help-36

[supple jolt]

i was talking about the first pack

you want to find how much it costs per kg

so you can compare it to the second pack

[novel walrus]

you would get

150 i think

1K- 150

@supple jolt

bro listen

im confusing myself

[supple jolt]

@novel walrus

what do you not understand

[novel walrus]

everything

[glass spade]

i need help with this integral

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[strange pike]

hi

[strange pike]

i need help with trigonometry

so basically ik if you have a shorter side that you have two answers

and i have a problem with understanding when to use that information in math problems

is there an easier way

eg. i have been given 3 heights of a triangle and there are 2 answers

[spice depot]

U talking about using law of sin?

[strange pike]

not sure what that is

probably

[void valley]

Probably it would be more clear if you gave as an example please

[strange pike]

.

[onyx peak]

are you calculating the sides or drawing the triangle?

[strange pike]

im calculating the angles

i did calculate them

[onyx peak]

ah

[strange pike]

but there is a second answers

answer*

[spice depot]

Give example

Like a problem

[strange pike]

im gonna give a second example

the information given is

a=32 cm

R=18cm

gamma=33

i need to find the leftover sides and angles

[spice depot]

What is R?

[strange pike]

second

[spice depot]

?

[strange pike]

a circle

[spice depot]

Oh

[strange pike]

outside of the triangle

idk what its called in english

[spice depot]

So we talking right angles?

Do u have a picture?

[strange pike]

both is no

so i did calculate one set of answers

but im not sure why there is another

[spice depot]

Ur not giving me much to go of off here

[strange pike]

yeah i dont have much info too

[spice depot]

Can u draw a picture and send it?

[strange pike]

nvm i got it

ty

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[ashen magnet]

Cos3xcosx

[ashen magnet]

How would I change it's form so I can integrate it. Idk integration by parts so I have to do it this way

[solar glade]

Rewrite cos(3x)

[ashen magnet]

I'll try

[solar glade]

Cos(x+2x)

[ashen magnet]

You can do that?

I didn't know

[solar glade]

What do you mean?

x+2x=3x

I imagine if you ask that you dont know cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

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[iron dagger]

what part do you need help with

[ashen magnet]

True

[subtle pier]

i did this in my exam

i didnt know how to simplify

so i took values for r and n

and i got 0

WAIT

yea i got 0

i remember i got same values for 2A10 and A8

but the answer key says its option 4

[iron dagger]

close

the values for A10 and A8 are the same

not 2A10 and A8

[final saddle]

@subtle pier Has your question been resolved?

[subtle pier]

oh 🥲

is there a method for this

to simplify the thing

[iron dagger]

?

you just expand terms out and see what cancels

matter of fact the best thing to do time-wise is to find only one determinant, Ar

[subtle pier]

oh i see

so theres no elemntary transformation or anything

[iron dagger]

well, you can

but I don't remember those and just computed the determinant by hand

[subtle pier]

ahk

thanks a lot

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[deft depot]

im having trouble using arcsin to figure out the angle in degrees

ik that arcsin has a domain between -1 and 1

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@deft depot Has your question been resolved?

[deft depot]

<@&286206848099549185>

[tranquil pine]

Reek weetqkiltkuyriktkitiikrjriruiukuuuu7eirkejukkèèqqelkkrkueirkqkklekrikwlekeiuieulekekekrkkieteiiaywkklkkrkeykyeykektiiwiueekkwekkkwukttwtkwtiteiwti

1+1 = 3

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[lyric lichen]

How do I solve this? Do I get a function from the lines?

I did this problem before hand

[spice depot]

Just look at the points

Its easy

[lyric lichen]

which points

any point?

[spice depot]

No the ones they are asking you to compute

[lyric lichen]

idk what you mean

[spice depot]

Bruh, wat is f(-8)?

Or g(-8)

Its in the graph

[lyric lichen]

okay so there two lines

f(x) g(x)

[spice depot]

Yes

[lyric lichen]

f+g

BUT

what are equations for f+g

[spice depot]

U dont need them

[lyric lichen]

OKAY

i need to add something SO

what is it

[spice depot]

(f+g)(x) = f(x) + g(x)

Its the same for the others

[lyric lichen]

OKAY

for the other problems there is an equation right

(6-9t)-(2t^2-8t-7)

r-f

[spice depot]

Okay?

[lyric lichen]

for THIS

WHAT

is the equation

[spice depot]

U dont need it, the information is in the graph

[lyric lichen]

HOW

[spice depot]

U need to compute a number

[lyric lichen]

HOW DO I GET THE GRAPH NUMBERS

[spice depot]

Bruh?

