help-10

[last crow]

I recognize that the inner bounds are a circle and I got 2cos(theta)

But i do not know how to get the near function exactly

The far function is 2cos(theta)

[rigid pine]

x = 1 will be required.

r cos(theta) = 1, r(theta) = sec(theta).

sec(theta) <= r <= 2 cos(theta), -pi/4 <= theta <= pi/4.

[obtuse pebble]

@last crow Has your question been resolved?

[last crow]

Why?

[rigid pine]

1 <= x <= 2 from the iterated integral.

[last crow]

So

X = 2 - 1?

Is that why x = 1

[rigid pine]

We have that x >= 1.

What?

[last crow]

Im confused why x = 1 is the lower bound

[last crow]

I can see that yes

Do we just do 2 - 1

Such that its x = 1

[rigid pine]

🧐

The bounds are x = 1 and x = 2.

[last crow]

I am confused regarding how to interpret the outer bounds

Okay the bounds are x = 1 and x = 2..

[rigid pine]

We have all x values between 1 and 2 but there's bounds on y too.

[last crow]

Oh i see so i guess we just don’t need to worry about x = 2 because thats the far bound

So we only need to take x = 1

[rigid pine]

the curve of the circle intersects and is tangent to x = 2 so it requires no further thought when going to polar coordinates.

[last crow]

Right

Exactly

Because thats covered by the inner bounds

[rigid pine]

Yes. It's dealt with there.

[last crow]

Alright thank you stabulo

I appreciate your patience

[rigid pine]

The bounds would have given us the full circle.

But the x bounds say which part of it we have.

Between x = 1 and x = 2.

[last crow]

.close

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[true barn]

They ask for the area of RQN, and they use the 0.5(base)(height) but it has to be the height of a straight line, so im trying to prove that NR is 90 degrees with PS for myself, what I want to know is if 1 means that they are 90 degrees, because the default math law is gradient 1 x gradient 2 = -1

[vocal valve]

Hello can someone help me with part b - 2 a and b

[true barn]

As far as I know it doesn't state that NR is perpendicular to PS
but they automatically assume that
since 0.5(base)(height) uses the perpendicular height of a triangle

@vocal valve send that in #help-37 and it will make a ticket for you

oh my word

I have missed this for an hour

.close

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[copper grove]

can someone help me got maths no calc paper 1 tmrw morning stuck on this question

[copper grove]

i understand how to get the segment 60/360 x pi x 25

not 100% sure on the triangle

but idk how to answer in the form its asking

[obtuse pebble]

@copper grove Has your question been resolved?

[copper grove]

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[daring tinsel]

i don't understand how
(a-b-c) = (a-(b+c))

[cedar lichen]

a(b + c) = ab + ac

-(b + c) = -1(b + c) = -b - c

[daring tinsel]

oh so you treat it as if there was a -1 in front of it

?

[cedar lichen]

Thats what - is

It's -1 *

[daring tinsel]

ok thank you

.close

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[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[upbeat island]

are outward facing brackets exclusive on the interval?

i haven't seen that notation before

@timid silo

i usually see this notation: https://en.wikipedia.org/wiki/Interval_(mathematics)#Including_or_excluding_endpoints

just clarifying that the outward-facing brackets are indicating that the endpoint is excluded from the interval

oh! i actually never noticed that they list both notations there nvm haha

what is opl mean here? opl(f(x) ≤ 0)

i think you're correct and that it's a typo here

[obtuse pebble]

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[wide hull]

3x-2y=6, how to find the coordinates of a unit square using this eqtn

[fierce lagoon]

I mean

You can choose any two points on that line such that their distance is 1

[obtuse pebble]

@wide hull Has your question been resolved?

[wide hull]

what does that mean

[fierce lagoon]

You have the equation of a line

Choose any two points, but the distance between them is 1

[wide hull]

how would I find the other 2 points

reflect the 2 points we already found?

on the y axis

[fierce lagoon]

No. You would create two perpendicular lines that each contain one of the points

Something like this

You're gonna repeat the process

[wide hull]

ok so we can find point a and c using this method but how does repeating the process find b and d

[fierce lagoon]

Draw a diagram

You're gonna find points on those two other lines

[wide hull]

so using the first line

find it's perpendicular

[fierce lagoon]

I was thinking about something like this

[wide hull]

and use the perpendicular to find b and d

[fierce lagoon]

This also works, but the distance between a and c have to be sqrt(2) then

Does it specify that the equation is the diagonal of the square?

Unless it specifies, there's literally an infinite number of answers

[wide hull]

well it's a unit square so it's original points at y=X are 00 10 01 and 11

but since we have 3x-2y=6

we have to transform the 4 points above

[fierce lagoon]

Because the equation could be a side

[wide hull]

It doesnt specify

[wide hull]

. close

.close

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[restive pilot]

Okay

What part are you struggling with?

