker]

niceee

[sacred zodiac]

so -3 times (oriented) area

[limber flicker]

and then it would swap sign because oritentation

yesss

nice

[sacred zodiac]

good job!

[limber flicker]

perfect

thanks guys

<3

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[woven bloom]

I know this relates to trigonometry and would require the use of sin, cos, and tan, but am stuck on how to solve the problem.

[fathom saddle]

Note that angle Q is 90°

[sacred zodiac]

Hint: Area of PQR is 1/2 PQ QR but is also 1/2 PQ PR cos P

*sin P i mean

[woven bloom]

yea i dont understand that

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@woven bloom Has your question been resolved?

[woven bloom]

<@&286206848099549185>

[mighty prairie]

THEY GIVE YOU THE AREA OF THE TRIANGLE
WHAT FORMULAS DO U KNOW FOR AREA OF A TRIANGLE

[woven bloom]

bh/2

[mighty prairie]

ANY OTHERS

[woven bloom]

not that i can think of...

[mighty prairie]

ALRIGHT

SO, HOW WOULD YOU FIND THE AREA OF THIS TRIANGLE

WHAT SIDES WOULD YOU WANT

THE POINT IS, YOU WANT TO MAKE EQUATIONS OUT OF THESE GUYS TO STEP CLOSER TO FINDING PR

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[ruby wolf]

3.42 3.43 and also

[ruby wolf]

4.34

[unkempt thorn]

for 3.42, factor a -1 out of the denominator of 9/(2-t) and then solve
for 3.43, square it and then it should solve pretty easy
for 4.34, you kinda just take out common factors until you cant anymore

[ruby wolf]

Oh tyy

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[mortal sigil]

You guys are never going to expect the answer

It's

[mortal sigil]

this is literally 5987 digits

how are they expecting us to calculate this in a math contest for 1 and a half hours 😭

Also it's 2019!-3

Idk why

[sharp thorn]

U can have g(x)=P(x)-3x which has 2018 roots

g(x)=(x-1)(x-2)....(x-2018)

[mortal sigil]

so the roots are x=1, 2, ..., 2018? right

[sharp thorn]

Yep

[mortal sigil]

And since the leading coefficient is 1, it can be written as this

[sharp thorn]

P(x)=g(x)+3x

Now substitute x=-1 to get p(-1)

[mortal sigil]

so it's basically (-1)^2018*(2019)!-3

Ohhhhh

I'm dumb

Thanks

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[compact widget]

What is this called?

[compact widget]

I am doing differentiation , but struggling with this last step , i cant find any questions like it

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[compact widget]

<@&286206848099549185>

[buoyant pewter]

Can you show what the example looks like from the beginning?

[paper kelp]

@compact widget

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@compact widget Has your question been resolved?

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[buoyant bluff]

1/4 of 7/8 forgot how to work theese out

[smoky wing]

of means times

[buoyant bluff]

cool thnks

[smoky wing]

[buoyant bluff]

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[static timber]

$|3x+1| = 4x$\newline
$3x+1 = 4x \vee -3x-1=4x$\newline
$1 = x \vee -1 = 7x$

[glossy valve]

Tomasz#5153

[static timber]

its not correct i dont know why

i suppose thats not how i should write absolute value

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[static timber]

pls help

[static timber]

.

[dapper bobcat]

are you looking to solve for x?

[static timber]

ye

$|3x+1| = 4x$\newline
$3x + 1 = 4x \vee -(3x+1) = 4x$

thats correct right?

[glossy valve]

Tomasz#5153

[dapper bobcat]

So how I always approached it was that the equation that is in the absolute value stays the same and the other portion of the problem becomes negative.
IE:
3x+1=4x and 3x+1=-4x

You would technically get the same thing, but that is how I would approach it.

So the x=1 is correct, but you need to finish solving for x in the second equation

[static timber]

i mean it says there is only one solution x = 1

i dont know why

its tragic dfont look at this ^

[dapper bobcat]

So the x=1 would be correct.
If you were to go through the process of solving for the other x, you would get that x=-1/7.

[static timber]

yes

that's what i got

[dapper bobcat]

Which when you go to check the answer, will not compute to equal one another.

I will write it down. It took me a second too but I forgot that we should check work to double check, especially when it comes to absolute values

[static timber]

wolframalpha says it has one solution

[dapper bobcat]

You have to take both of the x's and plug them into the original equation. If it works, then the answer is correct. If it does not work, then the x is incorrect.

