help-0

[vivid anvil]

Here's an example

1 is a factor of 4

cause 1 x 4 = 4

[alpine sable]

ye

[vivid anvil]

2 is another one

2 x 2 = 4

[alpine sable]

ye

[wary stream]

You should just read into factors of integers

[alpine sable]

ok

[median osprey]

Hi

[alpine sable]

hi

[median osprey]

can someone please help me to rearrange this equation

i want Cp0 to be the subject

and its not working out with me

[wary stream]

This is horribly typed

[median osprey]

what do you mean

[wary stream]

What did you try?

Is that supposed to be $Cp^{0}$? Or $Cp_{0}$?

[ocean seal]

dldh06

[vivid anvil]

Lol

It's an exponent

[vivid anvil]

oh

[median osprey]

i did this but its not working out

[vivid anvil]

So it's a subscript?

[wary stream]

The second line is good

That's it

[median osprey]

but i keep getting it wrong

[wary stream]

You made Cp0 the subject

[rich basin]

How would you do a limit question forexample

when x -> 0

for something like 2/x

what would it be

[median osprey]

thank you

[rich basin]

sorry for this really easy question

[warped phoenix]

ohhh right mb, tyy

[thorn hound]

<@&268886789983436800> can i get screen share so i can stream math

[dreamy cedar]

Role added

[alpine sable]

need help im legit just confused

I have a question

Is that from IXL

Pain if it is

IXL is pain

ixl?? idk what that is this is on webwork

Oh cool

ixl is a math practise site

Like webwork you could say

ahh cool

Anyways I remember something like that question from kumon

Of course that was a while ago

[vagrant vault]

IXL learning? Does it have English Science and Math sections? I think I may know what application your referring too

If not nvm

[alpine sable]

for y = 5, find the value of c given that the curves intersect

[hollow swift]

I want to know how to solve it too

[alpine sable]

i need some more explanation idk

well here's a graph with an arbitrary value of c

you want your function to be continuous, so you'd want to choose c such that they "intersect" (they dont really of course) at y=5 (or here x)

oh i see i see

[opal wing]

I am reviewing the FTC and was curious why F(4)=0

[alpine sable]

F(4) is constant so its derivative wrt x is 0

[tall wing]

namd

band

[stiff echo]

Can someone please help me on how to do section 3

It's a,b and c

[thorn kindle]

h=f(g(x))

[stiff echo]

But what is g(x)

Do I have to plug in anything

[soft belfry]

Mathematically

[opal wing]

I figured that out after I asked haha, thank you however for the reply!

[snow ravine]

hello

im new

can someone explain how this works

[opal wing]

You ask

Someone answers

[eternal spindle]

how do I do this?

<@&286206848099549185>

._.

[dark granite]

@alpine sable , @vale wigeon , and @alpine sable . In case this helps...you can count the answer to this combinatorially by counting the complement via the Principle of Inclusion and Exclusion and subtracting from the total. The total would be 4^3 since we're essentially finding the number of functions from {x_3, x_4, x_5} to {y_1, y_2, y_3, y_4}. Now we count the complement. So consider y_3 or y_4 isn't mapped to. Then we note that if y_4 isn't mapped to, we have 3^3 since that's the number of functions from {x_3, x_4, x_5} to {y_1, y_2, y_4}...similar can be said for y_3 not being mapped to. But then we realize we've overcounted the complement since it could've been that y_3 AND y_4 wasn't mapped to. When y_3 and y_4 isn't mapped to, we have 2^3 since that's the number of functions from {x_3, x_4, x_5} to {y_1, y_2}. Thus the answer is given by the following expression.

[ocean seal]

logician

[dark granite]

that^ is the value of n

[opal wing]

I am not sure what I did wrong here, I need to find the area of thay section

I got -60 but since it is above the X-axis, I thought it would be 60 but was wrong

<@&286206848099549185>

[alpine sable]

do you have the answer to this

[rigid smelt]

your integral was not set up properly

notice that relative to the y-axis, x=y^2-5y+20 is under the line x=3y+8

and not the opposite

so the integrand should be 3y-8 - (y^2-5y+20)

also channel seems occupied, you should move (referring to fwa)

[proven briar]

sorry!

[rigid smelt]

np 👍

[alpine sable]

this is what i got

wait sorry its ||2||

second last line is 2I

[opal wing]

Would you be able to show this on paper? I am not accustomed to Y variable of the acis

[alpine sable]

the line stays above the parabola relative to the y axis

so your order of subtraction is off

hi i need help

[dark granite]

hi i need sleep

in all seriousness tho, ask your question and explain what you've done to attempt to answer your question

[alpine sable]

If the weight of 12 sheets of thick paper is 40 grams, how many sheets of the same paper would weigh 2500 grams?

i havent done anything

[dark granite]

okay

this is all about proportions

so how I'd solve this is via the following equation

solve for x

[ocean seal]

logician

[alpine sable]

rule 3: stick to one channel and don't post same question on multiple channels

https://tenor.com/view/mr-potato-cat-nataon-dimden-gif-22528109

[dark granite]

https://tenor.com/view/dog-dawg-eoq-wut-wtf-gif-9118700

https://tenor.com/view/no-nooo-nope-eat-fingerwag-gif-4880183

lmaoo yea don't multipost your question in different channels

[opal wing]

Oh okay, so it is the same as the X-axis, just that I should rotate the graph?

