geometry-and-trigonometry

[upper karma]

and that equals c times c

right?

what i did on the second one was a-r plus b-r

to get c

c=b-r+a-r

=a+b-2r

no

[night karma]

Anyone willing to try this problem I made? If everything's correct, it shouldn't be too hard!

[upper karma]

@night karma can you spoil the answer

I mean not the method

[night karma]

Answer: ||72°||
Note: ||Knowing this answer might make the problem too obvious 😬 ||

[upper karma]

Pls milano use "||||" some ppl want to try still

:)

[night karma]

I'll do that!

[upper karma]

Ty!

I meant to explain xi but with ||||

[night karma]

OOooh 😂

[upper karma]

Thanks man

Wait wait wait

[night karma]

Xi, here's a small hint: ||Take a look at the equal sides||

[upper karma]

I mean for you to explain xi with |||| 😂

Ty

[night karma]

🤭

[upper karma]

xD

@upper karma

it's using cos because the graph starts from the bottom and rises

it's always a good idea to draw out the graph (like I did)

i said that it was -sine

it starts off at the vertical displacement

sin starts from 0, cos starts from 1

-sin means -0 🙂

but i started off somewhere different

thats why

-cos means -1

shit i meant

sine*

I started off at the vertical displacement

and then it goes up to 87

and down to 41

as for the period

I thought its pi/10

where did 10 come from?

sorry

the period itself is pi/10

the period is 12 months

i meant thats the value of B

I meant

the period is 20

there's no 10

well look

there's no 20 either 🙂

pause

let me explain

from month 2 -> month 12

thats 10 months

and only half a period has occured

it repeats every year... the period is 12 (12 months)

mmm

ur right

i was under the assumption that

its a sine function

but I did not know at what x value the vertical displacement would occur

so i assumed it was 45

sine would start from the midline

yea

the midline

but i didnt know where the midline would be

it's easy

add both values, and divide by two

yeah

but i didnt know what the min was

i assumed that it could go lower

the question never stated that

but cuz its 12 months

it should be obvious that its lowest is at 41

and highest is at 87

well, they tell you the temperatures for every month, and it doesn't go under 41, hence "41" is the min

ya

i just didnt fully interpret that problem

correctly

my bad

ty

sure thing

[upper karma]

@night karma

Why 1 tho? @upper karma

it would've worked with any number (used as a placeholder), but 1 was divided evenly by sin(90)

(which is also 1)

so sin(90)/1 = 1/1 = 1

it just simplifies things

Did you just made a sine from a 90°

Wow

Its smart and weird at the same time

and now that I think about it, it's more like 76, since alpha = sin^(-1)(0.7) is closer to 44 degrees

it's only for demo purposes 🙂

So you dont get the right answer, do you

sin^(-1)(0.707) is 45 deg

And the answer its supposed to be 72

Not 75

but without that extra 0.007, you get closer to 44

which results in 76

180-(44+60)

And the answer its supposed to be 72
.

I don't know what to tell you 🙂

I mean

numbers don't lie

[wanton lake]

anyone know the solution to this? i ended up with 35 but the correct answers i 45. can anyone help me out?

[upper karma]

and I'm 99% I did it right

So you assume your method is correct, but bc of so many approximations, you got 76 maybe?

I also got a similar approach to yours

🙂

Answer me lol

Maybe its bc of that

https://tenor.com/view/peter-griffin-family-guy-perhaps-maybe-gif-16249623

https://tenor.com/view/maybe-idontknow-gif-7667589

[night karma]

Dunno, while working out my problem, I found 72° 🤷‍♀️

[upper karma]

Lol

@wanton lake polygon angle-sum theorem

[wanton lake]

huh

can u help lead me through on dat

it is mainly the (3x+15) that stuffed me up

[upper karma]

Look it up

And you'll see

@upper karma @night karma

math checks out

@wanton lake then you come back

[wanton lake]

ahhh kk

[upper karma]

Got it?