Therefore the x=1 would be the only possible solution

[static timber]

there must be a better way than to plug values for x

[dapper bobcat]

It's one of the sure fire ways to always check your work.

[static timber]

ok thanks

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[full frost]

hello i seem to forgotten how to get the x in the picture i got the <A=180-50-90=40 so i have all of the corner sizes <A=40 <B=50 <C=90 but how do i get the x

[naive wadi]

x + 40 = 180

[full frost]

thank you.

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[waxen panther]

Hi guys this is a definiton from my script

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Please don't occupy multiple help channels.

[waxen panther]

in a ring R(+,*)

thats the proof

i dont get how they even start

from where comes the part after the plus?

[naive wadi]

do you mean 0∗a + 0∗a = (0+0)∗a

[waxen panther]

jeah

i want to prof 0*a= 0

so why do i start with 0a+0a

[naive wadi]

Well as for what's happening, that's the distributive property, (a+b)∗c = a∗c + b∗c
although I don't remember if it's actually true for all rings, it's been awhile since I studied rings

[waxen panther]

true

that i understand but

why can you start the proof with 0a+0a

[naive wadi]

We know 0 = 0 + 0
So 0∗a = (0+0)∗a = 0∗a + 0∗a is where the first part comes from, then, we have that by the distributive property this is equal to (0+0)∗a which equals 0∗a

So, we know that 0∗a + 0∗a = 0∗a

This can only be true if 0∗a = 0

I don't see where they prove a∗0 = 0 though

Oh, well the argument doesn't really depend on the order, so you can just the above argument but start with a∗0 = a∗(0+0) = a∗0 + a∗0 -> a∗0 = 0

All of this assumes that (a+b)∗c = a∗c + b∗c, and the second case, c∗(a+b) = c∗a + c∗b
Which I don't know if that's true or not for all rings, you'll have to figure that one out

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[shell kestrel]

I'm so tired

[shell kestrel]

idk how to find the intervals

I'm trying to work backwards with the example problem and nothing is making sense

why can't this just be straight forward

why does everything have to have some complicated formula and 6 different steps to find

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[south spindle]

let $\theta = y/2$

[glossy valve]

ταυταυ

[south spindle]

yeah

what's y in terms of theta ?

yeah

now replace y with 2 theta

so the 2 cancels out :)

see theorem 6 they say

sure

what's the triangle OAP ?

yeah

well

it's the definition of the sector using radians

you've learnt that the area of the circle is 2pi r^2, but it's only because the angle is 2pi

yeah

[south spindle]

my bad, it's pi not 2pi

because you have to devide by two lol, using calculus you can get why but idk if you're gonna get it

area is pi r^2

yeah

you divide by two

well

there's a calculus thing

but it's calc3 level

and you're doing precal which means you won't get it

idk the geometry of it

The reagion of the area is $$R = {(x, y) \mid x = \rho \cos{\varphi}, y = \rho \sin{\varphi} \mid 0 \leq \rho \leq r, 0 \leq \varphi \leq \theta }:.$$ So the area is the following double integral
$$\int_0^\theta \int_0^r \rho ,\mathrm{d}\rho, \mathrm{d}\varphi= \frac{\theta r^2}{2} :.$$

took me too long to write LOL

but you wouldn't get it anyway so i did it for no reason

[glossy valve]

ταυταυ

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[stuck surge]

Hello, nice to meet you. I would be very grateful if you could help me with this problem because I am having difficulty solving it.

[stuck surge]

What I have been able to deduce is that:
0.05*30=1.5mg/kg
30+1.5=31.5 mg/kg

[cursive dock]

so the legislation says that you can indeed add 1.5mg/kg of zinc per kg of soil per 100 years, right?

which means you can add X kg of sewage sludge per kg of soil, where X is the amount of kgs of sludge that has 1.5mg of zinc. follow?

@stuck surge

[stuck surge]

Yes

[cursive dock]

and 1kg of sludge contains ???? mg of zinc
which means in ????kg of sludge, there is 30mg of zinc

two different ????s to work out

[stuck surge]

But if that's the case, I can simply make the rule of three

Or am I making a mistake?