Okay, I also have another question

I am starting to assume the total area is 0 since they are symmetrical and one is above the x axis and one is below

<@&286206848099549185>

[gray gorge]

They're asking for the shaded area, and not the integral (sadly)

[opal wing]

I don't understand

Don't we find the area using the integral?

[gray gorge]

Integral means signed area, which is to say that anything below the x-axis is negative, potentially cancelling out anything above it

[quartz oxide]

I think he tried getting the area and it returned incorrect

[opal wing]

I did, yes

I multiplied the 2.837 by 2 since they are symmetrical

Since I am finding the shaded area and not the signed area, wouldn't my math be correct for the second attempt @gray gorge ?

[placid zinc]

CodeK

[night geyser]

@long magnet do not spam questions channels.

[placid zinc]

The picture of the graph is misleading - the area will be very large. 2.837 is wrong. It's hard to see what you are doing, what techniques are you using for the integral?

[opal wing]

Well, that was my second attempt. For my second attempt, I kept both functions separated and took the integral of both separately. In my first attempt, O combined the 2 functions when looking for intersections and integrated just one function and I got a very high number there

[placid zinc]

So you used u-sub or ibp?

[whole sluice]

,rccw

[whole sluice]

pls help solve question 5.i

Yess!

transpose 3 and 2 on the other side

We will take both 3 and 2 and put it on the other side of the equals to sign and change the signs of 2 and 3 from division to multiplication.

yes

3 & 2 with 1 + 2y and 3 with y - 1

yes

How?

yes

wait are you telling that we are multiplying the whole equation with 6?

and why

yes

i got a solution can i show you

[vale wigeon]

you are missing a ton of parentheses here.

and also something went horribly wrong with the multiplication itself.

[whole sluice]

in that y-1 portion i was about to tell

[vale wigeon]

what exactly did you attempt to do there as your first step?

multiply both sides by 6?

[whole sluice]

pls wait a bit

[vale wigeon]

...okay

then say whatever else you were going to say.

@whole sluice i'm waiting lol

[whole sluice]

thanks for waiting a "bit"

[vale wigeon]

i don't know how long "a bit" is supposed to be

[hoary citrus]

Kekw

[vale wigeon]

and you didn't say anything about what you were doing exactly

were you redoing the problem from scratch? were you trying to continue from the work you presented? you said nothing about that, so i had no way of knowing.

[sleek elbow]

#❓how-to-get-help

[fair slate]

What are u stuck on or confused by?

[glass hare]

I do not know if this is correct

The question is to symbolize the statement using quantifiers, predicates, and logical connectives

[glass hare]

<@&286206848099549185>

[vital pine]

what method am I supposed to use for this

[alpine sable]

looks like the change of base formula might help.

[vital pine]

can I not use partial fraction?

[alpine sable]

you don't have a polynomial at this moment, so no.

[vital pine]

it is a polynomial tho

[alpine sable]

in the numerator, i mean.

[vital pine]

yeah ex, e square x and e cube x

[alpine sable]

sure but it's in the base, so unless you simplify it into ax^n for natural n's, it's not a polynomial.

[vital pine]

yeah so I can write dx / xlogex.loge square x. log e cube x right?

so it won't be in the base anymore

[alpine sable]

not sure what this means, try using log_a for the base a in plain text

[vital pine]

basically what I'm saying is iwhatever is in numerator can be taken in the denominator other than dx, so that it can be converted into a polynomial without the bases

[alpine sable]

but yeah you can use $\log_b(a)=\frac{1}{\log_a(b)}$ into here

[ocean seal]

Al𝟛dium

[alpine sable]

yes, all i said it's that it wasn't a polynomial what you had at that moment, not that you couldn't simplify it into one

[vital pine]

ah ok thanks!

now that it's a polynomial I can use partial fraction and do it yeahM

?*

[eternal spindle]

@alpine sable thanks man how did I miss that lol

[floral imp]

having trouble on approaching how to solve this problem

you are given 5 numbers on a consecutive range from 15 to 19

how would you choose them such that the sum of all choices you make are 100

find longest and shortest solution

you can make a choice on a number you already chose before

[alpine sable]

look, what you have is $\int\frac{1}{x\ln(ex)\ln(ex²)\ln(ex³)}\dd x$ and by using the log laws $\int\frac{1}{x(1+\ln(x))(1+2\ln(x))(1+3\ln(x))}\dd x$

@vital pine

[ocean seal]

Al𝟛dium

[floral imp]

i'm sorry for invading

[alpine sable]

@floral imp this channel is occupied, please read #❓how-to-get-help .

no worries.

[deep scroll]

I have a question

If p,q are primes

Then prove pq>p+q

[alpine sable]

this channel is occupied, please read #❓how-to-get-help @deep scroll .

[deep scroll]

Uh ok

[vital pine]

yeah i got that thanks!

[alpine sable]

do you know how to continue from here?

[vital pine]

this is the solution I git

got*

thanks for your help! @alpine sable I was going nuts tryanna solve this since yesterday

[glass hare]

The question is to symbolize it by using quantifiers, predicates, and logical connectives.

Can someone help?

[alpine sable]

yw.

[vague coral]

what's the problem ? @glass hare

[glass hare]

Symbolize the statement

the statement Some Triangles are Scalene

Im not sure if my answer is correct

[prime badge]

I'd guess you need "in"

[glass hare]

in?

