help-17

[lyric fossil]

you should give it a shot

do not integrate by parts

[sinful nacelle]

oo why? i just glimpsed over it

[fluid pumice]

based off the syllabus, ibp isn't a thing

[sinful nacelle]

ah

[lyric fossil]

because it is overkill and you can do this directly from part i

dan note that sin^2 = 1 - cos^2

[fluid pumice]

you have to* do this directly

[fluid pumice]

mhm

[lyric fossil]

and 1 - cos^2 = (1 + cos)(1 - cos)

sub in and simplify

[fluid pumice]

2(1 - cosx)/(1 + cosx)(1-cosx)?

(2 - 2cosx)/(sin^2x) comes afterward

[lyric fossil]

hold on

what happened

[fluid pumice]

but then

i get 1 - cosx as the numerator not 1 + cosx

this was the main issue

[lyric fossil]

[\int\frac{1 +\cos x}{\sin^2x},\dd x= \int \frac{1 + \cos x}{(1+\cos x)(1-\cos x)},\dd x]

[twin meteor]

maximo¹

[glossy maple]

I suggest just subbing in y = sin(x)/(1 + cos(x)) instead

u-substitution

[fluid pumice]

okay so basically from the syllabus's generic solutions for any hence find this integral question

you use the derivative you find from (i) and equate that to the original y + C

[glossy maple]

??

[lyric fossil]

wait dan do you know the integral of tan(x)sec(x)?

[glossy maple]

This simplifies nicely, and I assume it's the intended solution

You are using the first part to find dx

[fluid pumice]

according to the syllabus nope

but it's secx + C

[ancient horizon]

but if you divide

wait nvm

[lyric fossil]

ok never mind what i said, redstone is right

[fluid pumice]

does redstone's solution use the answer from (i)?

[lyric fossil]

yes

[fluid pumice]

okay so i remove the 2 from the integral

idk how to use texit so bear with me

integral of 1/(1 + cosx) = sinx/(1 + cosx) + C

uhh

[sinful nacelle]

ok i had a sit down and did the question, i think the easiest route by far is to try to use something very similar to the original function to differentiate into 2/1-cosx

if you want i can show you pieces of my solution?

[fluid pumice]

sure

[sinful nacelle]

do you want the whole solution or just a hint?

[fluid pumice]

just a start

[sinful nacelle]

this looks very similar to what we got in part i, can we try and construct a function that would work for this based on what we know?

[fluid pumice]

very similar

but i don't see how you continue from here given the denominator is different

[sinful nacelle]

well work backwards, mess around with the minus signs a bit

you know that the denominator was v^2 via the product rule, and that v ended up cancelling out right?

so what could the new demoninator be based on that

[fluid pumice]

1 - 2cosx + cos^2x

[sinful nacelle]

however, due to the negative that cos gives when differentiation, you may want to make the numerator negative to account for that

[sinful nacelle]

dont expand it !

[fluid pumice]

(1 - cosx)^2..?

[sinful nacelle]

it's simpler when a fraction in this case

mhm

[fluid pumice]

looks like this so far

without the negative sign oops

[sinful nacelle]

no, dont change the intergal any further

use part i to help, you know what the integral of 2/1+cosx is, so you can really just mess with the minus signs

[sinful nacelle]

instead of integrating, try differentiating something similar to the original function, and see if you can get what you got here

[fluid pumice]

2/(1 - cosx)^2?

[sinful nacelle]

where did the sinx go?

[fluid pumice]

2sinx/(1 + cosx) this was the original f(x)

[sinful nacelle]

mhm

but how would getting rid of sinx help

if the only difference is the signs, why would u want to get rid of trig functions completely?

[fluid pumice]

to make life simpler i guess

okay so what's the follow-up after what we just did with the denominator

[sinful nacelle]

but then the denominator wouldnt cancel like how it did on the first time via sin^2x + cos^2x

do u want to see my full solution and then ask any questions if u have any?

[fluid pumice]

sure

[fluid pumice]

but how do we integrate that

[sinful nacelle]

well by fundamental theorem of calculus if by differentiating y you get dy/dx, then by integrating dy/dx you get y?

so you already have the answer

[fluid pumice]

so the answer is 1/(cosx - 1)?

[sinful nacelle]

no, the answer is -2sinx/1-cosx ....

we differentiated it to 2/1-cosx, yes

[fluid pumice]

we divided it by 2

[sinful nacelle]

and differentiating is the exact opposite of integrating

oh yes sorry

so -sinx/1-cosx

remove a linear factor

[fluid pumice]

and took out a negative sign?

[sinful nacelle]

no?