[lyric lichen]

bruh Ive asked several times

you're either not getting it or trolling

[spice depot]

Ok looking at the graph. What is f(-8)?

[lyric lichen]

f(-8) is the plug after I solve (f+g)

[spice depot]

No

[spice depot]

This is not the same problem, u dont need no equation

Do you know how to read a graph?

[lyric lichen]

what about the graph

i've been asking about the graph the whole time

you just say "you don't need it"

-8

[spice depot]

U dont need an equation

U need the graph

[lyric lichen]

okay what about the graph

[spice depot]

Looking at the graph wat is f(-8)?

[lyric lichen]

I don't know

what do you meaaaaaaan

[spice depot]

I cant explain how to read a graph to you on text

U need to review functions and reading graphs to solve this problem

Watch a video

[spice depot]

Ive been saying to look at the graph this whole time

[final saddle]

@lyric lichen Has your question been resolved?

[final saddle]

@lyric lichen Has your question been resolved?

[lyric lichen]

No

[spice depot]

Bruh go watch some videos

U cant be solving these problems without knowing how to read a graph

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[river flower]

.reopen

[grand niche]

how do i solve these 4?

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@grand niche Has your question been resolved?

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@grand niche Has your question been resolved?

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[prisma lantern]

How do I get x_1 and x_2 from the graph?

[wintry kindle]

what are x1 x2

I presume you refer to intersections and have some formula for area which includes some variables named x1 and x2

[prisma lantern]

How do I get them?

[young verge]

Are you talking about your integration bounds?

[prisma lantern]

Yeah

[young verge]

set f(x) = g(x)

[prisma lantern]

Ohh

But on the graph, g(x) is always higher

[young verge]

that would mean that it would be ∫g(x) - f(x)

[prisma lantern]

But wouldn't you need 2 integrals?

Since x=-2,0,2

@young verge

[young verge]

yea

[prisma lantern]

How would that work?

INTf(x)-INTg(x)?

[young verge]

well since g is on top of function f

This would be ur first integral

and then add the other integral

[prisma lantern]

What's the other integral?

[young verge]

the same thing except the bounds are 0 and 2

[prisma lantern]

Ohh ok ok thanks

[young verge]

You should get like 8.53 if im not wrong

[prisma lantern]

Yeah, I got it

Thank you so much

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[vital verge]

!hell

[vital verge]

<@&286206848099549185> how can i solve?

if 5 = 2x, is x sure to be 5/2?

[weak vortex]

It should be $\frac{5\sqrt{6}}{4}$

[soft zealot]

Bagchi234

[weak vortex]

You use trigonometry in the first triangle to find out is perpendicular, that is gonna be the hypotenuse of the second triangle

from there you solve it

[tranquil pine]

or this

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[tranquil pine]

how do you solve question 8

[wanton pine]

what do you know about the discriminant?

[tranquil pine]

b^2-4ac

[wanton pine]

yeah but specifically what does the discriminant tell us about a quadratic?

[tranquil pine]

how many solution

[wanton pine]

yeah

[tranquil pine]

what do I do with the k

[wanton pine]

so what do we need the disiminant to be for us to get 2 solutions?

[tranquil pine]

no clue

[wanton pine]

well think about it

if we have x= 2±√y as the solutions for x

if y is less then 0 what happens?

[tranquil pine]

no sloutions

[wanton pine]

yeah if its 0 then what happens

[tranquil pine]

1 solution

[wanton pine]

yeah so if its greater than 0 we must have 2 solutions right?

[tranquil pine]

yea

[wanton pine]

so if we want 2 solutions b^2-4ac>0

[tranquil pine]

what do I do after pluging in the numbers

[wanton pine]

solve for k

[tranquil pine]

i got k>-9 what do i do what that

[wanton pine]

let me see if its right

yeah no thats right

well that's it solved

[tranquil pine]

what is k

[wanton pine]

greater than -9

[tranquil pine]

can't i get a exact answer

[wanton pine]

it's asking for the 'values' of k, meaning there is more than 1

[tranquil pine]

so k and be anything higher than -9

[wanton pine]

yep

[tranquil pine]

it can be like -7,-8

[wanton pine]

yeah

[tranquil pine]

alright

why tho

arent there supposed to be 2 solutions

[wanton pine]

well when k>-9 the discriminat will always be greater than 0 and if the discriminant is greater than 0 we get 2 solutions

[tranquil pine]

ohh that makes so much sence now thanks

love you

[wanton pine]

all g

[tranquil pine]

.close

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[glad steeple]

i have to try solving this equation

[glad steeple]

using the famous formula

but i cant get , how i could see that 4 is a solution for t?