And what is your issue with it!

[violet sentinel]

this is close, but I think you multiplied by 1.04 rather than .04. The sales tax on $40 is $1.60, the total is $40 + $1.60 = $41.60

what's the other problem?

[obtuse pebble]

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[violet sentinel]

okay so how can we represent 125% as a ratio. Say if our 1st number is x and our 2nd number is y, how can we show that y is 125% of x?

would it be fair to say that y = 1.25x?

or in other words that $y = \frac{5}{4}x$?

[warm shale]

MellowDramaLlama

[violet sentinel]

okay, so then solve for x

then that shows that x is some fraction of y. If that fraction is 0.75, then Daren is correct

sorry what i mean is to isolate x (aka get x by itself)

so if we have (5/4)x, then what should we multiply by to get x by itself?

you mean 4/5?

if so then yep! 🙂

so then for $y = \frac{5}{4}x$, if we multiply both sides by $\frac{4}{5}$, we see that $\frac{4}{5}y = x$

[warm shale]

MellowDramaLlama

[violet sentinel]

it's always true that $\frac{a}{b} \cdot \frac{b}{a} = 1$ as long as $b$ and $a$ are not zero

[warm shale]

MellowDramaLlama

[violet sentinel]

r?

sorry what r are you talking about?

I don't see an r lol

uhhhh.... I'm not seeing an r sorry

not sure what you're talking about

oh no worries lol

but does that make sense?

no worries

can you tell me specifically where it doesn't make sense?

okay sure

so let's say we have 2 numbers

we don't know what those numbers are

so we'll call them x and y

yes correct!

they're called "variables"

for the sake of right now, yes. You can use any letter really

I could've called them a and b

c and d

z and y

it doesn't matter

they just represent numbers

Yes! 🙂

of course!

now by convention, you usually pick letters that are close by

so x and y

u and v

a and b

etc

you don't have to but that's what most people do

I'm going to switch letters

okay so we have 2 unknown numbers, a and b

Daren says that one of them (let's say b) is 125% greater than a

in other words, b = (125%) *a

how's that?

so far good?

no

it's part of the problem. Daren is saying that one number is 125% greater than the other

so then I'm just saying that we'll pick b to be 125% greater than a

well with percentages, when we are playing with percentages, then really we're taking our percentage amount, dividing by 100, and multiplying that value

so in other words

125 / 100 = 1.25, so b being 125% of a translates to b = 1.25a

yeah sure

nah you're fine. I appreciate your honesty.

math is one of those things the more you do it the better it gets

you start seeing a lot of patterns and can put the pieces together more easily

OH

I was doing 18

my apologies

no you're fine!

if you want to try

I can try to see if this can make intuitive understanding

if anything knowing how to do percentages will make sure you don't financially screw yourself down the line lol

okay

well let's see

do you know how to cross multiply?

have you ever heard of Kahn Academy?

this might be a great place to start. They really break everything down step by step from the beginning

https://www.khanacademy.org/math/arithmetic/fraction-arithmetic

yep!

they do the basics down to addition

like 3 + 5 = 8

you can literally start there knowing nothing about math

so like for example

https://www.khanacademy.org/math/arithmetic/fraction-arithmetic

you could start here at the 6th grade section since you said that you were having trouble with that stuff

and then skip what you do know and focus on stuff that's new or unfamiliar to you

for example, in the link above, there's a whole section on variables (the x and y we were talking about earlier)

plus the website is "gamified" meaning that you get points for completing videos so it can be fun 🙂

not really sure what to tell you then unfortunately. I would recommend that rather looking for the answer instead look for videos that go over what you're leaning, but if you really want to change the trajectory for math and get a basic to deeper understanding, then Kahn academy is about as good as it gets

then you got the answers but not an understanding as to why they got to the answer. You are correct that you can find answers all over the web for your problems, but if you don't understand how they got that answer then you're just hurting your future academic self. I would say getting the basics down for fractions, percentages, addition, multiplication, etc, is very important for day to day stuff.

yeah if you want to

.close is the command

best of luck to ya!

yeah my dude. I would really try to get through that stuff at least. You'll need to for the most basic of work (cashier, barista, etc) all use some form of math. I recommend Kahn Academy

to catch up

plus when you take your ACT and SAT, math is a big part of that

nah they cover similar topics but they're separate tests

[obtuse pebble]

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[obtuse pebble]

@mild spoke Has your question been resolved?