[fluid pumice]

we ended up with -(1 + cosx)/(sin^2x)

[sinful nacelle]

but you already have the integral of 2/1-cosx

[fluid pumice]

okay say that we have this expression that we constructed

[sinful nacelle]

look at this rearrangement here

[fluid pumice]

how does it use any answer from (i)?

[sinful nacelle]

becasue we used part i to be able to deduce how to construct this function

its just 2 sign changes from part i

[fluid pumice]

it's always the syllabus that restricts this from happening

[sinful nacelle]

we found the third line, thats the answer

[sinful nacelle]

what?

[fluid pumice]

but yes i would say it works

[sinful nacelle]

i didnt use any non elementary methods

i mean, idk ur syllabus but this is usually how hence questions are solved

[fluid pumice]

thanks though

.close

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[vast shale]

what

[vast shale]

how do you go about doing this

[lime gorge]

Do you know what critical points are

or using differentiation

@vast shale ?

[alpine night]

Differentiate using the chain rule, put it equal to 0

[viral copper]

you can really use chain rule here directly

[twin meteor]

²⁰Ne

[viral copper]

now you can

[vast shale]

that is what i thought

times 3 right?

so

the derivative of e^x is e^x

so

the derivative of that whole thing

e^xln(2x) * (d/dx(xln(2x))

[lime gorge]

yea

times 3

[vast shale]

so

3 (e^ln(2x)x * (x/2x + ln(2x)))

right?

[lime gorge]

ye

[vast shale]

simplify x/2x to 1/2

[lime gorge]

set this whole thing equal to 0 and solve

[vast shale]

that seems kinda stupid

divide both sides by 3 and the 3 goes away

then divide both sides by e^ln2x(x) and that goes away?

[lime gorge]

cant just do that

u have to set each factor equal to 0

and solve

[vast shale]

so the first factor is 3

so 3 = 0? wtf

[lime gorge]

yep, thats not true, so its not a solution

but we just divided that by both sides

i was more referring to non-constant factors

[vast shale]

is 3 a critical point?

[lime gorge]

so e^(x(ln(2x))) = 0 and (0.5 + ln(2x)) = 0

[lime gorge]

no, why would it be

we wanna find where x is 0

if i multiplied both sides by 6, would 6 be a critical point? No

[vast shale]

how the heck do you even solve for 0 with something as complicated as that

log both sides?

extract power?

[lime gorge]

well think about it logically

is there any number out there such that e^number = 0?

[vast shale]

uh

1?

[lime gorge]

Whats e^1 equal to

[vast shale]

e

[lime gorge]

right, so that cant be it

[vast shale]

e^0 == 1

[lime gorge]

thats correct

[vast shale]

so what the heck there is nothin

[lime gorge]

Correct

[vast shale]

ahh

[lime gorge]

So we just eliminate that completely and move on

now 0.5 + ln(2x) = 0

[vast shale]

other one subtract 1/2

and then..

move 2 outside?

ln(2) + ln(x)

[lime gorge]

keep going

[vast shale]

uh

i have to have exact answers

soo

[lime gorge]

ok well u havent solved for x yet

u have ln2 + lnx = -1/2

keep going

[vast shale]

subtract ln2 from both sides?

[lime gorge]

keep going

[vast shale]

then go

e = x^(-1/2 - ln2)

right?

[lime gorge]

e = x^(something)?

we are solve for x

not e

[vast shale]

you do that then you can go

(-1/2-ln2) root of e

[lime gorge]

no, this is an incorrect step

u need to backtrack

[vast shale]

im lost

hints?

[lime gorge]

So u got lnx = -0.5 - ln2

[vast shale]

right

[lime gorge]

now lets do this

lets make a new variable b

b = -0.5 - ln2

so now we get ln(x) = b

how do we solve for x

what must we do to both sides

[vast shale]

inverse ln?

[lime gorge]

Which is?

[vast shale]

e^x?

no

[lime gorge]

yes, we do e^both sides

[vast shale]

e^b = x

[lime gorge]

yes thats what it comes to

e^(ln x) = x

and then e^b is just e^b

so x = e^b

and whats our value of b?

[vast shale]

lnx?

so e^lnx?

[lime gorge]

what

ok i think i may have lost u

[lime gorge]

Back to this

look at th expoentital and logaritjmic property

lnx = -0.5 - ln2

what do we do to both sides

to get ‘x’ by itself

we just did it

just asking u to repeat

[vast shale]

so

e^ln(x) = -1/2 - ln(2)

[lime gorge]

no

remember

[vast shale]

(╯°□°)╯︵ ┻━┻

[lime gorge]

whatever we do to one side, we must do to the other side too

cant just do something on one side only

[vast shale]

but its equivalent

[lime gorge]

nope

[vast shale]

right?

it says right there

x = e^ln(x)

[lime gorge]

Thats showing that e^ is the functional inverse of lnx

[lime gorge]

This is a rule u cant break in math

ur on the right track, just try again

lnx = -0.5 - ln2

whatever u did to the left side, do it to the right side too

[vast shale]

e^ln(x) = e^(-1/2 - ln(2))

what?