[cursive dock]

not too sure what the rule of three is

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[smoky wing]

,tex \begin{theorem}
Let (X = (X_t){t \geq 0}) be a continuous local martingale with respect to a filtration ((\mathcal{F}t){t \geq 0}). If
[
\mathbb{E}\left[\sup{0 \leq s \leq t} |X_s|\right] < \infty \quad \forall t \geq 0,
]
then (X) is a martingale.
\end{theorem}

Ive shown the family $({|X_{t \wedge T_n}|})$ is uniformly integrable, using boundedness of the E(sup) and doob's inequality.

But Im a bit stuck on how to show it's a full martingale. I'll probably need the Dominated Convergence Theorem for taking limits but not really sure how to construct the proof.

[glossy valve]

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

[smoky wing]

this is so ugly wth

[glossy valve]

Pure

[idle kelp]

Don't you just need that $\mathbb{E}[X_t | \mathcal{F}s}] = \mathbb{E}[\lim{n\to \infty} X_{t\wedge T_n} |\mathcal{F}{s}] = X{s}$ by applying the dominated convergence theorem.

[glossy valve]

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

[smoky wing]

why do we have the second equality?

oh

[
\lim_{n \to \infty} \mathbb{E}[X_{t \wedge T_n} | \mathcal{F}s] = \lim{n \to \infty} X_{s \wedge T_n} = X_s.
]

[glossy valve]

Pure

[smoky wing]

its just this isnt it

[idle kelp]

I think so, but it's been a little while since I've used any stochastic calculus

[smoky wing]

yeah i think this is it i just forgot lol

thanks

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[smoky wing]

@layla

[rare dock]

:waves:

[smoky wing]

[rare dock]

[smoky wing]

oh u dont ahve niitro lmao

[rare dock]

T_T

[smoky wing]

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[torpid plaza]

hello

[torpid plaza]

i need help with the first question 😢

HELP!

[formal scarab]

pi radians = 180 degrees

[torpid plaza]

so 180 divided by whatever under?

[glacial pasture]

if x is in radians then x* 180/pi is the degree equivalent

the reverse is true

[torpid plaza]

huh

[glacial pasture]

if x is in degrees then x*pi/180 is the radian equivalent

[torpid plaza]

can you solve the first one for an example

[glacial pasture]

pi/3 * 180/pi = 180/3=60 degrees

[torpid plaza]

like this?

[glacial pasture]

not = no

you could maybe get away if you wrote rad on the left one

but i still wouldnt write it like that

$\frac{\pi}{3}\cdot \frac{180}{\pi}$

then =...

[glossy valve]

AℤØ

[torpid plaza]

HUH

[glacial pasture]

what?

[torpid plaza]

why is it written like that

😢

[glacial pasture]

its just what i wrote here

i dont see the issue

oops wrong one

[glacial pasture]

this one

[torpid plaza]

why is it multiplying

[glacial pasture]

because thats how you convert...

do you think 180/3 just pops out of thin air

[glacial pasture]

it comes from here

[torpid plaza]

why is there an A

[glacial pasture]

what?

i didnt write an A anywhere

[torpid plaza]

i mean the multiplying

why

[glacial pasture]

because im multiplying by the ratio of degree to radians

i really dont know what you want me to say

thats just how you convert it

[torpid plaza]

sorry for not understanding 😢💔

[glacial pasture]

If x is in degrees, then:
$$x\cdot \frac{\pi}{180}$$ is the radian equivalent angle. \newline
If x is in radians, then:
$$x\cdot \frac{180}{\pi}$$
is the degree equivalent angle.

[torpid plaza]

so basically the thing is cross multiplying

[glossy valve]

AℤØ

[torpid plaza]

oh

wait how do you show your work

[glacial pasture]

^

[torpid plaza]

what about the second question

i got this

how do you do c

HELP!

[glacial pasture]

ig that works, sure

there is nothing different about c

[glacial pasture]

just use this

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[twin wolf]

I’m confused what to do from here after factoring 7/4 out

[devout valley]

Might have helped to have factored out the 7/4 before deciding to make a substitution

[twin wolf]

Hm

okay

[twin wolf]

Like this?

[devout valley]

Yep, and then rerarrange that $\frac{ \pqty{x - \frac12 }^2 }{ \pqty{\frac74} }$ a tiny bit, which then should make another substitution that should make your life easier!