[obtuse pebble]

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[leaden hornet]

how to graph that to taht

[leaden hornet]

.close

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[celest whale]

i keep doing this but getting it wrong

first i found the area of the 2 squares which is 9x^2-42x+49

each

[violet sentinel]

good 🙂

[celest whale]

area of whole rectangle = 15x^2 + 5x+12

[tawny fog]

,w expand (3x+1)(5x+3)

[warm shale]

Results provided by WolframAlpha

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Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[tawny fog]

Wrong

[celest whale]

oh damn i forgot to do 3x times 3

bruh

[tawny fog]

Yeah bro

[celest whale]

lemme do that and see if it gives me the right answer

[tawny fog]

Yes ok

[celest whale]

yeah thats it

thanks

.close

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[late roost]

I tried mostly all of them

[obtuse pebble]

@late roost Has your question been resolved?

[timid silo]

should be all of them

wait

no

just f and h

cuz those are the continuous ones

EVT applies if continuous

[late roost]

Yeah i just figgured it out :D

thank you so much

So for this one it would just be f and g

since h and k are not continuous right

[obtuse pebble]

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[pastel jackal]

this is what i tried but i dont understand how to do it if r is onknown

[obtuse pebble]

@pastel jackal Has your question been resolved?

[tardy epoch]

Is percent change $\frac{\Delta F}{F}$

[warm shale]

riemann

[pastel jackal]

yeah

i think

but i dont know delta f or f

[tardy epoch]

your algebra should cancel the need for r

[pastel jackal]

hm

okay 1 sec

hm

[tardy epoch]

simplify this

[tardy epoch]

,calc 4 * .05

[warm shale]

Result:

0.2

[pastel jackal]

awesome

i had the right answer all alont lol

[tardy epoch]

you're off by a factor of 10

[pastel jackal]

i just misread my calculator

[tardy epoch]

should be this, but times 100 to get %

[pastel jackal]

cause its dark

yeah i got it right tysm

[tardy epoch]

[pastel jackal]

.close

[obtuse pebble]

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[glass sentinel]

Help pls:)

[tardy epoch]

too small

also

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

[glass sentinel]

What does Only 1 x intercept has a multiplicity >1

[pastel jackal]

it means it crosses at that point more than once

multiplicity 2 would mean it hits the point and then kinda bounces backj

[glass sentinel]

So it crosses the x axis at that intercept?

[glass sentinel]

Oh okay thank u

So if multiplicity is 1 it crosses

[pastel jackal]

if multiplicity 2 it crosses and then instandly crosses back up

[glass sentinel]

Would this graph meet the requirements?

[pastel jackal]

👍

[glass sentinel]

Thank you!

🙂

[pastel jackal]

np

[glass sentinel]

How would I start to figure out the equation

I have my x intercepts

Would it be something like this

F(x)=(x+3)^2(x+1)(x-3)-2

I’ve been trying to graph it on desmos but the y int isn’t positive

And it doesn’t bounce at the x internet

[pastel jackal]

F(x)=a(x+3)^2(x+1)(x-3)^2

that should be it

[glass sentinel]

Is this correct or nah

The range is so messed up

[pastel jackal]

yeah

a is basically the height multiplier

[glass sentinel]

Is there a way to make it smaller

To make my y int like 6 or something

[pastel jackal]

yeah

[glass sentinel]

What would A be

[pastel jackal]

a=0.049379

sorry that took so long i had to find it manually lol

[glass sentinel]

It’s okay

Thank u bro

[pastel jackal]

lol

np

that makes the yint =4 btw

[obtuse pebble]

@glass sentinel Has your question been resolved?

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[dusty spear]

what's the method to turn x^2-10x+16 into (x-2)(x-8)?

[manic escarp]

the simpler one would be to check every factor pair of 16 and see if any add up to -10. remember to check for negative factors too.

[earnest elk]

Splitting the middle could be one

[dusty spear]

wdym "simpler"

is there a harder but consistent method?

[manic escarp]

you could also use the quadratic formula to find the zeros of x

which is what youd have to do for non integer zeros

[dusty spear]

is this a way?

the person said x1 and x2 are roots but idk roots of what

[manic escarp]

ah i see

so the "root" refers in this case to the places where the equation = 0. the "zeros" of the equation. but thats just a generic formula, it doesnt tell you how to solve for the root

[dusty spear]

what about 2x^2-5x-3? Can you walk me through it?

[manic escarp]

sure

have you learned of the quadratic formula yet or no?

if not then i wont use it

unless you want me to

[dusty spear]

I'm learning to do it by factorization

[manic escarp]

ok. so first, the factors of -3 are +-1 and +-3

the equation has -3, so one of the two factors, and only one, must be negative

[dusty spear]

what if it was positive

[manic escarp]

then they would have to be either both positive or both negative

[dusty spear]

how do I know if it's both positive or negative?

[manic escarp]

you dont, so you just test for both

[dusty spear]

so trial and error?

okay

[manic escarp]

so the potential zeros here are either -1 and 3, or -3 and 1.

the x^2 term has a two in front, so the end result should look like (2x +- something)(x +- something)

again, if the zeros are integers

[dusty spear]

so what do I do with this information? Do I do trial and error as well?