[lime gorge]

correct

there u go

[vast shale]

how does that help though

[lime gorge]

now what is e^(ln(x)) equal to

simplify that

using the exponential logarithmoc property

[vast shale]

but wont i have to do it to both sides again?

I'm confused

[lime gorge]

Ur just simplifying, ur not doing e^ both sides again

[vast shale]

how can you simplify this?

e^ln(x) = -1/2 -ln(2)

ln(x) = -1/2-ln(2)

like that?

[lime gorge]

Nope thats wrong

[lime gorge]

stop going to this

[lime gorge]

this is correct

$a^{\log_a(b)} = b$

[vast shale]

why don't you need ln on the right side

[twin meteor]

Stephen

[lime gorge]

i think ur issue is understanding how the functional inverse works

e^ln(x) = e^(-1/2 - ln(2))

when u get here, youve already done e^ both sides

correct?

[vast shale]

ah i see

yeah

[lime gorge]

after that e^ln just cancels out

and it just becomes x

[vast shale]

what about the -1/2

[lime gorge]

what about it

[vast shale]

so

x=2?

[lime gorge]

wha

no dont change the right side up yet

[vast shale]

so

x=e^(-1/2-ln(2))

[lime gorge]

yes

[vast shale]

AHHH

[lime gorge]

This is a critical point

[vast shale]

wtf

weird

[lime gorge]

we can simplify it down a bit

using exponent properties

see anything u can use?

[vast shale]

negative exponent and fractional exponent

[lime gorge]

ok yes we can use those, but first we must use quotient rule

look at the a^{x-y}

see how its just like our e^(-0.5-ln2)?

[vast shale]

1/(\sqrt(e)e^ln(2))

right?

[lime gorge]

yes

now simplify that

e^ln2

[vast shale]

these little images with math stuff super helpful

[lime gorge]

fr

[vast shale]

so its just 2

[lime gorge]

Yep

so we end up with?

[vast shale]

1/(2\sqrt(e))

[lime gorge]

Yes

1/2sqrte

so now thats our critical point

but we have to do a few things now

[vast shale]

solve for maxima

[lime gorge]

we need all extrema, not just maxima

But first, check if 1/2sqrte is on our interval

otherwise, its invalid

and we cant use it for extremaa

[vast shale]

it is

.3032653299

so plug it in?

[lime gorge]

ok so this is in our interval, so we can potentially get an extreme value out of this

But hold on, we arent done yet

evaluate that into the calculator

and see what u get

[vast shale]

4.178361968

[lime gorge]

ok, now

look at our interval

is it including endpoints?

[vast shale]

ye

so plug those in

[lime gorge]

thus

Correct

what do u get from those

[vast shale]

1 gives 6

.1 gives 4.178361968

i mean

4.7649

so .1 is maximum?

1 is minimum?

[lime gorge]

u sure .1 gives that?

[vast shale]

nah 2.55401

so its my weird value

that is max

[lime gorge]

and what about this

is this correct?

check that again

[vast shale]

2.577911626

[lime gorge]

lol

what happened

[vast shale]

idk

[lime gorge]

xd

anyways

what are the extreme max and mins

[vast shale]

hard to type this in calculator

[lime gorge]

ye i gotchu

[vast shale]

max is 6 at 1

[lime gorge]

and min?

[vast shale]

uh

2.554019768 at .1

[lime gorge]

yep

so in the end

[vast shale]

but needs to be exact

[lime gorge]

We didnt end up using the critical point

[lime gorge]

Dont do the decimals

just do the plugged in version like u showed before

[vast shale]

this is wrong

oh crap .1

[lime gorge]

Lol

[vast shale]

so if i replace x = with .1 its right?