[prime badge]

like set membership, but that doesn't sound right

I don't know, do you have a similar example where you know the answer?

[prime badge]

aha

no I think you got it right

unless you assume x is predicate it's the only option

[harsh cargo]

hi uh can someone help please

its study for an exam and i don't know how to do it

like at all

[placid zinc]

Have you set up the differential equation?

[harsh cargo]

but

[placid zinc]

Remember that acceleration is dv/dt

[harsh cargo]

i don't even know what i'm supposed to differentiate

yeaaa i do know that

[placid zinc]

So the particle is fully described by:

[harsh cargo]

but velocity is just v

[placid zinc]

dv/dt = k(4 - v^2)

[harsh cargo]

so

is that

the accel/

[placid zinc]

Yes, I suppose both sides are a way to represent the acceleration

How much differential equations do you know? This is an example of one.

In that the derivative of v is equal to a function of v itself

[harsh cargo]

ok

that kinda does make sense

but i still don't understand

the answers

say this

i know why they use half

cause you need v not 2v

[placid zinc]

I am not understanding it haha. wtf is x here? What does d(1/2 v^2)/dx get?

[harsh cargo]

wait i kinda get it

ok ok

idk why tf x is there

which is v stupid

but anywaysssssss

they're using 1/2 because

when u differentiate v^2

it becomes 2v

so they half it

[placid zinc]

Right. So the left becomes v(dv/dx)

[harsh cargo]

uh huh

i just

idk

why

x

unless theyre tryna get displacement

[placid zinc]

Oh kek. I forgor that v(dv/dx) = a

Where x is indeed displacement

So all is good, the left side is indeed a

[harsh cargo]

yeaaaaah

wait

do you mind if i dm'd you more

later?

[placid zinc]

And v^2 = 4 + Ae^(-2kx) does indeed fit the condition in the question

[harsh cargo]

yeaaa i get thatttt

[placid zinc]

Sorry I'm not great with dms. Feel free to ping though

[harsh cargo]

oohhhh

okkkk

das okkk

are you

any good

w vectors

by chance

i'm struggling with those too

[placid zinc]

Ye

[harsh cargo]

i never understood vectors tbh

sooo

uhhh

yea

does vector L's tail at point L?

or is the arrow there

[placid zinc]

I have the best paint skills.

[harsh cargo]

period.

[placid zinc]

Green represents vector b.
Red represents vector c.

It's ideal to generally think of all vectors as coming from the origin

You might then wonder why I drew grey as not coming from the origin

[harsh cargo]

ohhh so the vector is classified by the point the arrow is at

but grey

is the resultant

vector

right?

[placid zinc]

Very nice you know where I'm going with this

Specifically, you get this equation:
Red + Grey = Green

[harsh cargo]

how does that work

whaaat

oh wait

nvm

green is resultant

[placid zinc]

Rearranging that, and putting labels:
Grey = b - c

[harsh cargo]

ohhhh

ok kinda kinda get it

[placid zinc]

Now, I want to go up to point C,
Then go only half-way to point B.

Which I can do with:
C + Grey/2

That gets me L. With me on that one?

[harsh cargo]

yesss

i actually do get that

[placid zinc]

Of course, I have a better label for grey:
L = c + Grey/2
L = c + (b - c)/2
L = (b + c)/2

So, it's neat to find that the midpoint of B and C is given by the average of b and c

[harsh cargo]

i seeee

i seeee

[alpine sable]

hey guys i just finished my math exam and i have a lil question

[placid zinc]

Just generally handy to draw and label all of the arrows you'll need, all of then either start at the origin or at other vectors

Then write any loops you have as an eqaution

[alpine sable]

a rectangular wire of length and breadth 7 cm and 9 cm respectively is made into a square so what is the side of the square

i have written square root 63 because side of square is square rooted area is this correct?

[harsh cargo]

ohhh

ok

[placid zinc]

What is the breadth of a wire? haha

[alpine sable]

wait

length is 9 cm

sry

and breadth is 7

anyone?

[rigid wind]

yeah root 63 sounds right

[alpine sable]

yes but 2(l+b) = 2(9+7) = 16*2 = 32 and a square as 4 sides so 32/4 = 8. sooo are both right coz in what i wrote its root 63 but in the second one its root 64

[slow inlet]

Yes

[rigid wind]

oh yeah

[alpine sable]

is what i wrote correct

[fringe turret]

guys whats 1+1

[errant tangle]

2

[icy trail]

10

[light sleet]

Alright so Im doin this problem right? and the angle for c is clearly 45 degrees, but it doesnt show the degrees for B. Since it's a right angle I thought "hmm it's clearly 90 degrees plus 45 degrees (which equals 45) then 180 - 45 (which equals 135) and the answer I submitted was wrong. Why?

[shell widget]

You forgot the 90

180 - 90 - 45 = 45

[light sleet]

wait-

OHHHHH

Oh my gosh I'm stupid, I completely missed that

thanks for pointing that out

[shell widget]

np

[alpine sable]

i need help

[alpine sable]

please?

is that i wrote correct

[cloud mirage]

a

[alpine sable]

yes

thnks

[light sleet]

now this angle i don't understand, what is this

i just need to know the degrees on the angle for A I'll do the rest

[alpine sable]

90 degrees

look at the angle

its a kind of squarey instead of the normal curve when showing an angle

means its 90 degrees

soo

[light sleet]

so: 180 - 90 - 34.. right?