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[boreal basin]

Can someone explain how the answer is positive

I’m having trouble finding where I did smth wrong

[steep tendon]

what was the original question?

[steep tendon]

this is 4?

[boreal basin]

Yeah

[steep tendon]

mm if u just think about it, the answer is going to be positive because it is area

[boreal basin]

But there is no algebraic way to prove it?

[steep tendon]

well,im not sure because i haven't revised on integrals, but i would expect something small has happened along the way

[boreal basin]

yeah prob

thanks though

can anyone else spot my mistake?

i think i just put my upper and lower limit wrong

[steep tendon]

just the integral for the 1-x^2 isn't negative

u see, all of that 2/3 area is above the x axis

[boreal basin]

yeah but when i simplify

its left with the negative

[steep tendon]

weird

[boreal basin]

yeah i just put my limits wrong

wait

no i didnt

[steep tendon]

$x-x^3/3$

[soft zealot]

shark in a pool

[steep tendon]

that's the antidiff right?

[boreal basin]

no

antideriv for x^1/2?

[steep tendon]

oh nah not that one

1-x^2

[boreal basin]

yeah

still the same thing

[steep tendon]

$1/2sqrt(x)$ is diff for that iirc

[soft zealot]

shark in a pool

[steep tendon]

inputting 1 = 2/3 and inputting 0 is 0 iirc

[boreal basin]

iirc?

[steep tendon]

dont worry about that i just put it on random sentences

it means if i recall/remember correctly

[boreal basin]

yeah

[steep tendon]

2/3 - 0 is 2/3

[boreal basin]

but there is still the negative

its times though

got the answer

i put the functions in the wrong spots

1-x^2 is the upper function

.close

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[steep tendon]

ah perfect

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[haughty spindle]

how to approach this problem

[haughty spindle]

nvm i figured it out

.close

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[south wave]

Can some1 please tell me if my answer is correct or not

[south wave]

<@&286206848099549185>

[gilded spade]

not right according too my calculations

each curve of the semi circle is 10pi and you got 3 so 30pi

and then you got 2 radi which add up to 20

so should be 30pi + 20 which is 114cm

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[south wave]

Ty bro

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[south wave]

Is this correct

[south wave]

The answer i got was 32.7

<@&286206848099549185>

[frank fulcrum]

Looks correct, just double checking

you're correct up to 12.466

so yeah, looks good to me

[south wave]

.close

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[olive socket]

need someone to help me check if my calculation of the determinate ins right

[pliant shore]

Sure

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@olive socket Has your question been resolved?

[olive socket]

my answer I got is -p^2sin^3(∂)cos^2(theta)-p^2sin^3(∂ )sin^2(theta)-p^2 * cos^2(∂ )*sin(∂ )

[pliant shore]

You should be able to use the identity cos^2 + sin^2 = 1

For both theta and phi

If you do it right you should end up with -r^2 sin phi

And you take the absolute value of that ofc

[pliant shore]

Factor stuff out to use the identity

[olive socket]

wdym?

idk how I could use the identity to reduce it down that much

[pliant shore]

-p^2sin^3(∂)cos^2(theta)-p^2sin^3(∂ )sin^2(theta)

You can factor out a -rho^2 sin^3 phi here

Then the other bracket will be just cos^2 theta + sin^2 theta = 1

[pliant shore]

Btw

[olive socket]

oh I'm dumb I was looking at my orginal matrix lmao

thanks mate

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[dim laurel]

I am so close to finishing this understanding I just need a few more few hints and I should be able to get it

[dim laurel]

I’m not sure where to go after figuring out it’s factors must be 2 and 0

Oh btw we are using the assumption that for any <I> over the a field Q[x] if I is irreducible in Q[x] then <I> Is a maximal ideal

So I’m trying to prove <I> is irreducible in Q[x]

Which should show that <I> is maximal

Cause you can use 1/n and n for some n in Q[x] as the constant

Since the units of Q[x] are it’s constants

It sucks cause I feel like the answer is just within reach

Like I’m right there

It’s probably due to the fact that n*1/n is just 1 but I’m not sure if I can just say that any ((1/n)x^2-(1/n)2) n ⇒ (x^2-2)1 for any n in Q[x] such that n is constant

Or can I?

<@&286206848099549185>

I should’nt be using 1/n I should use n^-1

It’s bad practice

Or wait does that even matter

Either way it’s still irreducible in Q[x] right?

Regardless what n is

<@&286206848099549185>

Wait is it cause Q[x] is a PID

Meaning the ideal can be generated by only 1 element

Which in this case must be itself?