[lime gorge]

well ehats the question asking for

Max or min

[vast shale]

nope it says this

wtf is that

[lime gorge]

what is that for

max or min

[vast shale]

that isn't even the critical point we found

[lime gorge]

lets see

[vast shale]

i feel better now that someone way smarter than me also is confused

[lime gorge]

well lets see

im p sure thats the same answer lol

hold on

$3(2(\frac 1{2\sqrt e}))^{\frac 1{2\sqrt e}} = 3(\frac 1{\sqrt e})^{\frac 1{2\sqrt e}} = 3(\sqrt {\frac 1e})^{\frac 1{2\sqrt e}}$

[twin meteor]

Stephen

[vast shale]

what the heck

gotta do it all over again

[lime gorge]

yea hold on im still trying to understand

bear with me ive only a phone keyboard lol

im on discord web version on phone

i see exactly where we went wrong

[lime gorge]

Right here

shoulda caught this earlier on

try that part again

thats crazy bro

[vast shale]

the question is different now

anways lets solve the original one

[lime gorge]

Well at least ya learned something

good job for persisting

[vast shale]

i can solve it now on my own i think

but what is wrong with how I did it?

[lime gorge]

many people just quit

[vast shale]

i dont see

is the x/2x wrong?

[lime gorge]

yes

we need chain rule on the 2x

so it becomes x/2x * 2

which is just 1

[vast shale]

ahh

[lime gorge]

man i wish i caught that

[vast shale]

yeah that makes it a lot easier

i think i can solve the new one on my own

let me try

[lime gorge]

sure, lmk if u need help

[vast shale]

so its

1/(2\sqrt(e))

thats the other critical point

[lime gorge]

Is that ur new problem

ok lets see

[vast shale]

its wrong

[lime gorge]

do u have a new problem now

[vast shale]

factors i got are

2

e^xln(2x)

(1+ 2ln(2x))

its almost exactly the same as before

solving for 1+2ln(2x) = 0 gives me

x = 1/2\sqrt(e)

sorry if thats hard to read

[lime gorge]

I didnt get that

[vast shale]

where did i mess up

[vast shale]

why no 2 in front of the ln?

[lime gorge]

on which line

[vast shale]

3rd

isn't that chain rule so it goes outside

oh

that's what i did

so ts just 1/2e

same mistake i made before

😄

thanks so much

.solve

.solved

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[lime gorge]

Have a good one bro

keep working at it

u have good perseverance

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[noble umbra]

Sorry my last help channel dissappeared can someone help with my data man assignment, thanks!

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Please don't occupy multiple help channels.

[noble umbra]

a quick check over to see if everything looks good would be appreciated

<@&286206848099549185>

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@noble umbra Has your question been resolved?

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@noble umbra Has your question been resolved?

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@noble umbra Has your question been resolved?

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[carmine mesa]

Can someone explain to me how the answer is B

[heavy yoke]

which one do you think it should be?

[carmine mesa]

I know the answer is B. I just hope someone can explain this to me because this is the only thing on the study guide I don't know how to solve and I don't really know how to do these "Areas in Polar Coordinates" things from out of everything in the unit

wait okay now I see why

but can you explain why it's B and not D? @heavy yoke

[heavy yoke]

you want to pick the right bounds so that the curve goes all the way back to where it started

[carmine mesa]

AHHHHHHHHH I see

and if we picked D, it would've been half of that

right?

[heavy yoke]

in this case it would integrate only half of the curve, yes

[carmine mesa]

gotcha

thank you so much

.done

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[prime cradle]

help meee, somebody helpppp

[prime cradle]

if i was required to learn this topic in less than an hour

[tidal dock]

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[prime cradle]

1

I have a general idea

I thought I had the topic but when I checked my work I found out I was wrong

[tidal dock]

show your work

[prime cradle]

I just assumed from the way it looked

[tidal dock]

can you identify the equation of this line? without the shaded area, just the line at the boundary?

[prime cradle]

I don't think I can

[tidal dock]

do you know what is the equation of the blue line, roughly?

replace all $\geq$ and $\leq$ in your options with $=$ and you get two equations of lines

[twin meteor]

artemetra

[tidal dock]

one of them will be the blue line

[prime cradle]

oh

I'll try that

thank you

y=4x+3

[tidal dock]

yep

so first two options are out

[prime cradle]

I'd say y is less than or equal to

wait no greater than or equal to

[tidal dock]

correct

[prime cradle]

o

[tidal dock]

no

less then

[prime cradle]

can you explain why

I thought because it went to the right

[tidal dock]

why? think about it like this: the shaded area includes all values of y for which it is smaller or equal to 4x+3

[prime cradle]

OH

that makes so much sense

thank you

[tidal dock]

no problem

any other question?

[prime cradle]

I don't think so

thanks for all your help

[tidal dock]

no problem, if you are done, type ".close"

[prime cradle]

.close

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[brittle shell]

Been stuck on this question

[subtle blaze]

What have you tried so far?