[alpine sable]

90+x+34 = 180

x = 180-90-34

yes

[light sleet]

ah ok

so the way i did it? or ur way

[alpine sable]

both are right

[light sleet]

ah ok

thank you

[alpine sable]

but the way i wrote it just explains more

ur decicion

decision

[light sleet]

i got 66

is that correct

[alpine sable]

its 56

[light sleet]

hm, I must've did something wrong

[alpine sable]

180-90 = 90 ...... 90 - 30 = 60 ....... 60 - 4 = 56 ................basic math dude

[light sleet]

i use calc wrong das why

[unique fjord]

what's the value of b here?

is it a=1 b=0 c=225

[alpine sable]

u mean x

[unique fjord]

?

[unique fjord]

if im right the value of x is just 1

[alpine sable]

x^2 = 0+255 .............. x^2 = 255.................. x= square root of 255 = 15.9

[unique fjord]

hm so u transposed it?

[alpine sable]

yup

[unique fjord]

we have to solve this with quadratic equation tho-

hm

[rigid wind]

225, in which case it's +-15

[alpine sable]

oh-

[unique fjord]

yea that was my answer (with the big help of itsjustnai) with completing the square thingy

[alpine sable]

yup

[rigid wind]

when it's simple as x^2 = a constant you don't bring quadratic formula with a,b,c in

just take +- root 225

[unique fjord]

i see

[alpine sable]

OH WAIT

[unique fjord]

huh-

[alpine sable]

ROOT OF 225 is 15 . sorry i read the 225 as 255

[unique fjord]

its ok XDD

[unique fjord]

hm so technically my answer is just the same? x=15 and x=-15?

[alpine sable]

yea

[unique fjord]

cuz we had to answer the same thing but solve it byy completing the square

[unique fjord]

i see

[alpine sable]

coz -15*-15 = 225 and 15*15 is also 225

square root propertiez

[unique fjord]

yeaaaaa

yep i checked an online calculator-

[alpine sable]

do you know how to square root anyways just askin

[unique fjord]

yep

[alpine sable]

cool

[unique fjord]

wait crap i didnt see number 2-

[alpine sable]

soo

here

(x+3)^2 = 16

its a whole square

i think...

(x+3)(x+3) = 16

[cinder mantle]

nvm

go ahead

[alpine sable]

u can

[cinder mantle]

ok ty

[alpine sable]

go ahead

[cinder mantle]

expanded (x+3)^2 is as its griz said (x+3)(x+3)

it will then look like

(x+3)(x+3) = 16

[unique fjord]

i see

[cinder mantle]

we then expand the bracket

if we expand the brakcet what iwll it be?

[rigid wind]

there's an easier solution to expanding

[cinder mantle]

@unique fjord

[unique fjord]

uh

idk what expanding the bracket means-

[cinder mantle]

oh shi-

ok

[unique fjord]

sorry

[alpine sable]

@unique fjord heres a hint distributive method

[cinder mantle]

nah its fine

[rigid wind]

again, square of something is a constant, you take root of that, so x+3 = +- 4

[alpine sable]

what a smart guys vee

[unique fjord]

oh

o h

wait but isnt that completing square?

idk

[alpine sable]

umm @cinder mantle continue..

[unique fjord]

but like i cant even use the quadratic formula?

[rigid wind]

you can, a lot more work tho

[cinder mantle]

wait

well

[unique fjord]

https://tenor.com/view/aaaa-cat-shout-yell-shaky-gif-17266666

[cinder mantle]

u expand it so it becomes (x^2+6x+9)=16

with expanding u do
(a+b)(a+b)

it will look like

a*a + a * b + b * a + b * b

ultiamtely resulting in a^2 +2ab+b^2

we can apply this to (x+3) ^2 where a=x b=3

hence the result (x+6x+9)

@unique fjord does this make sense

[unique fjord]

uh-

[alpine sable]

i have an easier method -

[cinder mantle]

pls go ahead

i realsied im shit at teaching

[alpine sable]

(x-3)^2 = 16

lol

(x-3)(x-3) = 16

so we will use the distributive formula in another way

so

x(x-3) - 3(x-3) = 16

wait guys

nothing

so then we will just use the normal distributive formula

2x-3x - 3x+9 = 16

5 x - 3x +9 = 16

2x+9 = 16

2x = 16-9

2x = 7

x = 7/1

7/2

i think

[rigid wind]

same method, but you made a mistake, x^2 not 2x

[alpine sable]

ur actually not shit : P

oops sry

[unique fjord]

wait- my dad is explaining it to me at the same time as you-

[cinder mantle]

ahahahhahahah

[alpine sable]

yea : PPP

my dad also taught me

[cinder mantle]

that is actually a really good way of explaining it

[alpine sable]

yea

i also know a lil trignomety if anyone got any questions

[woven shale]

Anyone free to help

[cinder mantle]

what u got cuz

[woven shale]

my question is in 3

[unique fjord]

my dad just multiplied x+3 with x+3 in a grade 1 style like
. x+3
. x+3
. -----
. 3x+9
x^2+3x

=x^2+6x+9

[rigid wind]

that works too

same principle

[unique fjord]

and then transposed the 16 to the other side-

[rigid wind]

i think you should try the x+3 = +- 4 tho :p

2 lines solution

[unique fjord]

i see

[alpine sable]

transposing is like my fav thing in maths

so frickin cool

[unique fjord]

i like math but its just hard to memorize

its easy to understand, i just cant remember anything right after that

[alpine sable]

thats where ur wrong dont memorize!