[glossy valve]

@devout valley

[twin wolf]

I get this

But I’m just confused with the substitution for arc tan because I need u^2 alone but I can’t do that when I have (4u^2)/7

[devout valley]

You can "absorb" everything inside the square root

[twin wolf]

I’m not sure I totally understand

[devout valley]

You know, for example, that $4 = 2^2$, so $4\pqty{x - \frac12}^2 = (2x - 1)^2$

[glossy valve]

@devout valley

[devout valley]

Can you think of something similar for that 7?

[twin wolf]

wait wat 4(x-1/2)^2 = (2x-1)^2?

Did you expand that out

to find that

Or was there some sort of cancellation I didn’t see

[devout valley]

Cause you have $2^2 \pqty{x -\frac12 }^2 = \pqty{ 2\pqty{x - \frac12} }^2$

[glossy valve]

@devout valley

[twin wolf]

Hm I have never seen this algebra manipulation trick before

[devout valley]

You haven't?

Damn, it's basically a consequence of rules of exponents though, that $(ab)^n = a^n b^n$

[glossy valve]

@devout valley

[devout valley]

I need to stop changing my mind to use tex halfway through a message

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[twin wolf]

Well I’ve seen that but I haven’t seen it applied in like quadratics here

[devout valley]

Awww, well the aim of doing it here is that you just need to make only one substitution

That then makes your life easier

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[devout valley]

.reopen

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[static phoenix]

Any idea whats wrong?

[static phoenix]

Im supposed to just divide so i dont know how my answer is wrong

[buoyant pewter]

$3a^2+2a+5-\frac{19}{a+2}$

[glossy valve]

Joanna Angel

[static phoenix]

Oh

Right

[buoyant pewter]

you did it

[static phoenix]

?

[buoyant pewter]

$3a^2+2a+5\text{ is a quotient}\\-19\text{ is a remainder}\\a+2\text{ is a divisor}\\3a^3+8a^2+9a-9\text{ is a dividend}\\\text{in same way you do in next problem}$

[glossy valve]

Joanna Angel

[static phoenix]

Then -5x+11/x+2?

Or -5x+10+1/x+2

[buoyant pewter]

quotient is -5x + 10

x+ 2 is a divisor

[static phoenix]

So where does 19 appear

[buoyant pewter]

-19 is a remainder

[static phoenix]

But 1 is the remainder of this equation?

[buoyant pewter]

remainder of the division

[grim skiff]

The expression is -5x^2 + 1, when you do the division, you should add placeholders. Because there is 0x, you should have written -5 0 1

Not -5 1

[static phoenix]

But there wasnt an x to begin with?

Oh waiut

wait

hang on

[grim skiff]

And that's what I'm saying, put place holders for the 0 terms

So -5x^2 + 1 would look like -5x^2 + 0x + 1

So that way you write -5 0 1

[static phoenix]

Then that just gives me 10x+0+1

[grim skiff]

You should have something like this

The arrow with -2 is to indicate that it got multiplied by -2

[static phoenix]

Oh wait

I multiplied 10 by 0

instead of adding

Ok well i see now thank you

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[coarse sinew]

I know the first one uses sine rule but the second one how ?

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[torn jolt]

How in it value of tan calculated

From 44.1/20 to 65.6

[strange basalt]

i sense calculator

[torn jolt]

There's isn't any way?

[strange basalt]

like artan ( 44.1/20)

most prob no

[craggy tapir]

you use a calculator

[strange basalt]

and you can leave the answer as tan invese

[craggy tapir]

since this isnt a special case or anything

[strange basalt]

aint no way that will come in the mcq

dw

[torn jolt]

It did

Came

[strange basalt]

IT ASKED FOR BETA?

[torn jolt]

And the answer was in degree

That's why I asked

[strange basalt]

where calculator was not allowed?

[torn jolt]

No

Who allows calculator in exam

[strange basalt]

some countries idk

[torn jolt]

Lol

Ok

[strange basalt]

thats crazy tho..