<@&286206848099549185>

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@dim laurel Has your question been resolved?

[dim laurel]

<@&286206848099549185>

[dim laurel]

<@&286206848099549185>

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@dim laurel Has your question been resolved?

[dim laurel]

<@&286206848099549185>

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@dim laurel Has your question been resolved?

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[tranquil pine]

I have an equation at the top of the screen and I was wondering if I could simplify it into the equation at the bottom ?

[dusty canyon]

you cant divide throughout by it but you can factor it out

[solar glade]

To visualize it better, you can subsitute y-2=x

[tranquil pine]

Factorise it like this ?

[dusty canyon]

uh

last term is not factored right

[tranquil pine]

Ohhh yes

[solar glade]

Neither second

[tranquil pine]

Fixed it

Any ideas on how to further simplify or is that all we can do ?

[solar glade]

Based on the product u got, you can already find 2 solutions

[tranquil pine]

Y=2, y=2

[solar glade]

Now the other one you can rewrite (y-2)^4 like ((y-2)^2)^2

[dusty canyon]

just take (y-2)^2 = t as mentioned earlier

then solve the quadratic

[tranquil pine]

Like this ?

[tranquil pine]

Also why are we subtitling t ?

[solar glade]

No

I didnt say factor

I said rewrite

You have 3((y-2)^2)^2-17(y-2)^2-6

[solar glade]

To make things clearer

That way you would have 3t^2-17t-6

[tranquil pine]

OOOOOOH I SEE BYE UTILISING PSEUDO QUADRATICS

Very smart

I get it now

[solar glade]

Remember to sub back when u solve in t

To get all solutions

[tranquil pine]

Do you mind if I go and give this a go real quick ?

So where do I sub my values of y into ?

So the answer is in the part b section so what did I do wrong here ?

Never mind I realised where I went wrong

Thanks again everybody for your help!

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[feral siren]

Can some tell me how this is supposed to look like

[feral siren]

i know how the walls are

its mainly where the ball is going where i have a problem tho

[fringe nova]

is this a infinitely small sphere problem

[feral siren]

no

its the ones where u have sphere where it bounces off a wall twice

im not to sure where the second bounce goes tho

[fringe nova]

yeh the system would be symmetric

by symmetry the ball should only have 1 bounce

badly phrased problem or smth

[feral siren]

oh ok

[fringe nova]

or...

the two surfaces are like a ramp

but that would be weird

[feral siren]

i think its like that kinda

since the wall is bent i think

[fringe nova]

maybe "towards the intersection" does not mean towards the intersection line

but in the direction of the intersection

[fringe nova]

or maybe you consider the walls infinitly large

in which case it will have 4 bounces

no matter where it strikes

[fringe nova]

graphical solve, dont take my word for it

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[fathom saffron]

Given AB = 24 inches, and AC = 60 inches, I was able to figure out that BC is ~54.99. I can't seem to figure out how to find AD?

[abstract bramble]

arent they similar triangles

ADB and ABC

[sage pasture]

They are

A is a shared angle and both have a 90 degree angle

[abstract bramble]

well it was more of a question for them to ponder but yeah it is similar by AA

[sage pasture]

Fair enough

[fathom saffron]

Yes

[abstract bramble]

ok so what do you know about similar triangles

and the (ratio of their) side lengths

[fathom saffron]

Oh my God this makes sense

So AD would be 9.6
Because AD / AB = AB / AC
AD = *x*
= x / 24 = 24 / 60
= 60x = 576
= 60x / 60 = 576 / 60
x = 9.6

[fathom saffron]

And this would be the Geometric Mean Theorem, right?

[abstract bramble]

the what

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[fathom saffron]

The theorem that states that a right triangle with an altitude from the right angle creates two smaller triangles which are similar to each other and the primary triangle; isn't that the Geometric Mean Theorem?

[abstract bramble]

idk looks like it from google

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[crimson lichen]

can someone show me how to solve this problem with a example or show me how to solve it

[tulip latch]

yea... pretty ez

[crimson lichen]

thx

[scarlet sequoia]

first, the integral from -8 to 1 is that of a linear function, so how would you proceed?

[crimson lichen]

i mean all i have to do is find the area under the graph in the range -8,5 but is bad at gemonety and i cant find the area

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@crimson lichen Has your question been resolved?

[distant ridge]

from -8 to -4 there is a signed negative triangle

from -4 to 1 there is a signed positive triangle

from 1 to 5 you have the area of a rectangle but you substract half the area of a circle

hope this helps

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[tawny venture]

for these statements are my answers correct or wrong?