[manic escarp]

now hold on. we're getting there. first, let me say that you can factor the two out of the brackets, but in this case that would make the numbers hard to work with so i would leave it.

now, we know that when the equation gets expanded, the 2x will multiply with the constant term (the something) on the right

+-1 and 3 are the potential answers for what the "something" could be btw

so itll be either (2 * 1) + (-3), (2 * -1) + (3), (2 * 3) + (-1), or (2 * -3) + (1), that results in -5x

then you trial and error through those 4 options to see which results in -5

[dusty spear]

so the trick is to just try everything until you get a correct answer?

[manic escarp]

figure out what the potential answers could be through factoring the last term, and then yeah, just trial and error

the more rigorous way is the quadratic formula

[dusty spear]

appreciate the help

thank you

[manic escarp]

np

[dusty spear]

.close

[obtuse pebble]

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[ripe geyser]

I need help factoring

[ripe geyser]

2x^4 -16x

[pastel jackal]

you can pull out a 2x

[ripe geyser]

2^4x ?

[pastel jackal]

no

you can pull out 2x

and it becomes

2x(x^3 - 8)

[ripe geyser]

ohh that's true

than I do the x^3 - 2^3

[earnest elk]

Yes

[ripe geyser]

( x -8) ( x^2 + 8x + 4)

[earnest elk]

No?

[daring rock]

That 8 in the middle isn't right

in the middle of x^2+8x+4

[earnest elk]

Also the (x-8)

[daring rock]

oh

yeah that too

[ripe geyser]

I'm trying to follow the rule

ohh

it's 2 instead?

( x -2) ( x^2 + 2x + 4)

where should I put the 2x that I pull out

[daring rock]

It's just another factor

2x(x-2)(x^2+2x+4)

[ripe geyser]

ohh ok thank you sm

.close

[obtuse pebble]

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[spring frost]

Can someone explain this?

[obtuse pebble]

@spring frost Has your question been resolved?

[spring frost]

<@&286206848099549185> can you help?

[untold fiber]

hey im pretty sure its b

[spring frost]

ah ok

I just need to know why tho

[untold fiber]

ok actually i have no clue sorry

but if u take the derivative and then graph that, there are two critcial points

(0.177, 1,256)

[spring frost]

ok thanks

.close

[obtuse pebble]

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[cloud nimbus]

!claim

[cloud nimbus]

this is the correct tree diagram

this is mine

why for the normal weight cans do they do it without replacement?

as the question just says the 2 cans from the 10 are tested without replacement and doesnt specify for the rest

[opaque flicker]

yes i see what you mean...

what do you mean by do it without replacement?

[cloud nimbus]

as in they do the normal cans without replacement

[pseudo ledge]

Maybe change the 1/5 and 4/5 to 2/10 and 8/10
That might help you understand the diagram better bc “not replacing” would change 10 to 9 which logically makes sense yknow

[cloud nimbus]

thats a good idea, but i was just checking if i missed something in the question or its just poorly worded

because there is no way of telling without looking at the answers if the normal cans are done with or without replacement

[pseudo ledge]

I think the question just requires you to think about it for a while

If the probability of picking a normal can is 8/10 the first time, then picking a normal can the second time would be 7/9 cuz you -1 from top and bottom

[cloud nimbus]

good point

[pseudo ledge]

I think you misunderstood the question as saying “The underweight cans are not replaced” but they’re talking about all the cans

[cloud nimbus]

well it does say 2 cans from the 10 are tested without replacement

but i appreciate the help

.close

[obtuse pebble]

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[timid silo]

how?

[sage geode]

Notice that you can solve for d^2y/dx^2 at x = 1

[timid silo]

so d^2y/dx^2 = 1?

[sage geode]

Yes

You can also solve for d^3y/dx^3 by differentiating the given equation

d^3y/dx^3 + (dy/dx)^2 + y d^2y/dx^2 = 1

And with y, y', y'' and y''' known at x = 1, you can evaluate the series solution

[timid silo]

ok i think i get it

let me try

[timid silo]

how do you find f(0)?

if y=f(x)

[sage geode]

It's given that y = 0 when x = 1

[timid silo]

yeah that means that f(1) = 0

[sage geode]

Yes

[timid silo]

so how do i get f(0)?

idk if this is obvious lol

[sage geode]

You don't need f(0)

The polynomial should have ascending powers of (x - 1)

Not x

[timid silo]

ohhhhh

right

[timid silo]

so how would the expansion look like?

generally

[sage geode]

f(1) + f'(1)(x - 1) + f''(1)(x - 1)^2/2 + f'''(1)(x - 1)^3/6

[timid silo]

yeah

[timid silo]

so f'''(1) = 1 - 2^2?