[torn jolt]

Yes

[strange basalt]

like the optiuons had 65.6

and not just like tan inverse 40.1/20

[torn jolt]

No 66

Was right answer

[strange basalt]

what were other options

ig we can use aproximation

[torn jolt]

66 35 27 72

It said approx

[strange basalt]

i see

we know tan 60

tan 30

[torn jolt]

Yea

[strange basalt]

so ig we just have to like kinda guese which is closest

[torn jolt]

One is 15 not 35

[strange basalt]

ah ok thats better

we atleast know that

[torn jolt]

Yea

[strange basalt]

i cant think of any way other than extreme aproximation , which is again kinda wild to ask

[torn jolt]

Yeah

Anyways thanks

[strange basalt]

damn

[torn jolt]

Closing this

.close

[strange basalt]

cool gl

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[patent sierra]

For the fifth part i tried looking for some pattern

Made sets like
When the sum is 5
(1,4) (2,3) (3,2) (4,1), 4 sets

likewise when 10
(1,9) (2,8) (3,7) (4,6) (5,5) (6,4) then repeating so ,9 sets

now did 4+9+14...99 which is 1030 but the no of favourable cases is 520 whats wrong in this

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@patent sierra Has your question been resolved?

[slate violet]

when you get to 55 the number of cases starts decreasing

you start from 4 + 51, 5 + 50, 6 + 49, ...., 51 + 4

there are 51 - 4 + 1 = 48 cases here

[patent sierra]

Damn

really missed that

[slate violet]

yeah

and 60 has 51 - 9 + 1 = 43 cases

[patent sierra]

i seee

[slate violet]

continuing that pattern does give 520

[patent sierra]

so its decreasing in the same apttern

[slate violet]

yes

so one pattern from 5 to 50

[patent sierra]

ah thanks

yea

[slate violet]

another one from 55 until 100

no worries

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[peak ledge]

help pls

[peak ledge]

what do i do now

[worn matrix]

what kind of triangle is ABC?

[peak ledge]

looks equalatrieral

[worn matrix]

in maths you dont judge something by looking at it lmao

AB=BC tells u what?

[peak ledge]

those segments are congruent

[worn matrix]

no i mean about the ABC triangle

[peak ledge]

those sides are equal

[worn matrix]

bruh

do you know what are isosceles triangle?

[peak ledge]

yes triangle with 2 congruent sides

[worn matrix]

so is this also a isosceles trinagle?

[peak ledge]

yes

[worn matrix]

what can you say about angle A and C now?

[peak ledge]

they are equal

[worn matrix]

right

so you're done with the proof now

[peak ledge]

oh god

now i have to reorder proof steps

😭

[worn matrix]

lol

[peak ledge]

where do i even start

im so confused

[worn matrix]

angle GBC equal to angle GCB gives you GB=GC

just the converse of what ive told

you

above

[peak ledge]

no i mean it wants me to reorder it

to be in the logical ordere

but idk what the logical order is

[worn matrix]

no idea about that

i can help you only with the proof

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[heavy gulch]

Is there another way to find G? This is for reflection angles of light rays defined as a vector field of V. The reflected vector is R, the T is the surface tangent, and n is the surface normal. X is the X-axis. I got the table from just logically thinking about what the reflection angles would be when the light ray hits a surface at specified angles just to get a sense of what the graph would look like for G. I can only find the reflection angle defined as 360-G which G is some function of a angle. Such a angle I will call XZ. XZ is between the surface normal to the X-axis. The reflection angle is also taking in the input of such angle, but I am having trouble finding the G function. I feel like what I have is correct to a sense, given that three of the values match what I expect G to be at the given angle XZ as shown by the table.

[heavy gulch]

Though I highly doubt its a coincidence that two or three of the values I got for a calculated G, match what I expect G to output at three XZ angles. If that makes sense?

Basically I have doubt its a coincidence that the G function I calculated, isn't in some way, related to the expected G function. This is because three of the 5 values it gives me, match

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[heavy gulch]

Back

[heavy gulch]

Is there another way to find G? This is for reflection angles of light rays defined as a vector field of V. The reflected vector is R, the T is the surface tangent, and n is the surface normal. X is the X-axis. I got the table from just logically thinking about what the reflection angles would be when the light ray hits a surface at specified angles just to get a sense of what the graph would look like for G. I can only find the reflection angle defined as 360-G which G is some function of a angle. Such a angle I will call XZ. XZ is between the surface normal to the X-axis. The reflection angle is also taking in the input of such angle, but I am having trouble finding the G function. I feel like what I have is correct to a sense, given that three of the values match what I expect G to be at the given angle XZ as shown by the table.

Basically I have doubt its a coincidence that the G function I calculated, isn't in some way, related to the expected G function. This is because three of the 5 values it gives me, match

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@heavy gulch Has your question been resolved?

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@heavy gulch Has your question been resolved?