[tawny venture]

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[tawny venture]

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[tawny venture]

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[tawny venture]

is this right????

16. The cost function for producing x units of a product is 𝐶(𝑥) = −0.002𝑥^2 + 5.4𝑥 + 2800. The revenue function is 𝑅(𝑥) = 150𝑥 − 0.16𝑥^2 Determine the marginal profit for the sale of 150 units of the product.

Is the answer suppose to be 150?

not 150

97.2?

[tawny venture]

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[tawny venture]

how would I do this

I really need hlep I haven't gottenhelp in a while please

[gritty drift]

What’s the relationship between position and velocity?

[tawny venture]

position is the place moved and velocity is the speed at which it moves at

btw this is a position time graph

[tawny venture]

would it move away from the origin between a to C and E to F???

can you please at least verify my answer if possible

Does the object change direction at A, C, and E

I'm very confused

on that

<@&286206848099549185>

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[tranquil pine]

two things, for A I wanted to check if my method of doing it was right, I chose to just do div(Curl G) found it didnt = 0 and said that because of that it cant be said P,Q,R of G have continuous second order partial derivatives and there are no vector fields possible to make some vector field G = curl G

[tranquil pine]

the other thing is uh not really sure what to do with B

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[timid tulip]

wouldnt the $t = \frac{D}{v}$ part require integration

[soft zealot]

Shah2044UwU

[timid tulip]

if so

how do i do it

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[magic ether]

x=e^2y how can i find the revolution in the x axis

[final saddle]

Please don't occupy multiple help channels.

[magic ether]

do i need to put it so y=...

[magic ether]

ok yeah that was all i needed

thank you

[full ravine]

u think you can help me

[magic ether]

with?

[full ravine]

A modern-day zeppelin holds 9,510 m3 of helium. Compute its maximum payload at sea level. (Assume the helium and air to be at 0°C and 1 atm.)

[magic ether]

how do we find the payload?

at sea level

[full ravine]

i dont know

lol

[magic ether]

this is more of a physics question

[full ravine]

oh okie

[magic ether]

The zeppelin is subject to two forces: gravity and boyance.
The force of gravity is due to its mass m plus the mass of the helium:
g(m + Vh)

The force of boyance is due to the weight of air displaced by the zeppelin:
gVa

The maximum m is achieved when the two forces are in balance:
g(m + Vh) = gVa
m = V(a - h)

Substitute actual numbers

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[dense bane]

how does ln|1/9| rearrange into -2ln|3|

[dense bane]

nvm i get it, 3^-2 is 1/9 and exponents within a natural log can be brought to the front as a product

i c

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[slow maple]

Can someone tell me why my answer is wrong

[grand hinge]

hm i don’t see an error

[slow maple]

it was wrong when i submitted it

[grand hinge]

can you show what you submitted? maybe you typed it incorrectly

[slow maple]

yea sure

this is a diff question

is this right?

i made a mistakes its no +C

but i fixed it and its still wrong

gotit

nvm

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[vestal oyster]

Hi How much would a hollow tube of polystyrene weigh if the tube is 7 metres long and the polystyrene is 0.823m thick and the diameter is 8.23 metres

[vestal oyster]

We can use a rough estimate for teh weight of polystrene per metre cubed from the internet or something

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[long wraith]

What have u tried?

[timid flare]

i dont know where to start

[long wraith]

U need a table

Every row is another step of h

[timid flare]

but how?

[long wraith]

For the first row

x_1 = x_0 + 0.5

The subs it in y'

Then

y_1 = y_0 + y'(x_1)

Repeat for every row

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[proven chasm]

I've pretty much solved it but

when I checked the solutions, there was an extra step that I don't understand

I assumed that all the points on the segment [AB]

would be of minimal cost

However

It seems it's only between 15 and 25 for c_a

Why is this so?

oh wait i have a wrong constraint

woops

💀

[long wraith]

Probably it acts just like ur constraint section

[proven chasm]

yeah it does

i just fucked up 1 constraint

😭

it's ok now

I'm stupid 💀

[long wraith]

Since mathematically speaking, if half of the 0.5*TV gives out smaller amount, then its still valid

[proven chasm]

yeah

[long wraith]

The constraint k stops it

[proven chasm]

that makes sense yeah

[long wraith]

Hell, even -2TV could be valid too

[proven chasm]

I realised i fucked up here:

[long wraith]

💀

[proven chasm]

i had a constraint of

0 < c_a < 30

instead of 25

but ye ig it makes sense

welp ty 😭

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[sonic crystal]

A soccer ball is kicked (from ground level on a flat field) and after being in flight for 2 seconds is found to have a velocity of 15m/s parallel with the ground.