[glass lichen]

moved the 16 to the other side would be better phrasing, transposed has an actual meaning

[unique fjord]

oooh

wait so how then

internalize?

[alpine sable]

the thing about math is u dont need to memorize u need to remember

like u gtta remember that + changes to - after tranposing

and vice versa

[teal turtle]

when you practice enough you naturally know how to do things

[alpine sable]

i agree with u

im lost

how does a score from b??

[rigid wind]

row i column j is 1 if i connects to j

[alpine sable]

row i?

oh ij princiblw

[glass lichen]

B goes to A, so the B row and the A column get a 1 where they meet

[alpine sable]

ok but look at d and e

d doesnt get a score for e

nor does c score from b

[rigid wind]

oh true

[alpine sable]

any <@&286206848099549185>

anybody got any upto high school level questions

[woven shale]

Are u available rn?

[alpine sable]

ye

[alpine sable]

you really gotta find that with a scietific calc

[woven shale]

I got one

[alpine sable]

or trignometry table

OR ur phone

anyways

[woven shale]

I need working out

[alpine sable]

1st one is 0.86

second one is 0.70

working that out...

[woven shale]

apparently its 3 root 2

[alpine sable]

yea i guess

[woven shale]

Apparently its root 3 over 2

for a

[alpine sable]

heres a table

[woven shale]

yep

[alpine sable]

in college u get to use a scientific calc i think

[royal elm]

when u say apparently, did you find out how to do it?

[alpine sable]

he got it from the internet therefore he has the ability to use apparently

[royal elm]

No i mean does he still need help with figuring it out?

[alpine sable]

do you @woven shale

[woven shale]

yeah

whithout that table how do u get root 3 over 2

[alpine sable]

let me guess

60 +60 = 120

[woven shale]

?

[cinder mantle]

do u not use the unit circle?

[woven shale]

yea im pretty sure u gotta use the unit circle]

[alpine sable]

thats a guess

n its wrong

[left spear]

If it asks for exact values don't use approximations

[cinder mantle]

i thinks its the unit circle
so first quad is theta=alpha where alpha is the related acute angle
2nd quad is alpha = 180˚-theta
3rd quad is alpha = 180˚+theta
4th quad is alpha = 360˚-therta

theta is the angle given in this case, 120

[woven shale]

yep

[cinder mantle]

and 120 falls in which quadrant?

[woven shale]

2nd

[cinder mantle]

correct

now u do alpha = 180˚-120

which gives u 60

[woven shale]

yep

[cinder mantle]

so the related acute angle for sin 120 is sin 60

which gives u root3/2

[woven shale]

how is sin 60 root3/2

thats what im confused about

[alpine sable]

sin 60 degrees

[cinder mantle]

i got a proof somewhere

let me whip it out

[woven shale]

k

[cinder mantle]

i like to memorise the trig triangle instead of the table

table is a fat trek

[woven shale]

OHHH

Now I remember

Thank you

[cinder mantle]

yw

[fringe robin]

Consider a square with vertices at (1, 1), (-1, 1), (-1, -1) & (1, -1) Let S be the region consisting of all points inside the square which are nearer to the origin than to any edge Sketch the region S & find its area

How do i sketch this?

[glass lichen]

plot the points

then draw the square..

[fringe robin]

bruv

Let S be the region consisting of all points inside the square which are nearer to the origin than to any edge Sketch the region S & find its area

this part

[glass lichen]

Oh yeah, sorry I just woke up lol

[fringe robin]

Lmao np

[glass lichen]

https://math.stackexchange.com/questions/2515367/area-of-points-in-a-square-closer-to-the-center-than-the-edge

[woven shale]

<@&286206848099549185>

[fringe robin]

Occupied

[fringe robin]

ill see thanks

[cinder mantle]

this is using the cosine rule

what is the cosine rule?

soz i didnt mean to ping helpers

mybad

[woven shale]

a^2+b^2-2(a)(b)Cos(c)?

@cinder mantle

[cinder mantle]

lets goooo

gj

cos C is the given angle

[woven shale]

yep

[cinder mantle]

a and be are the two sides either side of them which is 8 and 7

chuck that all in the formula u get ur c^2

[woven shale]

Ok then root

[cinder mantle]

correct

[woven shale]

and answer

[cinder mantle]

thats ur answer

:)

[woven shale]

Ty once again

[cinder mantle]

yw :)

[fringe robin]

?

[obsidian wagon]

hey so here I don't understand how we multiplied cos(theta)sin(theta) with [cos(theta)/sin(theta). + sin(theta)/cos(theta)] to get cos^2(theta)+sin^2(theta)=1

is it because we multiple cos and sin both sides to give this?

[rigid smelt]

no, they just rewritten cot(x) and tan(x) in terms of sin(x) and cos(x), thats all

[obsidian wagon]

in regards to the final answer

i understand the cot and tan rewritten

what i dont understand is the ones in the bracket of second part :/

[rigid smelt]

which ones in the brackets?