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@heavy gulch Has your question been resolved?

[full forum]

@heavy gulch Has your question been resolved?

[full forum]

@heavy gulch Has your question been resolved?

[quaint prawn]

I respect your dedication

good luck again

[full forum]

@heavy gulch Has your question been resolved?

[heavy gulch]

Thanks @quaint prawn and @torn jolt

[heavy gulch]

This is what I've noticed as of now

I need to make Ge = G from 0 to pi somehow

[full forum]

@heavy gulch Has your question been resolved?

[heavy gulch]

I think I may have solved it

By analysis and two lines of mathematical manipulation, θR = 270+2*θxz

There is a weird thing where the reflection angle will go past 360 but I don't think the math in the future will care

@austere cove

[austere cove]

Hi, is this the same problem from before?

[heavy gulch]

Yes

[austere cove]

I never really took the time to understand how your system of variables worked, so I can't speak to how correct it is, but it looks reasonable assuming that you are having to add 90 to the degrees in order to take it from a normal vector to a plane.

[heavy gulch]

I didn't realize the equation I got was the correct answer until I tried like, 6 different angles and they all got the same answer I expected, but a few were rotate a full 360° but it still got the same spot

[austere cove]

I would recommend just simply trying an angle for each quadrant

[heavy gulch]

I tried more than that

I'll refine the work some time in the future

But for now it's just disorganized chaos in my notebook

[austere cove]

No worries. The end result is very simple

which agrees with my result

[heavy gulch]

Surprisingly

[austere cove]

and is not the evil arccos shit you had before 🙂

[heavy gulch]

And your idea of shifting the perspective made it possible

[austere cove]

I'm glad to have been of help

[heavy gulch]

Here is the short work I did after my analysis but its a bit difficult to explain. I was getting two different equations that satisfied what the expected reflected angle got. The two equations made up the domain of θxz but each equation had its own subdomain within the domain of θxz. For example, this equation I was trying to get, was apart of 0° <= θxz <= 45° and another equation I got was from 90°<= θxz<=180°. This was just seeming like another dead end to me though it was helping with my analysis of how the reflection angle is effected by θxz. However the first equation I was using, I refined a bit more as shown in the work I sent, and it turned out to solve the entire problem I was having, which was unexpected and it's simple as you said

Sorry for my crappy handwriting for G'(θxz)

But basically G'(θxz) = 90 + 2*θxz

[heavy gulch]

G'(θxz) is prime because it's what I calculated based on your idea of using a perspective shift. G(θxz) is what I expected

Of course you can subtract my equation I got for the reflection angle by 360 and it still doesn't really matter because a negative angle is still mathematically the same when it's calculatrd with different trig functions

I've never struggled for almost a month on a math problem until now that turned out to be so simple but a lot of it was thanks to your help and my intuition

Therefore it definitely was not a coincidence that some of the values matched (G') to what I expected on both the reflection angle and it's complementary angle (G). Of which made up 360° when added together

[torn jolt]

help!

!help

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[elfin stream]

@heavy gulch is this channel done?

[heavy gulch]

Yeah

[elfin stream]

pls close it

nvm me gon do it

.close

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[tepid valve]

may i know how to do this?

[cerulean cargo]

well how mant digits does this number have to have

[tepid valve]

4

[open igloo]

Pi

Jk I'm bored

[cerulean cargo]

how 4

all the numbers except two must appear twice

[tepid valve]

3?

[cerulean cargo]

give me an example of a number

that would obey those rules set by the problem

[tepid valve]

5572

correct?

[cerulean cargo]

but then 7 doesn't appear twice

and 9 doesn't appear at all

you need them to appear twice

[tepid valve]

5577992?

[cerulean cargo]

yes

so how many digits

[tepid valve]

7

wth

[cerulean cargo]

yes

well to me it seems you've correctly identified

that

the 2 must go in the last spot right

so

we can ignore the 2

and then the problem becomes

how many numbers can we form with the digits 5,7,9 appearing exactly twice

do you see how that is true

[tepid valve]

yes

how to calculate tho?

[cerulean cargo]

ok

[cerulean cargo]

this is our new problem that gives the same answer

so how to answer this one

well

this is like a similar problem

how many distinct 6 letter words can you form with the letters

AABBCC

do you know how to solve that one?

[tepid valve]

6!