[sonic crystal]

i dont get the parallel with the ground part

what does that exactly mean

[craggy plank]

Literally

[sonic crystal]

wat

[craggy plank]

It means it has reached the peak

[sonic crystal]

oh

so its not touching the ground?

[craggy plank]

No

As soon as it touches the ground, the velocity would be zero

[sonic crystal]

so if i were to find initial horizontal velocity how would i find that

cuz it doesnt say whether 15m/s is horizontal or vertical

[craggy plank]

You can start from dividing the force

"2 seconds to reach the peak"

it means with the divided force facing upwards

It will take 2 seconds to reach the peak

[sonic crystal]

yes

[craggy plank]

So, calculate the initial velocity

[sonic crystal]

the answer says 15m/s is its horizontal initial

but how

[craggy plank]

The horizontal velocity remains the same

[split acorn]

horizontal velocity doesn't change in a projectile motion

[sonic crystal]

yeah i get that but

[split acorn]

but?

[sonic crystal]

where in the question does it imply that the 15 m/s is horizontal

[split acorn]

parallel to the ground

[craggy plank]

..

[pliant shore]

So parallel with the ground (if the ground is flat) means that the 15 m/s is only in the x-direction

[split acorn]

parallel to the ground -> horizontal

[pliant shore]

It's 15 m/s horizontally ye

[sonic crystal]

o

ok that makes sense

[craggy plank]

Do you still get any confusion?

[sonic crystal]

nop

thanks guys

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[dry lance]

Hi i need help with part B

is there any given points in this that i can use to construct the rule

do we have the points (0,0) and (900,s)?

(0,0) since when the taxi hasnt been moving due to t=0 then it cant have travelled and moved any distance so s=0 aswell

nevermind

s is the distance remaining before it loses charge

well we still know that when t=900 it loses charge

so a point must still be (900,0)

wait maybe I was right, is the other point still (0,0) since maybe... you cant measure the meters it has left since it hasnt even started driving yet?

[tiny iris]

hii i dont get any motivation to study discrete mathematics

why discrete mathematics need ?

[dry lance]

im currently in this channel

you gotta go to one of the ones that isnt occupied

[tiny iris]

me?

[dry lance]

yes

[tiny iris]

okok

[dry lance]

can anybody help please?

<@&286206848099549185>

part b please

quite stuck

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[fluid nacelle]

found this image

[fluid nacelle]

i learnt the formula on the right today

one of left looks easier to write

however i have no clue what the matrix thingo means

[rich tide]

well matrix is a more efficient way of representing certain formulas

[fluid nacelle]

yeah i just dont know how to read it

is it easy to understand?

[rich tide]

the matrix would open as:\
$x_{1}\left(y_{2}\cdot1-y_{3}\cdot1\right)-y_{1}\left(x_{2}\cdot1-x_{3}\cdot1\right)+1\left(x_{2}y_{3}-x_{3}y_{2}\right)$

[fluid nacelle]

this is the formula we did in class

[soft zealot]

B-eard

[rich tide]

that is the same thing opened a different way

I'd recommend you to look on a video explaining on how to expand a matrix

[fluid nacelle]

👍

right-o cya later

thanks

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[rich tide]

welcome

[fluid nacelle]

ok i get it now

one you sent used top row

example used left row

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[cobalt spoke]

can anybody give me a hand whith this?

[tranquil pine]

take u = 2x^2 + 3x -4, what derivative do you get from that

[cobalt spoke]

I alreadu tried u-sub and trigonometry sub, but stuck in half calculations, answer is another...

[tranquil pine]

right, but can you answer my question

[cobalt spoke]

4x+3

[tranquil pine]

yeah, so you want your numerator to be equal to 4x+3, What operations, which retain the equality, can get you 4x+3 in the numerator?

[young bridge]

manipulate the numerator to be a multiple of (4x+3) + a constant

more rigorously we can write $(x+1)=\alpha (4x+3) +\beta$

[soft zealot]

Obotron

[young bridge]

comparing coefficients will allow you to find alpha and beta

[cobalt spoke]

multiple by 1/4??

and then separate for 2 part integrals??

[tranquil pine]

are u saying multiplying x + 1 by 1/4 gets u 4x + 3?

[cobalt spoke]

yeah, i actually ment this you right

[tranquil pine]

Huh? Sorry I don't quite get what you are saying. Can you be clear with what your process is?

[cobalt spoke]

here what i tried, but why my answer differs from that from the book?

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@cobalt spoke Has your question been resolved?

[cobalt spoke]

mb mb

50 50

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[misty sequoia]

What did I do wrong?