= -3

[sage geode]

Yes

[timid silo]

nice

i got the ans

ty!

.close

[obtuse pebble]

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[worldly cypress]

Need help

[worldly cypress]

Why did he do sel[1]

[bitter quarry]

I feel like this is either a very well hidden Discrete Maths question, or one for the computer science server

[obtuse pebble]

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[hexed kindle]

I have a product which costs £162.50, a customer wants to buy 3 of the exact same product. That totals to £487.50. There is a 20% tax which totals the entire amount to £585.00. The customer wants a discount where they end up paying £400 altogether. The payment system being used is a contact less payment system which also allows me to add a discount as a number. My question is, how can I calculate the amount to enter to total the entire cost with the VAT to £400 without just doing trial and error.

[hexed kindle]

Note: the reason why I'm finding this difficult is because when I try subtracting the total of £585 by a £185 discount. It doesn't total to my desired amount of £400 due to the tax.

I figured it out! So my thought process of removing £185 from the entire total was the first step. However, due to there being tax involved. What I had to do was deduct 20% (tax) from £185 which makes it £154.17 and deducting this amount from the total of £585 gives me £400

.close

[obtuse pebble]

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[royal basin]

@wary bear have you made progress thus far?

[wary bear]

no

@royal basin

[royal basin]

do you know in general how to solve linear homogeneous ODEs with constant coefficients?

[wary bear]

not really

wait

i know how to solve DEs

but this one is confusing to me

[royal basin]

have you solved DEs of the kind i just described before?

[obtuse pebble]

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[obtuse pebble]

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[normal dagger]

how can I calculate the value from a table

[normal dagger]

does the question mean the normal distribution table

[royal basin]

no, this is a binomial distribution

those usually are not tabulated, to my knowledge.

[normal dagger]

using the bionomial formula?

[obtuse pebble]

@normal dagger Has your question been resolved?

[royal basin]

can't say i understand what you mean by "the binomial formula"

[normal dagger]

I mean this

[royal basin]

this is the probability mass function

[normal dagger]

it's said this one won't work and you need to use a calculator

but I don't know how to

[obtuse pebble]

@normal dagger Has your question been resolved?

[obtuse pebble]

@normal dagger Has your question been resolved?

[obtuse pebble]

@normal dagger Has your question been resolved?

[tardy epoch]

https://www.sjsu.edu/people/saul.cohn/courses/stats/s0/BinomialProbabTable.pdf table source

oh you need n=30. that's not in the pdf. my guess is you use a calculator for all 13 values P(X=0) + P(X=1) + ... + P(X=12)

[tardy epoch]

n = 30, X = 0, 1, 2, ..., 12, p = 0.4, q = 1-p.

[obtuse pebble]

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[sand crystal]

how can move a circle to a point so that the edge of the circle is touching it

[kind hawk]

context?

[sand crystal]

its for a game im making and i need it to be in realtime

[tardy epoch]

you'd have to specify more given information like eqns of each with center and radius, and how you want to "move"

[candid moth]

Is this like a smooth translation or a teleport

And how do you decide where to put the circle. Just move it straight to the point until it touches?

You can check whether they're touching easily by just checking if the distance between the point and the center of the circle is equal to its radius

So the question is just what kind of motion you want

[sand crystal]

teleportation

the problem is if i move the circle up right (1,1) only the corner of the square that covers the circle will touch the point

not the circle itself

while if i move the circle up (0,1) the circle will touch the point

[drifting wraith]

what if you move by 1.2071

[sand crystal]

in what direction

[drifting wraith]

same direction?

you just need to get closer?

[sand crystal]

why 1.2071

hello?

[drifting wraith]

i don't know i was hoping you'd say if it's too far still

[sand crystal]

i need an equation that will give me the right result for any point i want to use

[drifting wraith]

and i want you to try random numbers until it works then i know what the equation could be

try 1.64359

[obtuse pebble]

@sand crystal Has your question been resolved?

[obtuse pebble]

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[smoky panther]

hey by any chance have you guys done tech design?

[smoky panther]

I need help with missing lines sheets

<@&286206848099549185>

[obtuse pebble]

@smoky panther Has your question been resolved?

[smoky panther]

no

[hexed agate]

https://www.youtube.com/watch?v=v32P5CAJaWM&ab_channel=KentEddy

[obtuse pebble]

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[obtuse pebble]

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[pale mortar]

I did sometjing wrong that i dont know, someone can help me out?

[pale mortar]

,rotate

[obtuse pebble]

@pale mortar Has your question been resolved?

[pale mortar]

<@&286206848099549185>

[obtuse pebble]

@pale mortar Has your question been resolved?

[obtuse pebble]

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[warm olive]

ik this is probably really stupid

[warm olive]

but if i^2=-1

is -i^2=1

?