[wanton lake]

im working on it

[upper karma]

@upper karma it might be you correct and him wrong lol imagine

But that'd be weird bc his way seems correct as well

also for 76 if you're declaring angle alpha closer to 44 deg:

so, yeah

@wanton lake lmk if you got anymore doubts

[wanton lake]

ok i will thx!

[upper karma]

Np!

[night karma]

Exactly, that's why it's interesting. You think the figure might be indefinite?

[upper karma]

no, it just depends on how much precision you want

Hmm

sin^-1(0.7)

[night karma]

It's quite close.. but still

[upper karma]

Yeah the notation is fucked up @upper karma

With the ^ and _

[somber coyote]

DSpider:
Compile Error! Click the reaction for details. (You may edit your message)

[upper karma]

$sin^(^-^1^)(0.7)$

[somber coyote]

Al3dium:
Compile Error! Click the reaction for details. (You may edit your message)

[upper karma]

Omg it worked

error

But it gets what you want

compile error

eh

[night karma]

asin(0.7) ~= 44.427...

[upper karma]

yeah, see? it's closer to 44

so you end up with 76

or you could keep the decimals and write the "true" angle

@upper karma just use the inverse. Which is csc

nah, it should be 44

this is a bad approximation

it's closer to 44, I should've jotted it down as 44

oh, well 😛

[night karma]

This is the figure, made in Geogebra

[upper karma]

Omg

[night karma]

🤷‍♀️

[upper karma]

Wow

I dont really know who is right/wrong

Should we just post it as a question. And see if anyone made a wrong assumption?

[night karma]

No clue. The figure doesn't lie, unless it's indefinite

Sounds good 😏

#help-1 ?

[upper karma]

Should we just post it as a question. And see if anyone made a wrong assumption?
@upper karma do you want to?

#help-1 ?
@night karma lets wait for him to accede

go for it

[wanton lake]

hmmm

[upper karma]

Les goo

@upper karma you could post yours in #help-1 and @night karma yours as well with all the context

(Sorry for the tag)

[night karma]

Still here, so no worries

[upper karma]

@wanton lake whats your problem?

[wanton lake]

i just cant figure it out 😭 sorry

[upper karma]

Np, but ill help you to do so

Dont put it in the other channel

[wanton lake]

ahh ok

[upper karma]

Erase it, bc im already helping you :)

Okay so

[wanton lake]

yes

[upper karma]

Did you understood the theorem i gave you?

Ya I did

Wut

[wanton lake]

the polygon angle sum theorem right?

[upper karma]

Yes

[wanton lake]

yes i understand that

[upper karma]

So

[wanton lake]

i did a few question like this before hand and got them correct as well

it mainly the (3x+15) that is stuffing me up a bit

[upper karma]

From the theorem you should have gotten that the total sum of all the angles on a hexagon is ....?

180(n-2) right

Where n is the amount of sides

But

[wanton lake]

what he said

[upper karma]

Yes but you arent him

Ok

I like geometry

So you both want to learn it

I already learned this

Lol

So dont answer

Ok sorry

I want him to do so

Ok

Yes

I’ll just watch silently

:)

[wanton lake]

ye i did 180(6-2) which = 720 i believe

[upper karma]

Yes

And that gives us the total sum of all the angles on a hexagon

So

How can we make an equation, which contains/mentions the sum of all the angles?

How the equation should be?