[final tangle]

you tried to do this with a single integral

and subtracting the two functions like that isn't the way to go about this

what you've calculated was essentially

instead you'd want to set up an integral to caculate the areas of region 1 and 2 separately

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[misty sequoia]

I didn't really get your point here

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[misty sequoia]

I didn't get what you meantioned here

[final tangle]

are you aware that definite integrals give signed area over a certain interval?

[misty sequoia]

Yes

[final tangle]

considering the lines you have intersect at $x=3/2$, the integral you have can be split into
$$=\blue{\int_0^{\frac32} 3 -x - x \dd{x}} + \red{\int_{\frac 32}^{3} 3 -x - x \dd{x}}$$

[soft zealot]

ℝαμΩℕωⅤ

[final tangle]

the blue integral gives the signed area of the region i've indicated with the blue + sign

note that for values of x between 3/2 and 3,
x >=3 - x
(observable from the graph by which line is higher)

$$\red{\int_{\frac 32}^{3} 3 -x - x \dd{x}} = -\int_{\frac 32}^{3} x- (3-x) \dd{x}$$
which will contribute negatively to the signed area signed

[soft zealot]

ℝαμΩℕωⅤ

[final tangle]

as i've indicated with the red -

you graph isn't to scale, but those will cancel out producing the result of 0

which you've calculated

[misty sequoia]

Could you clarify those parts:

[misty sequoia]

.

[misty sequoia]

.

[final tangle]

$$\blue{\int_0^{\frac32} (3 -x) - x \dd{x}}$$
gives the signed area of the region bounded by the lines $y = 3-x, y=x$ from $x=0$ to $\frac 32$

[soft zealot]

ℝαμΩℕωⅤ

[misty sequoia]

What do you mean by signed?

[final tangle]

you said yes when i asked you about signed area and definite integrals

[misty sequoia]

I don't get the word "signed " in the context

[final tangle]

positive/negative area

depending on whether one curve is above/below another

[misty sequoia]

You see, in computer engineering we have signed and unsigned values

signed is standing for values <= 0 or > 0

[final tangle]

same here

[misty sequoia]

Okay, so the meaning is the same

[misty sequoia]

And what about that?

[final tangle]

which part of that?

[misty sequoia]

The part in the parentheses

[final tangle]

look on your graph

specifically the part past x=1.5

which line is above the other one

[misty sequoia]

y = x

[final tangle]

yes

thus in that interval i specificed
x >= 3- x

[misty sequoia]

And that's integral from (x - 3 - x)

So that's integral from -3

[final tangle]

wdym by from -3

no

you're missing ()

[misty sequoia]

Oh

exactly

[final tangle]

x minus the whole 3-x
i.e. x - (3-x)

[misty sequoia]

2x - 3

x^2 - 3x

[final tangle]

but again, i'm describing what you actually calculated
which is irrelevant to what the question is asking for

[misty sequoia]

That's right

Okay, but there's a thing

[misty sequoia]

The value of the first integral is 9/4

[final tangle]

yes

[misty sequoia]

The value of this one is 1/2

9/4 + 1/2 = 11/4

[final tangle]

no

show how you're getting 1/2

[misty sequoia]

-((3^2 - 3 * 3) - (9/4 - 9/2))

[misty sequoia]

5/2

[final tangle]

you messed up your simplification

keep showing your work after this

[final tangle]

yes

which isn't the value you obtained

[misty sequoia]

My bad, 9/4

[final tangle]

anyway in the say you've set it up, your subtraction was the other way around

[misty sequoia]

9/4 - 9/4 = 0

[final tangle]

yes

[misty sequoia]

This answer is wrong

[final tangle]

yes

[misty sequoia]

Then how to solve that?

[final tangle]

#help-36 message

[misty sequoia]

I literally did it 2 mins ago

[final tangle]

did what 2 minutes ago

[misty sequoia]

.

[final tangle]

you have not

[misty sequoia]

.

[misty sequoia]

The area of that part is 9/4

[final tangle]

that was an explanation of what you were actually calcuating

[misty sequoia]

The value of that is -(9/4)

[final tangle]

the areas of those + and - signs

which i've reminded you on multiple occasions throughout this convo that these were irrelevant to what the question was asking for

and an exaplanation of why you got 0 when you used a signle integral ealier

that your integral doesn't have anything to do with the area they want from you

[misty sequoia]

Anyway, I didn't get what I have to do

[misty sequoia]

I see that

[final tangle]

which is basically area under a line/curve twice (once for each region)

[misty sequoia]

It seems like you're trying to explain why I'm getting 0 as the value

And I'm asking for how to solve

[final tangle]

yes

i provided you with both

[misty sequoia]

Even if I'll be looking at the picture for 3 hours I won't get how to solve it

[final tangle]

you asked what you did wrong

i explained that

and i've also told you how to approach it properly

do you know how to calculate area under a curve (area between a curve and the x-axis)?