[brittle shell]

Um using 6 as the hypotenuse and then doing a2+b2=c2

I dont know where im supposed to start because Im only given height and a base and a hypotenuse

The formula is SA=Ph+2B

I got 4.8 but now I have no idea what to do with it

[mild flower]

surface area is just a matter of finding all of the areas of the shapes and adding them up

in this case we have two triangles (front, back) and three rectangles (bottom, left, top)

what did you get 4.8 for?

[brittle shell]

A2+b2=c2

C=36
B=12.96

Then I got 23.04 from subtracting them

[mild flower]

okay, so a = 4.8; what does that mean on your image? can you label the side that is 4.8 (unit??) long?

[brittle shell]

Then I square rooted it to get 4.8

[mental falcon]

yep, now find the area of each surface like hailey said

[mild flower]

(that's not the right edge...)

[brittle shell]

Ohhhhh…

[mild flower]

that one you already know how long it is

the pythagorean theorem applies to right triangles. can you label a right triangle in that image? color it in green for me.

[mild flower]

those are right angles

which is a good start

but we want a right triangle

which is a 2d shape

[mild flower]

yes good

now, do you know the lengths of any of the sides of that triangle?

[brittle shell]

Um 4.8 sorry im bad at labeling

[mild flower]

which side is 4.8?

feel free to use more colors to make yourself clear

[mild flower]

that's not even part of the triangle!

didn't you use the pythagorean theorem to calculate 4.8 inches?

[brittle shell]

Yes

[mild flower]

the pythagorean theorem applies to right triangles

considering your right triangle that you've labeled -- the green one -- which side lengths do you know?

one of them is labeled already

[brittle shell]

3.6

[mild flower]

which side is 3.6?

[mild flower]

okay, you could have said "the blue one" but that works

are there any other side lengths in that triangle that you know?

[brittle shell]

I dont think so maybe 6 but what would be for the other triangle

[mild flower]

the two triangles are going to be the same

since this is a prism

[brittle shell]

Ohhh

[mild flower]

so knowing this, are there any other side lengths in that triangle that you know?

okay, using your words, which side lengths do you know in that triangle?

[brittle shell]

3.6

[mild flower]

using your words

"the purple side is 7.8 inches long"

is what your answer should look like

but with different colors and numbers

[brittle shell]

How did you get 7.8?

[mild flower]

i didn't

there's also no purple side

this is an example answer, to show you the format that i'd like you to answer in

[brittle shell]

The purple side is 3.6in

[mild flower]

(that's blue but ok sure)

any others?

[brittle shell]

And another 3.6 because they are the same?

[mild flower]

which other is 3.6?

[brittle shell]

We are given 8.5 but that would be the height?

[mild flower]

okay, i think that black number you've written is referring to the vertical line in the back? that's great, but what we care about is the triangle in front

you very excitedly circled the "6 in" in the back -- why?

[brittle shell]

Because you said that it would be to same for the other triangle since

But I thought that 6 was the hypotenuse and not a side

[mild flower]

the hypotenuse is also a side

[brittle shell]

Ohhhh

[mild flower]

it's not a leg but it's a side

[brittle shell]

I got taught that in what ever direction the square is facing that would be the hypotenuse for the problem

[mild flower]

well, yes, the side opposite the right angle is the hypotenuse

[brittle shell]

So 6 would be another side then

[mild flower]

yes... one of the sides of the front triangle is 6 inches long

which side?

blue, orange, or green?

[brittle shell]

Orange

[mild flower]

okay. so which side lengths do you know, in that front triangle?

the blue-orange-green one

[brittle shell]

6 and 3.6

[mild flower]

the blue side is ___ long
the orange side is ___ long
the green side is ___ long

please fill this out

or put in ??? if you don't know

[brittle shell]

the blue side is 3.6 long
the orange side is 6 long
the green side is ? long

[mild flower]

okay cool

now can you use that information to find the length of the green side?

[brittle shell]

Would I use the Pythagorean theorem?

[mild flower]

perhaps

the pythagorean theorem is for right triangles

do you have a right triangle?

[brittle shell]

Yes

So I would do 6squared which is 36 then

3.6 squared to get 12.96

[mild flower]

okay, keep going...

[brittle shell]

Then add both to get 48.96

[mild flower]

i don't think you want to do that

[brittle shell]

Subtract

Subtract 36-12.96

[mild flower]

,calc 36 - 12.96

[twin meteor]

Result:

23.04

[brittle shell]

Which is 23.04

Then square root it to get 4.8

So circumstance = 4.8

[mild flower]

circumstance?

[brittle shell]

Base?

[mild flower]

shrug

which side is 4.8?

[brittle shell]

So I would add 4.8+6+3.6 to get 14.4

[mild flower]

uh

why

also

you didn't answer my question

which side is 4.8?