[velvet cloud]

Do you mean (-i)^2 or -(i^2)?

[warm olive]

1st one

(-i)^2

[velvet cloud]

According to Google: -1.

[warm olive]

ahh ok

[velvet cloud]

It's also that according to wolfram alpha, so I think we're ok.

[warm olive]

klkl

thakn u

.close

[obtuse pebble]

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[timid silo]

he¿i

[timid silo]

hi

can someone help me in this one?

[restive pilot]

Whats your issue?

[timid silo]

is that I need to find the area and perimeter of this shape

like this

some of them have colors

[restive pilot]

Sure. Where would you start?

[timid silo]

I want to know how to like look for the area

I need to find the area of the semi circle right?

[restive pilot]

Yes you have 2 quarters of a circle

[timid silo]

yes

[restive pilot]

You able to do that?

[timid silo]

but I can like move them right?

[restive pilot]

Move?

[timid silo]

yeah

my teacher said to like move the colors to another side

you get me?

[restive pilot]

Not really

[timid silo]

xd

dont worry

[restive pilot]

I assume he means that to visualise the areas better

[timid silo]

yes

[restive pilot]

Okay what about the little triangles

[timid silo]

I dont know really, beacuse im starting yet

[restive pilot]

Okay lets just look at the bottom 2 marked areas

[timid silo]

ok

[restive pilot]

By what lines are there limits, what are the borders

What shapes are important to those marked areas

[restive pilot]

The bottom triangle

[timid silo]

the circles and triangles

[restive pilot]

Right.

[timid silo]

isnt that right?

ok good

ok wait

[restive pilot]

Yeye are they within the circle or within the triangle?

[timid silo]

so if there are 4 triangles

I need to get the answer and then multiply it by 4?

only for triangle

[restive pilot]

No we talk about the marked areas that are close to triangles

[timid silo]

circles?

umm hi?

hello?

[restive pilot]

Hi

[timid silo]

ok ok

[restive pilot]

Just the bottom ones

[timid silo]

those are triangles?

[restive pilot]

Not really

They are close though

[timid silo]

yeah

[restive pilot]

Notice the curved site

[timid silo]

but now I find the area of the circle?

[restive pilot]

We got the diameter

[timid silo]

yes

and then?

[restive pilot]

You know formula?

[timid silo]

of cirlce?

yes I know formula

its loke

[restive pilot]

Go ahead

[timid silo]

like*

A = π r²

that?

[restive pilot]

Yes

[timid silo]

ok good

radios I dont know 🙂

[restive pilot]

What is radius to diameter ratio?

[timid silo]

4?

[restive pilot]

Hlaf

[timid silo]

2?

[restive pilot]

Radius = diameter/2

[timid silo]

ok

is it 2?

but wait I dont get the part of diameter

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

hello?

are u there my guy?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[mental plaza]

find slope

and use the same formula

we used just now

[sacred peak]

i did

it said it was wrong

[mental plaza]

what did u try

got a picture?

[sacred peak]

no it was on the problem before this

it deletes when i get it wrong

[mental plaza]

u don’t write down ur work?

do u calculate in ur head

[sacred peak]

the cords were (7,4) (2,1)

[mental plaza]

what

[sacred peak]

sometimes or i just send them in a dm

[mental plaza]

where did (7,4) come from

and (2,1)

[sacred peak]

those were from my last question

[mental plaza]

oh okay

what did u try for this qn

choose any 2 coords

[sacred peak]

first two

slope is -2

so

y--9 = -2 (x-2)

[mental plaza]

eh

what did u type

[sacred peak]

y--9 = -2 (x-2)

[mental plaza]

did u type it in the form y=mx+c

[mental plaza]

bruh HAHAHA

simplify

[sacred peak]

HOW

the last problem didnt need me too

[mental plaza]

y-(-9) =-2(x-2)

what’s -(-9)

[sacred peak]

9

[mental plaza]

mhm

so u have y+9=-2(x-2)

expand -2(x-2)

[sacred peak]

-2x+4 ?

[mental plaza]

yes

so y+9=-2x+4

now on the right side we have y+9

but for point slope form

we only want y on the right

left**

[mental plaza]

left**

so we move 9 to the right side

and what happens to the sign

yea I rly do lmaoo

[sacred peak]

-?

[mental plaza]

yep

so

[sacred peak]

yeppie

[mental plaza]

what’s ur final eqn

[sacred peak]

uh

9=-2x+4? 💀

[mental plaza]

no

where did the y go?

[sacred peak]

uh

y=9+2x+4

?

or am i going completely bonkers

[mental plaza]

why did the 2x become positive

[sacred peak]

y=9-2x+4

wish i knew

[mental plaza]

nope

u had y+9=-2x+4

[sacred peak]

okay

[mental plaza]

how do u remove the 9 on the left side

[sacred peak]

ur lefts and rights r confusing me

u subtract

[mental plaza]

left side means in the left of the equal sign

commonly known as LHS

[mental plaza]

yea

subtract 9 on both sides

so y+9-9=-2x+4-9

and what do u get

[sacred peak]

y=-2x-5

?