[misty sequoia]

That's the explanation of what I did wrong

[final tangle]

no

that's what i said what you should be doing

as described by that text accompanying it

the image with the + and - is what you effectively calculated / did wrong

[misty sequoia]

I have to add a minus sign

[final tangle]

NO

you're NOT reading what i'm saying

you should NOT be doing any subtraction between the curves
y=x and y=3-x

you're not being asked for the areas between them

[misty sequoia]

I suppose no then

[final tangle]

well here its just the area between two curves where one curve is above the other
and the x-axis just has the equation y=0

the lines intersect at x=3/2

area of region 1 is the area between
y=x and y=0 from x=0 and 3/2

same idea for region 2

[misty sequoia]

The two curves are represented by y = 3 - x and y = x?

[final tangle]

no

read the message directly above

[misty sequoia]

.

[final tangle]

after that

[misty sequoia]

.

[final tangle]

yes

[misty sequoia]

What are the two curves?

[final tangle]

read the messages after that

like the whole text block

all the info is there

[misty sequoia]

The x-axis just has the equation y = 0

[final tangle]

the WHOLE thing

[misty sequoia]

The thing is

I understand all the information you said here, and what I don't understand is what you're trying to lead me to by telling me that information

[final tangle]

do you know how to set up an integral to calculate the area between two curves/lines?

technically that's not even needed as you can do this geometrically with area of a triangle
but its a good excercise to properly set up your integrals

[misty sequoia]

Well, I don't

[final tangle]

its what you attempted to do in your intial work

look up arae between two curves and area under a curve

[misty sequoia]

I propose to begin from the very start

[final tangle]

start with

look up area between two curves and area under a curve

[misty sequoia]

I need the explanation of what are two curves again

[final tangle]

a curve is the graphical representation of an equation on the coordinate plane

[misty sequoia]

Okay, so y = 0, y = x, y = 3 - x are curves

[final tangle]

yes

[misty sequoia]

Actually, there are 2 limited areas

This triangle

And that triangle

[final tangle]

you don't care about the first triangle you posted

only the second

[misty sequoia]

So those are called curves

[final tangle]

they're equations

their graphical representation are curves

[misty sequoia]

Exactly

And what's next?

[final tangle]

i've said this multiple times already

[misty sequoia]

I'm trying to get a fresh look

[final tangle]

i'm only going to say things once now and not going to repeat myself again

[misty sequoia]

Like I won't understand if I read it multiple times

[final tangle]

i've labelled the lines,, x-axis has the equation y=0, forgot to label that but that's trivial

the region bounded by all 3 is shaded in purple as i've indicated

do you have any issue with that?

[misty sequoia]

With that - completely no

[final tangle]

here's a good time to repeat what i mentioned earlier

technically that's not even needed as you can do this geometrically with area of a triangle
but its a good exercise to properly set up your integrals

do you still with to try do this using integrals

[misty sequoia]

Do I will?

[final tangle]

to calculate the total area, you can calculate the area of the blue and green region separately then add them up

do you have an issue with that?

[misty sequoia]

In geometry - absolutely no

But with integrals it seems a little tricky

[final tangle]

now lets focus on the first region

moved the label

area of region 1 is the area between
y=x and y=0 from x=0 and 3/2

[misty sequoia]

The area is 9/4

No, hold on

Yes, 9/4

[final tangle]

how are you getting that

[misty sequoia]

We got the first region

The second left

[final tangle]

how did you set up your integral
what specific calculations did you perform leading to the value of 9/4 for the blue region

[misty sequoia]

The second integral is from 3/2 to 3 from (3 - x)

[final tangle]

was you value of 9/4 supposed to be for the blue or the whole region

[misty sequoia]

For the blue

[final tangle]

ok, show your work
pic would be preferable

how did you set up your integral for the blue

[misty sequoia]

The not defined integral of (x) = (x^2)

[final tangle]

what

again

for the last time that is irrelevant

hecne why you shouldn't be doing it at all

oh edited, still wrong

integrating x doesn't give x^2

[misty sequoia]

The integral from 0 to 1.5 from (x - 0)

[misty sequoia]

Damn, (x^2)/2

[final tangle]

yes

[misty sequoia]

I've got mistaken

(9 / 4) / 2 = 9/8

[final tangle]

yes