[brittle shell]

And then height =8.5

[mild flower]

meyvis

[brittle shell]

Yes

[mild flower]

i don't know which side that is

[mild flower]

okay yes good

here

what is the length of each color? you should label them on your piece of paper

[brittle shell]

Blue =3.6
Orange =6
Green=4.8

[mild flower]

okay, and yellow?

[brittle shell]

Yellow is 8.5

[mild flower]

now do you have enough information to compute the area of each side?

at this point i'd like you to start keeping track of your units

blue = 3.6 inches
orange = 6 inches
green = 4.8 inches
yellow = 8.5 inches

[brittle shell]

Ok gotcha

[mild flower]

keeping track of your units will make sure that you don't make an error

[brittle shell]

The equation for problem is SA=Ph+2B

[mild flower]

i don't know what any of those variables mean

[brittle shell]

Now im assuming that I would be adding 3.6+6+4.8 to get 14.4

P stands for perimeter and h stands for height and B is base

[mild flower]

okay... sure that seems fine to me

[brittle shell]

Thats the equation I received

[mild flower]

14.4 what? 14.4 school buses? 14.4 antelopes?

[brittle shell]

Inches

[mild flower]

okay

[brittle shell]

So I would do B=1/2(3.6)(6)

To get 10.8

[mild flower]

units

[brittle shell]

Inches

[mild flower]

redo it

with units

B = ?

[brittle shell]

B=10.8 units

[mild flower]

no that's not at all what i meant

B = what

[brittle shell]

Huh???

B = base

[mild flower]

you have something like B = 1/2 b h right?

[brittle shell]

Yes

[mild flower]

okay

what is b?

what is h?

[brittle shell]

B=3.6

H=6

[mild flower]

you've made a distinction between B and b

so please keep it that way

but also

UNITSS

b = 3.6 what

3.6 jellybeans??

[brittle shell]

3.6 inches

[mild flower]

we're trying to find the area of this triangle, right?

[brittle shell]

Yes

[mild flower]

what are b and h?

base and height?

[brittle shell]

base is the green part right?

[mild flower]

yes

[brittle shell]

4.8

Ohhhh I see now

[mild flower]

stop giving me numbers without corresponding units

they mean nothing

you can use "in" for inches

[brittle shell]

4.8in

[mild flower]

okay

so what is the area of this triangle?

[brittle shell]

Just making sure h=6in because we are solving for a right triangle

[mild flower]

which side is 6in?

[brittle shell]

Orange

[mild flower]

that is not the height of the triangle

the height of the triangle is perpendicular to the base

do you know why the area of a triangle is 1/2 b h?

[brittle shell]

Wait no sorry 8.5in

[mild flower]

it's because it's half of a rectangle

[mild flower]

isn't that the yellow side??

[brittle shell]

Wait no then blue would be the height

[mild flower]

yes

[brittle shell]

B=1/2(4.8in)(3.6in)

[mild flower]

cough

i see numbers

without units

great

now what is B?

[brittle shell]

B=8.64

[mild flower]

that's a number

without a unit

[brittle shell]

B=8.64in

[mild flower]

you can calculate the unit

notice

that you are multiplying in times in

which will not produce in

but rather in²

which is a fundamentally different unit than in, since in² measures area and in measures length

[brittle shell]

Wait so is my answer right for Base im confused?

[mild flower]

what's Base supposed to be?

[brittle shell]

8.64in

[mild flower]

that's a measurement of length

what's Base supposed to be? like, use your words

[brittle shell]

I got 8.64 by multiply 4.8 x 3.6 then halfing it

To get 8.64in for Base

[mild flower]

the next time you give me numbers without units i'm leaving for 10 minutes

you multiplied 4.8 in x 3.6 in

,calc 4.8 * 3.6

[twin meteor]

Result:

17.28

[mild flower]

which would give you 17.28 in²

[brittle shell]

Sorry I just keep forgetting to add them

[mild flower]

do you see the little 2 up there?

[brittle shell]

Yes

[mild flower]

in² is also called square inches

[brittle shell]

Ohhhh

[mild flower]

it's a unit of area

and yes we'll still need to multiply by 1/2

[brittle shell]

Ohhhh so thats what you we’re talking about

[mild flower]

you can also write sq in if writing ² is hard

[brittle shell]

B=8.65sq

[mild flower]

great ok

so

[brittle shell]

Alright good good

[mild flower]

where are we, let's go back to the formula you had for surface area

SA = Ph + 2B

[brittle shell]

Yes

Perimeter you would add all parts

[mild flower]

yes, you found P earlier

[brittle shell]

3.6in+6in+4.8in= 14.4in

[mild flower]

good

[brittle shell]

Ok ok great

[mild flower]

and what's h? i think you got confused earlier because this should be the height (really the length in this case) of the prism, not the height of the triangle

[brittle shell]

Wait would it be 8.5in?