[mental plaza]

nice

[sacred peak]

yayy

wanna do the next 1 with me 💀

ok so slope is -2

so

uh

y--1 = -2 (x-1)

then u do the annoying simplifying thing

[mental plaza]

yep

try

and lemme know ur ans

[sacred peak]

ima go search 4 a paper hol up

[mental plaza]

yeye

[sacred peak]

y=−2x+1 ?

[mental plaza]

nice

[sacred peak]

yeppie

thank you, again 😢

.close

[obtuse pebble]

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[timid silo]

how can i prove that a | b is transitive?

[timid silo]

i have said that if a | b and b | c exist in R, then a | c exists in R as well but im not sure how to prove it

[civic zealot]

write down the defintion of divides for both of them, see if you can maybe do something with that

[timid silo]

well i have a = bq, b = cq and then c = aq

but im not really sure what to do with them to prove that c = aq exists

and im not sure if i should make q a different variable for each one as well

[civic zealot]

your definition is backwards

a | b implies b = aq

and yes, you should use a different variable for each one.

[timid silo]

oh right so it should be like b = ax, c = by, and c = az

[civic zealot]

yeah.

can you do something with b=ax and c=by to get to something like c=az?

[timid silo]

ohhhh

so what i did then is i made ax = c/y and then rearranged the equation to get axy = c, finally i said let xy = z since xy is an arbitrary integer and so i got az = c

so then that would prove that if a | b and b | c exist, then a | c exists as well

[wispy wadi]

a | b means there exists an integer k such that b = ak
b | c means ... c = bq
so c = akq, a | c

'cause kq is obviously an integer

[timid silo]

oh lol well ig that would be the easier way lol 😭

was the other method of thinking i did still correct tho

[civic zealot]

that's ok,
writing b = c/y gets weird because division is not guaranteed to maintain integers (it does here because you know y divides c)
so doing direct substitution is usually a better route

[timid silo]

Okay thank you that makes sense

I also wanted to ask if this is correct if anyone has the time

Here I tried to prove it was anti symmetric

[wispy wadi]

b = ax and a = by for some integers x and y yeah
but now you want to prove a = b
b = b(yx) where y and x are two integers such that product is 1
so y = 1 and x = 1, so a = b

[civic zealot]

are a and b both given to be positive integers? otherwise it's not antisymmetric

[wispy wadi]

I assumed that yes, since it seems he has to prove it is antisymmetric

but if you define it on Z, you can obviously have -a divides a and a divides -a

[timid silo]

that makes sense

thank you guys!!!!

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[vocal cosmos]

hello i've completed part a and am not sure what to do for part b & c

[vocal cosmos]

i know that the below is my answer i just need help getting there please

[obtuse pebble]

@vocal cosmos Has your question been resolved?

[vocal cosmos]

i think i've done this right, but i don't know how to turn into fraction form

<@&286206848099549185>

<@&286206848099549185>

it's been almost 40 mins 😦 gotta finish my study guide rip

.close

[obtuse pebble]

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[nimble harbor]

Can yall help me

[obtuse pebble]

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[nimble harbor]

Nvm

[obtuse pebble]

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[cyan sinew]

hi how would i find the derivative of this $\sqrt{1+\sqrt{1+\sqrt{x}}}$

[warm shale]

artichokes

[fierce lagoon]

Multiple chain rules

[cyan sinew]

would it be easier to do it in exponential form or to leave it as is?

[timid silo]

Exponential form yeah

[fierce lagoon]

Up to you

If you're forced to simplify, go ahead and use exponential form

[cyan sinew]

ok ty

.close

[obtuse pebble]

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[forest cove]

can someone explain this I thought it was c cos

i thought it was c coz
if u apply an arbitrary value for voltage, like 10V
then use ohms law to get V/R = I
and the R1 has 1A
and R2 has 0.25A

but thats wrong

[obtuse pebble]

@forest cove Has your question been resolved?

[forest cove]

<@&286206848099549185>

[obtuse pebble]

@forest cove Has your question been resolved?

[tardy epoch]

#old-network for physics server

Or EE server

[drifting wraith]

I would also say c

must be a typo

[forest cove]

ok thanks

.close

[obtuse pebble]

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[balmy flame]

I am trying to find the derivative of the function using part 1 of the fundamental theorem of calculus.

[balmy flame]

After simplifying it, I got to this part but I am not sure how I can simplify it further. The answer is (-sin^4(1/x))/2, but I'm not sure to simplify it to that.

[obtuse pebble]

@balmy flame Has your question been resolved?