[rigid smelt]

cos(theta)/sin(theta) + sin(theta)/cos(theta)

then again

like i have said

they just rewrote cot(theta) and tan(theta) thats all

and then they expanded the thing

[obsidian wagon]

how did this give cos^2 theta + sin^2

[rigid smelt]

by expanding the brackets

notice that cos(theta)/sin(theta) cancels out with the sin(theta) in cos(theta)sin(theta)

equivalently for sin(theta)/cos(theta)

[obsidian wagon]

ohh wait i see

but how does it square up if they both cancel out

*or expand to cos^2theta+sin^2theta=1

[ocean seal]

waler

[rigid smelt]

similarly for the second term

[obsidian wagon]

ohhh!

i get it now

this makes sense thanks @rigid smelt

[alpine sable]

FFFFF this makes a lot of sense, Thank you so much!!

[magic warren]

this cant be factorised right? a(b+c)-d(b+c)

[candid sluice]

it can

take (b+c) as a common factor

[magic warren]

oh wait im so dumb

thanks @candid sluice

[glass lichen]

what info are you given?

[rich basin]

Just that

[glass lichen]

No tables of data you can use?

ie density / bulk moduli?

[rich basin]

no

[glass lichen]

Then you cant solve it

[rich basin]

just that question and i'm pretty sure the information relating to that would just be in like a data table you can search up

[glass lichen]

Yeah, so search up the density and bulk modulus of each material

then plug into the formula

[rich basin]

so what exactly is the formula?

[glass lichen]

$v=\sqrt{\frac{B}{\rho}}$

[ocean seal]

Mosh

[glass lichen]

for bulk B and density rho

[rich basin]

is there like a formula sheet where i can get like all the formulas for a certain topic?

[uneven sorrel]

.

[rich basin]

Like it is just all over the place, and some of them I have no idea without a formula sheet

[glass lichen]

Google

or, use the one for the course...

[rich basin]

Not google, but like a data sheet all

[glass lichen]

Im not going to do googling for you

[alpine sable]

Lol

[rich basin]

@glass lichen Wait how do you know the formulas of by heart?

[glass lichen]

I have a formula sheet

from my physics course

[rich basin]

Could I please have your formula sheet from your physics course

I don't even hve one

[glass lichen]

Ask your prof about a formula sheet, or check the textbook

[urban gust]

If there is a multivariable function which is not continuous at a point will its partial derivatives exist?

[ionic jewel]

partial derivatives don't imply continuity, but continuity does imply partial derivatives

[glass lichen]

if m+n is even, then m and n have the same parity

[restive hill]

Hi everyone Good evening

Can someone help me this?

[glass lichen]

even and odd

is the parity of the integer

yes

though m and n dont have to be the same number

[vernal dragon]

i tried to substitute sint=u and then integrate why is that wrong?

in the correction they prefer i take into account the trigonometric rules

[glass lichen]

but yeah, m+n is even means either m and n are odd, or they are even. however if m and n are odd, mn is odd

so m and n need to be even

[alpine sable]

did you change the upper and lower bounds

[glass lichen]

And did you account for the absolute value bars?

[alpine sable]

i got the answer as 1 i maybe wrong though

wait mb

I havent done integration in school yet so please check my answer

its coming

,r

,rotate

[split oriole]

Is that modulus?

[glass lichen]

abs value

[vernal dragon]

why do we need to change the interval when we change the trig identity

?

[split oriole]

Multiply divide by 2 then make 2sinx cosx to sin2x , them split integral in postitive and negative part

[vernal dragon]

allahs goat this server is busy

[woven shale]

oh

[glass lichen]

you dont

you change the bounds

[vernal dragon]

bounds?

[glass lichen]

however you havent done a sub there, so nothing changes

[vernal dragon]

they change the interval when they change the trig identities

do you agree with that statement?

@glass lichen

[gray isle]

they skipped steps

[vernal dragon]

damn that makes sense

but i am still not getting the same resuly

resuly

result

this is how i did it

[gray isle]

you can't ditch the abs value bars

[vernal dragon]

oookey damn

[gray isle]

it may help to look at a graph of the function

[vernal dragon]

still when i integrate sin ( ) i get minus cos. the abs bar makes it plus cos. cos 2pi - cos 0 is still 0

[gray isle]

you can't really integrate over the absolute value like that

[gray isle]

$\int |f(x)| \dd{x} \neq \big|\int f(x) \dd{x}\big|$

[vernal dragon]

damn

[ocean seal]

ℝamonov

[gray isle]

it may help to look at a graph of the function

and consider what's happening geometrically

that is a graph of y = |sin(2x)| from 0 to 2pi

[gray isle]

what's ds

[vernal dragon]

the period is 2 times shorther

[vernal dragon]

ds is length

[upbeat helm]

It's not possible to find all the angles here right?

Like you can't do it.

[gray isle]

the period is 2 times shorther
wdym

[alpine sable]

there are other channels

[upbeat helm]

uh

idk

[vernal dragon]

sin (2x) compared to sin x is has a shorter period

agree?

[gray isle]

yeh.

[vernal dragon]

how do i integrate this equation

the abs bars are confusing me

[gray isle]

do you have any issue accepting that the above graph is of what I said it is

[alpine sable]

May I ask why not do a u=sin t substitution?