?

idk man

[open igloo]

Yes you're right abt the first part, but all the letters are repeated twice
So if it's just 6!, The word
AACCBB would be counted twice, you see?

[tepid valve]

6! / 2

[cerulean cargo]

it would actually be counted 8 times I believe

[open igloo]

Ye

That's coz of the other letters too, right...

[cerulean cargo]

yea

[open igloo]

But the dividing by 2 is just for one letter
You gotta do it a few more times, why?

[tepid valve]

oh should i divide by 6

cuz theres 6 letters?

[open igloo]

No

[cerulean cargo]

pretending that each of the letters are different

we have

6!

$A_1A_2B_1B_2C_1C_2$

[glossy valve]

WOWWA

[cerulean cargo]

if we pretend they are different

like this

then there are 6! ways to arrange

but

$A_2A_1B_1B_2C_1C_2$

[glossy valve]

WOWWA

[cerulean cargo]

is the same word as

$A_1A_2B_1B_2C_1C_2$

[glossy valve]

WOWWA

[cerulean cargo]

in terms of what we are actually looking for

and you can see

that for each "word" if we pretend the letters are different

we are overcounting by 8

because the As can be arranged in 2! ways

and the Bs can be arranged in 2! ways

and the Cs can be arranged in 2! ways

for each word

[tepid valve]

so divide by 8?

[cerulean cargo]

so for each word there are 2!*2!*2! copies if we consider each letter different

yes

[tepid valve]

ooo

[open igloo]

The only difference is that instead of 2x3, you do 2³

[tepid valve]

so its 6! / 8

[cerulean cargo]

yes

6!/8

[tepid valve]

ok

thx

[open igloo]

Nice explanation man

[cerulean cargo]

thanks

[tepid valve]

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[empty venture]

hello i got some issues in the first question and i need some help

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[tiny drum]

any idea on how to solve this

[tiny drum]

do i subtract both runaround

equations

[dusky locust]

yes

[minor crater]

yes

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[honest tusk]

So this is a solids of revolution problem. I am not sure how to use the washer formula because I can't think of either function for the inner and outer radius to plug in, or the bounds. I think it being rotated around the y-axis is specifically what is throwing me off the most. Here is the problem:
Find the volume of the solid generated by revolving the region defined by the triangle with vertices [(1, 0), (2, 1), (1, 1)] by the y-axis

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Please don't occupy multiple help channels.

[honest tusk]

dang it

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[errant bison]

Question 1b

Why is the permanent order label 2 on node B shouldn’t it be of order 3

<@&286206848099549185>

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[torn jolt]

I took C as 100% then B = 75% and A = 89%
Then A’s income is 11 % less than C’s income
(11/100 * 100)

But the answer is 10%
Where am I going wrong?

[wild sleet]

the 89 is wrong

calculate again

[torn jolt]

yeah

it is 90

messed up with mental calculation

thx btw

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[keen canyon]

If one section has 10% students who got A+, In approximation how many students in % in 8 sections ?

[keen canyon]

I know it's 10% but how do I explain someone it is ?

He isn't taking the logic can anyone logically put it for him to understand ?

[wild sleet]

what does he say instead?

[keen canyon]

he says it's 80%

help a homie out

anyone ?😭

[wild sleet]

try to explain it?

it's a math exercise, don't give up

[keen canyon]

I can't

help meeee

I tried for half a hour

[wild sleet]

why would they listen for half an hour

logically, they are having fun

[torn jolt]

He's confusing amount and percentage

He likely understands that 8 sections will have in total approximatively 8 times more students and 8 times more A+ grades than a single section

What he doesn't realise is that this means the total percentage of students with A+ will remain the same

[warped crystal]

if one section has 100 people then ten people would have an A+. if we add all the sections together we would have 800 people with and 80 people in each section with an A+ so that is still 10%

[torn jolt]

If there's 8 times more students and 8 times more A+ grades then the total fraction of A+ grades to students stays the same

There's likely quite a bit more ways to explain this

I don't know which is best

[warped crystal]

it would be 80/800 which simplifies to 1/10 or 10%

[torn jolt]

I like Fox's explanation because people tend to be bad at abstract thinking so you often have to provide a concrete example

[keen canyon]

thanks

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[coarse tundra]

i need to check my answer on this question
Proof directly that,
“If m is an even integer and n is an odd integer then m+n is an odd”

[sharp thorn]

Ok