[mild flower]

yep

[brittle shell]

Ok great

[mild flower]

and you know B = 8.65 in²

[brittle shell]

Yes

[mild flower]

so can you use all of that to find SA?

[brittle shell]

SA=14.4in(8.5sq)+2(14.4in)

[mild flower]

i think you substituted in the wrong spots

are you using pencil and paper at all to keep track of what's going on here?

i hope you're not constantly scrolling through the chat

[brittle shell]

Im just keeping it in my head 🫠

[mild flower]

don't do that

[brittle shell]

But I understand where I got

[mild flower]

short pencil > long memory

i usually just keep a stack of printer paper near me to doodle whatever

[brittle shell]

Ok ill do that

[mild flower]

anyway yes you seem to have substituted incorrectly, in particular i see two copies of "14.4 in"

which i don't think should be the case

[brittle shell]

Oh wait was that supposed to be 8.5sq

SA=14.4in(8.5sq)+2(8.64sq)

[mild flower]

you're having so much trouble keeping track of these things

[brittle shell]

Im going to just say them
P=14.4in
h=8.5in
B=8.64sq

[mild flower]

good

[brittle shell]

SA=Ph+2B

So SA=14.4in(8.5in)+2(8.64sq)

[mild flower]

where did 8.5 sq in come from?

[brittle shell]

Height

[mild flower]

ah you corrected it

ok

now you can continue to calculate

[brittle shell]

So then the final answer would be 139.68sq

SA=139.68sq

[mild flower]

note that you can add inches and inches, and you can add in² and in² together, but if you ended up in a situation where you had like 4in + 7.3sq then you'd know something went wrong

[brittle shell]

Alright well I submitted the answer and it was correct 👍

[mild flower]

right

in the future, use paper! or a whiteboard or something

[brittle shell]

Thank you for your time I really appreciate it!

[mild flower]

just write stuff down

[brittle shell]

Will do

[mild flower]

no one's going to look at it but it can help you keep track and figure out what went wrong

[brittle shell]

Ok thankyou

How do I close this?

[mild flower]

type .close

[brittle shell]

Ok alright thank you again

.close

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[weary gale]

Kind of confused on this

i think she messed up on step 2

since if it was perpendicular wouldnt it be opposite reciprocals ?

[lone linden]

yeah

[weary gale]

so i would just switch them out

like

the wrong slope with the right slope

and redo the equation with the correct slope this time

[lone linden]

yeah

[weary gale]

Thats all i gotta do ?

[lone linden]

where's the x

[weary gale]

oops

so y = -2/4x - 4 ?

[lone linden]

you can't just copy the incorrect y intercept over

also -2/4 can be simplified to -1/2

[weary gale]

i forgot to simplify

mb

wouldnt we have to change the -4 to a + 4 ?

since its a negative now

[lone linden]

just use point slope form...

$y-(-2)=-\frac{1}{2}(x-4)$

[twin meteor]

Civil Service Pigeon

[lone linden]

I gotta dip but you should get ||y = -x/2||

[weary gale]

i struggle with that

alright

thanks for the help

.close

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[opaque trench]

how do I figure out how to make -2 rad be in the domain 0 and 2pi

[opaque trench]

,tex for z conjugate it would be $8\cos(2)-i8\sin(2)$

[twin meteor]

jacky

[opaque trench]

but then you can make the angle -2

,tex $8\cos(-2)+i8\sin(-2)$

[twin meteor]

jacky

[opaque trench]

I feel like I just can't visualize where -2 on the unit circle is and to try to find the angle on right domain

[crystal gazelle]

How many radians are there in a circle?

[opaque trench]

2pi?

[crystal gazelle]

Indeed

So

If you travel a full revolution of a circle to the point where you started

You've travelled 2pi rads

[opaque trench]

yes

[crystal gazelle]

So if you add any integer multiple of 2pi to the argument, it produces the same angle

[opaque trench]

ohhhh

so 2pi - 2

?

[crystal gazelle]

Exactly

[opaque trench]

thanks i don't think i could've thought that way

[crystal gazelle]

Np!

[opaque trench]

.close

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[twin meteor]

nox💫

[livid oxide]

what

they all have x-1

so that narrows it down to 1st and second one

correct?

so confused

so which one would it be

like how do i find out

mane

can you show@me how to do it

[vocal sleet]

@livid oxide Has your question been resolved?