[gray isle]

you still need to deal with the abs val bars in one way or another

[vernal dragon]

in the correction they are not prefering it but i tried it and got 0 as anwser and thats wrong

omg

i got it

because of the abs bar i to add when i normally should subtract

right?

[gray isle]

very bad wording

[vernal dragon]

can u attach where it is bad wording so i can rewrite

[gray isle]

because of the abs bar i to add when i normally should subtract
the whole thing is bad and unclear

and it may also seem in one interpretation you are still misunderstanding

[vernal dragon]

usually sin 2t looks like this

because of out abs bar it looks like your attachment

in my exercise they are asking me to determinate the area of the function, where the interval is t € (0,2pi)

[alpine sable]

im pretty stupid but why not break the intervals from 0 to pi to pi to 2pi

[vernal dragon]

if u let me finsh i was going to say that i promise

[alpine sable]

sorry

[vernal dragon]

after breaking up the intervals and add them together i got 4

(3a *4)/2=6a

[vernal dragon]

its all love brother

[gray isle]

it looks like you're still fudging it

[vernal dragon]

where

i am asking so i can learn

i agree

[gray isle]

the worked solution is supposed to be a relatively clear way to solve it

you're making a crap ton of random arbitrary splitting and illegal sign adjustments

that i can't even begin to describe

[vernal dragon]

ok damn

[gray isle]

and i was in the process of guiding you to what they did

but you went off on your own

[vernal dragon]

ok think i need to go back to basics but thanks for your help

[gray isle]

don't try and integrate over absolute values like that

it doesn't end well

going back, do you have any issue accepting that my graph is of what I said it is

[vernal dragon]

i dont have any issue that your graph represent (sin (2x))

[gray isle]

are you familiar with the relation between the the definite integral and the signed area between a curve and the x-axis?

or what people refer to as the signed area under a curve

[vernal dragon]

i am not familiar with signed area under the curve

[gray isle]

what about just area under the curve

[vernal dragon]

i am familiar with area under the curve

[gray isle]

when calculating the area from the integral, stuff above the x-axis is counted as positive, under the x-axis is counted as negative

[vernal dragon]

i 100% agree with that statement

[gray isle]

e.g. if you were to integrate
y = -1 from x=0 to 1,
you'd get -1 which is the signed area

instead of 1 which would be the actual area

[vernal dragon]

yes i agree

[gray isle]

that's more or less the difference

anyway moving on

note that the period of |sin(2x)| is pi/2

as indicated from the graph above

[vernal dragon]

i agree that the half period is from t €(0,pi/2)

[gray isle]

what half period

why did you add the word half in there

[vernal dragon]

are u telling the that from 0 to pi/2 its a whole period?

[gray isle]

the period of sin(x) is 2pi
the period of sin(2x) is pi
the period of |sin(2x)| is pi/2

[vernal dragon]

or should i forget the term

[gray isle]

because as indicated by the graph i made,

[vernal dragon]

i agree with that

[gray isle]

the graphed function |sin(2x)| repeats every pi/2

[vernal dragon]

yes

[gray isle]

and hence you have 4 regions with the same area under the curve there

from 0 to 2pi

[vernal dragon]

agree

[gray isle]

so to calculate one of those regions you can integrate |sin(2x)| from x=0 to pi/2

[vernal dragon]

-cos (2x)*1/2+c

if we integrate sin (2x)

[gray isle]

since you're considering 0 < x < pi/2, then 0 < 2x < pi
the sin of 2x will be positive and the absolute value bars can be removed now

only then should you actually do the integration

[vernal dragon]

thanks for your patience with me

[buoyant willow]

Hey ho, don't need much help, just need to clarify a concept.

Usually functions like this start with a y in the beginning, but this time it's in the middle, do I still solve it the same way?

[ancient saddle]

Hello. Yes, it represents the same relation between x and y, it's just been rearranged

[vale gorge]

Help: Find positive intergers a,b such that: (a+1)^2-4b and (a-1)^2-4(b-a+1) are square numbers.

[buoyant willow]

Thank you!

[vital snow]

can anyone help me?

[shell widget]

@vital snow Use the formula for a straight line, y - y_1 = m(x - x_1) where (x_1,y_1) is a point on the line/a point the line goes through.

and m is the slope

[vital snow]

then would i put it back into y=mx+b?

[shell widget]

no

[vital snow]

@shell widget

[shell widget]

They tell you that the line you want is parallel to y=mx+c so that you can know what the slope of your line will be

[vital snow]

the slope will be the same on the parallel line right?

[shell widget]

@vital snow The slope of the line required(which passes through the point (a,b)) will be same as the slope of the line y=mx+c

And that slope is "m"

[vital snow]

i don’t understand what point (a,b) is for

[shell widget]

Plug it in the equation y - y_1 = m(x - x_1)

This is the equation for a line with slope m and goes through point (x_1, y_1)

[vital snow]

so y-b=m(x-a)

[shell widget]

Yes

[vital snow]

and then isolate y?

[shell widget]

Yes

not necessary but sure

[vital snow]

because i can’t change the

y=

when answering

[shell widget]

ok

[vital snow]

so for the perpendicular

since perpendicular is -x/m

[shell widget]

?

[vital snow]

perpendicular line’s slope is -x/m

[shell widget]

slope will be negative reciprocal of m

-1/m

and then same thing, plug the point and the slope in y - y_1 = m(x - x_1)

[vital snow]

so what would that look like in point slope?

y-y1”-1/m(x-x1)?