[cursive peak]

?

@glad warren

[obtuse pebble]

@balmy flame Has your question been resolved?

[teal ledge]

what is f(x)?

[obtuse pebble]

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[bright scaffold]

Hi, just a theoretical question about Fourier series with periods other then 2pi

[bright scaffold]

How do they determine the expressions for even/odd so quickly?

[feral sedge]

by recognizing that cos pi = -1 and cos 2 pi = 1

In your case, if n is odd, then 1-cos n pi = 2
If n is even, then 1 - cos n pi = 0

@bright scaffold

[bright scaffold]

AHAAA

Brilliant, tysm kelvin

!close

[merry geode]

.close

[bright scaffold]

.close

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[timid silo]

how do I do this

[timid silo]

It's not possible because numerator would be DNE right

after you factor it

[sage geode]

You first need to check if the limit of f at -1 exists

[timid silo]

No

doesn't exist

I factored it to lim as x->1 (x-3/2)/x+1

[sage geode]

Then no such value for f(-1) exists

[timid silo]

why cant i check left and right limits

and what about the first 1

why does it exist

[sage geode]

You can check the left and right limits, they don't exist too

[timid silo]

ye

[sage geode]

The limit of (x - 3/2)/(x + 1) as x -> -1 doesn't exist because the top approaches -5/2 and the bottom approaches 0

[timid silo]

ok

the top exists because after factoring limit

0 wont be in denominator after plug?

[sage geode]

Wdym by that?

[timid silo]

nvm

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[jolly dome]

Is this correct?

[obtuse pebble]

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.close

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[prisma sail]

someone can pls help me with thios

[prisma sail]

idk which is true or false

i think first one is true statement

[obtuse pebble]

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[true lily]

hello, how can I solve this?

[prime yacht]

?

Can you convert from millimeter to centimeter and vice versa?

that would be my hint

[true lily]

yes I converted 50mm^2 to 0.5cm^2

but that was wrong

[prime yacht]

How many millimeter are in one centimeter

[true lily]

10

[prime yacht]

Lovely! So that would mean the first equation is 4500/10 right?

[true lily]

but there is 100 square millimeter in 1 square cm

[prime yacht]

oh right

I forgot that

wouldn't that mean it's 4500/100

[true lily]

so I figured I would divide by 100 to convert

[prime yacht]

right

so you'd have 4500/100

and the other one is 50/1

which means the second one right?

[fading quest]

you are pretty close but again remember that the cm^2 will be in the denominator, so remember what happens when you divide by something in the denominator?

[true lily]

but this is not fraction

so how can it be in the denominator

[fading quest]

it is, its $$50\frac{\text{bacteria}}{mm^2}$$

[warm shale]

Duh Hello

[fading quest]

bacteria per mm^2

[true lily]

oh yeah

50 bacteria / 0.01 cm2

[fading quest]

so $$50\frac{\text{bacteria}}{mm^2}=50\frac{\text{bacteria}}{\frac{1}{100}cm^2}=?$$

[warm shale]

Duh Hello

[fading quest]

yes exactly

[true lily]

so it should be 0.5 bacteria / cm2

[fading quest]

not quite

[prime yacht]

reciprocal!

[true lily]

is there a method I can use to do the unit convertion? For the past 2 weeks I have constantly failed at these same issues

[prime yacht]

dw wait Duh Hello will explain

[fading quest]

remember when you did this

[prime yacht]

why you need to times both the denum and numerator with 100

[fading quest]

its exactly the same from here

[true lily]

yeah I remember that

[fading quest]

$$\frac{50\text{bacteria}}{\frac{1}{100}cm^2}=\frac{50\text{bacteria}}{1}\div \frac{1}{100}cm^2$$

[warm shale]

Duh Hello

[true lily]

I don't understand how you get to this point

[fading quest]

remember that you just got confused at the notation

[true lily]

yes this makes sense

but I don't know how to get to that

[fading quest]

same way you got there here

note that the one on the bottom is the first one and the one on the top is the one you got to after changing notation. its in the wrong order chronologically

[true lily]

yeah if I can just get it into that form it wouldn't be any issue now since I know that trick

but now I have 50bacteria / square millimeter

[fading quest]

this is the same form

can just change $0.01$ to $\frac{1}{100}$

[warm shale]

Duh Hello

[fading quest]

should be 50, not 5, other than that yes its correct

but i think if you just carefully read through this exact conversation again whilst keeping in mind what you did the last time then i think ull be able to get this one on your own

[true lily]

It's the same kind of problem as my last ones but I don't seem to learn the method

[fading quest]

go back to here and carefully read and see if you can apply the logic from this screenshot

[true lily]

I don't understand why I can't just convert 50 mm2 to cm2

[true lily]

I already understand that

I know how to divide 2 fractions