[livid oxide]

how would i get the residual for 4 hours

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@livid oxide Has your question been resolved?

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[vast shale]

How should I approach this type of questions?

Is there something like f(0)×f(1)<0, saw it in answer can't really understand

Or maybe another way to approach this question

[bronze osprey]

It's a quadratic so there can be maximum two roots

[vast shale]

Yes

[bronze osprey]

So check the sign of that function when x = 0 and x = 1

If you get positive, positive or negative, negative

There must be a turning point in between, or the function was simply positive/negative all the time there

So if you get positive, negative or negative, positive

This implies there is exactly one root in [0, 1)

Oh you also have to check if x = 0 is a root cause of the closed bracket

[vast shale]

Ok

I will try

[bronze osprey]

No worries

[vast shale]

thanks @bronze osprey

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No worries!

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[coral glen]

I understand that we are using the pigeonhole principle, however is it expecting us to write the answer in some sort of equation..?

[edgy sapphire]

nah you'd just write out a set and apply the principle

[twin citrus]

how is that ?

[edgy sapphire]

like the blueprint for these proofs is: there's a set {a,b,c,d,...} of size n, and by pigeonhole if you pick n+1 from this set there's overlap

[twin citrus]

yes that's pigeon principle but how that would apply to this problem

[edgy sapphire]

the set is {1,2,3,...9} for people to know

[twin citrus]

why not {1,2,3...10}

a to know

yes of course

then?

[edgy sapphire]

9 is max because that's how many other people there are

then 10>9 so pigeonhole and you're done

[twin citrus]

sure

[twin citrus]

can you elaborate?

[edgy sapphire]

idk how to, pigeonhole is usually like a 1 step proof

each person has a number 1-9

[twin citrus]

like in terms of sets how would you describe this problem

[edgy sapphire]

I mean I said the set already

there's a set and it's smaller than the number of people

[twin citrus]

so how could you do that?

i am 5 years old

5,5 ok

so approximately 6

[edgy sapphire]

I subscribe to feynmann quotes instead sry

[twin citrus]

what are they?

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[solemn falcon]

Hoping for help on this question because in class we haven’t looked at graphs yet nor intervals on integrals

[flint idol]

how do we usually look for global and absolute maxima?

[solemn falcon]

take a derivative?

[flint idol]

yep

so we need to find $\frac{d}{dx} \int_{0}^{x} f(x) dx$

[twin meteor]

y0shi

[flint idol]

are you familiar with the fundamental theorem of calculus?

[solemn falcon]

not overly derivatives were my weakness and we just moved to integrals

[flint idol]

so basically the fundamental theorem states that:

$\frac{d}{dt}\int_{0}^{t} f(x) dx=f(t)$

!occupied

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[solemn falcon]

I have seen that formula before

[twin meteor]

y0shi

[flint idol]

yep

we need to use this to find the derivative

so all we really have to do is plug in that t for x

and in our case, we’re just going to replace the t in f(x) with x

and that’ll be our F’(x)

[solemn falcon]

an x at the top of the integral sign?

where the t is?

[flint idol]

so basically i’m trying to say $\frac{d}{dx} \int_{0}^{x} f(t)dt=f(x)$

[twin meteor]

y0shi

[flint idol]

by the fundamental theorem of calc

[solemn falcon]

okay okay

[flint idol]

so yeah basically replace the t with the x

and to find our critical points we usually set our derivative equal to?

[solemn falcon]

0

[flint idol]

yep

so what would be our critical points

[solemn falcon]

0

?

[flint idol]

yep that’s one

there’s a few more

[solemn falcon]

but theres no numbers just variables?

[flint idol]

if we set $F’(x)=f(x)=0$

[twin meteor]

y0shi

[flint idol]

what value of x gives us 0?

the graph f(x) is given to us

basically look for the x intercepts of the graph

[solemn falcon]

-6,4?

[flint idol]

-6, 4, 0

and one more

we need to include the endpoints

[solemn falcon]

6

[flint idol]

yep

and now we have to observe how the sign changes around that point

so let’s start at -6

how does the sign change near that point?

[solemn falcon]

-6 is up

[flint idol]

what do you mean by up?

increasing near -6?

[solemn falcon]

yes sorry

[flint idol]

yeah alright

so basically for endpoints

we just need to look at the right or the left of the endpoint

if F’(x) is positive to the right that means that at x=-6, there is a?

[solemn falcon]

positive?

inf?

sorry brains at all time low rn lol

[flint idol]

i meant to ask if it was a relative min or max

[flint idol]

all good

[solemn falcon]

min

[flint idol]

yep that’s correct