[shell widget]

you forgot that (x_1, y_1) is (a,b)

[vital snow]

so y-b=-1m(x-a)

-1/m

[shell widget]

y-b=(-1/m)(x-a)

[vital snow]

so would i foil it out

the right side

[shell widget]

If u want

[vital snow]

what did i do wrong for the top?

[shell widget]

y-b=m(x-a)

y-b=mx-ma

y=mx-ma+b

[vital snow]

ohhh

thank you

[hybrid moon]

Are rectangles inscribed in a right triangle usually solved using optimization

[rare adder]

747 digits were used to number the pages of the book. how many pages were there in the book? can anyone help me with this?

[hybrid moon]

How would I solve this particular YouTube problem if they didn’t ask for the largest rectangle area

https://m.youtube.com/watch?v=UzftAa_sXYs

[fading zephyr]

dude it says on the thumbnail

"largest"

[hybrid moon]

No the goal is to solve not the largest

[fading zephyr]

so yes, you'd make a function that incorporates the dimensions of the triangle into the computation and then take the derivative

[hybrid moon]

I have a similar hw question

[fading zephyr]

it says it there

[hybrid moon]

But I don’t want to post that one

[fading zephyr]

if in your HW you are not looking for the largest, it isn't the same problem

[gray isle]

if no size restriction is stated, anything goes

[fading zephyr]

what ramonov said, which is the same as i told you in discussion

[hybrid moon]

What I’m asking is how do I solve this question if they didn’t have a size restriction

[fading zephyr]

just use geometry and trig

[hybrid moon]

I forgot that lol

That’s what I’m asking

[fading zephyr]

we can't say anything else if you don't show the actual problem

[gray isle]

if no size restriction is stated, anything goes

[fading zephyr]

cuz we don't know what you are given nor what you need

[gray isle]

the best you'd get is an expression for area in terms of variables

[fading zephyr]

"trig and geom" is the best we can tell you

[hybrid moon]

Damn

Cuz I’m not allowed to post hw questions

[gray isle]

which is more or less what i suspect they'd start with in the vedior before optimisation

are they copyrighted by your school or something?

[hybrid moon]

No don’t want to get in trouble for cheating

[gray isle]

homework is usually ok. tests are not

[hybrid moon]

Yeah but my school is wack

[fading zephyr]

you could make a drawing in which you indicate which quantities are known and which ones are sought after, without any numbers

just assign them letters

[hybrid moon]

Alright I’ll do that

[fading zephyr]

then people here can help you set up your equations

[hybrid moon]

That’s actually a good idea

[gray isle]

,rotate

[gray isle]

apply properties of similar figures to express y in terms of x

[hybrid moon]

Essentially:

16-y/x=16/10

[gray isle]

why 16-y/x

[hybrid moon]

The upper triangle length

[gray isle]

not sure what what you're doing

by \verb|16-y/x| did you mean $\frac{16-y}{x}$

[ocean seal]

ℝamonov

[gray isle]

it seems like you're trying to take the ratio of those two sides

[hybrid moon]

Yes I am

[gray isle]

which is pretty much meaningless

[hybrid moon]

Y

[gray isle]

why did you think that gives a ratio of 16/10

and how are similar figure being applied

the blue segment and green segment aren't even sides in the same shape

instead you should simply consider

the red triangle will be similar to the largest triangle

[hybrid moon]

Wait okay so proportion should be 16-y/y=10-x/x

[gray isle]

overcomplicating it

[hybrid moon]

Ok

16-y/y=16/10

[gray isle]

why not just focus on the red triangle instead of wandering to the triangle in the top left for some strange reason

no

and also please use parentheses around your numerators and denominators to make them clearer

why not just focus on the red triangle instead of wandering to the triangle in the top left for some strange reason

[hybrid moon]

Y/x red

Correct

[gray isle]

yes, the ratio of the legs in the red triangle is y/x

what about the ratio of the legs in the largest triangle

[hybrid moon]

Omg I’m an idiot

16/10=y/x

[gray isle]

yeh

[hybrid moon]

Ok then what do I do next

[gray isle]

express y in terms of x

[hybrid moon]

10x/16

=y

[gray isle]

no

you didn't manipulate your equation properly

[hybrid moon]

Omg

Ok just tell me once I manipulate it

What is my next move

[gray isle]

well you have one side of the rectangle in terms of x

get the other side in terms of x

their product will be the area

division by 0

[eternal spindle]

pls help

[cinder finch]

The triangle is equivalent to (y * 10) + {(x * y)/ 2} + {(16 - y) * 10)} = The Area of the triangle

You would be required to find equivalent sides on each end which will give you an expression of what the area of the triangle would be

[real quartz]

did you find out fx and gx

[eternal spindle]

fx is given

and gx is the inverse

i know that a = 1

but idk b and c

[real quartz]

uhmm hmm

b is equal to 0

[hybrid moon]

What

[gray isle]

what what?

[real quartz]

the form ax+b / cx^2 + x + d doesnt have a horizontal asymptote unless c is equal to 0

[gray isle]

@hybrid moon

[real quartz]

that is, it is make sense that c=0, and that we have a form ax+ b / cx + d

[eternal spindle]

but this one has a horizontal asymptote

y=1

[real quartz]

oh