[wanton lake]

well, im not sure it is correct. But normally/before i added up all the angles which i could get (100+140+90+120+120 = 570) and subtract it from the total sum of all angles on the hexagon

[upper karma]

Can you repost the drawing

I dont remember the angles lol

[upper karma]

well, im not sure it is correct. But normally/before i added up all the angles which i could get (100+140+90+120+120 = 570) and subtract it from the total sum of all angles on the hexagon
@wanton lake and yes that its what you have to do, but with the hexagon

But this time

You know the total sum

And you are missing one part of one angle

[wanton lake]

yes

[upper karma]

So

The equation generalised is:
a+b+c+d+e+f= total sum of all the angles.
Where from a to f are the angles

So apply that equation, to our problem

Hey? @wanton lake

[wanton lake]

ok

[upper karma]

Ill wait

[wanton lake]

so x = 150

i beleive

[upper karma]

@wanton lake
(UPDATE: this is incorrect)

Wait

Thank you for stealing it

And ruining it

You just fucked up the learning process @upper karma

[wanton lake]

let my brain do it for a sec

[upper karma]

https://tenor.com/view/dr-house-oops-gif-6118887

[high zephyr]

did dspider posted the solutions?

[upper karma]

Yes

[wanton lake]

the answer in the math book i am meant to get is 45

[upper karma]

Straight away

While i was explaining

[high zephyr]

@upper karma we dont give away answers, we help them get to the answer themselves

[upper karma]

am I in trouble?

[high zephyr]

no

[upper karma]

But its really rude

[high zephyr]

^

[upper karma]

https://tenor.com/view/ok-sorry-homer-simpsons-gif-9894844

[high zephyr]

unless u put it in spoiler than it's fine ig

[wanton lake]

so it would be (3x +1 5) where x = 150?

[upper karma]

Kinda

Wait

I missed what you were saying

[wanton lake]

ok

so

100+140+90+120+120 = 570

720 - 570 =150

so it would be (3x + 15) where x = 150?

[upper karma]

But

Where did the 6th angle went to

[wanton lake]

wait

dats wht i need help for

[upper karma]

100+140+90+120+120
I count 5 angles

Not 6

[wanton lake]

yeah

i need help to calculate the 6th angle

[upper karma]

The sixth one is (3x+15)

Dont spider dont

[wanton lake]

yep

[upper karma]

So

[wanton lake]

and how i get x?

[upper karma]

Remember

Dont skip things

[high zephyr]

i believe this is the question https://cdn.discordapp.com/attachments/326138757474680852/704105253804703834/unknown.png

[upper karma]

The equation generalised is:
a+b+c+d+e+f= total sum of all the angles.
Where from a to f are the angles
@upper karma

@high zephyr yes

@wanton lake please use the equation up there

Apply it to the problem

[wanton lake]

ok

[upper karma]

Dont be scared of the x

[wanton lake]

100+140+90+120+120+(3x+15)?

[upper karma]

=720

Yes

So now we have an equation with one variable

[wanton lake]

yep

[upper karma]

Its pretty straight forward now

[wanton lake]

indeed, just gotta find x

[upper karma]

Lmk if you still need help

Yep

@upper karma also your method is completely wrong lol

[wanton lake]

how do i get x? 😂

[upper karma]

Lmao

Simplify it first

[high zephyr]

$3x+15=150$

[somber coyote]

nighty:

[wanton lake]

ahhhh

[upper karma]

Yes

[high zephyr]

u want to make x the subject here

[upper karma]

That is the simplified form

[wanton lake]

so x = 45?

[upper karma]

Yeah

If nighty didnt missed the summing process, yeah

xD

[high zephyr]

@upper karma u're supposed to subtract all other angles that is obvious, not angle (3x+15)

which leaves u with the expression for 6th angle

nah i dont think i missed any summing process

[wanton lake]

ahhh i get it

yesss

[high zephyr]

👍

[upper karma]

Np! :)

I didn't check 😦

am sawy 😦

[high zephyr]

no worries

now u know

[wanton lake]

dabs in needs a bit more help

[upper karma]

The same thing

But with an additional issue

[wanton lake]

i think i gpt it thx

[upper karma]

Oh lol

Np

[wanton lake]

hmm

but this time there is 2 pronumerals..

[upper karma]

Didnt got it

It doesnt matter

Look

Did you got the the (x+100) part

In which you need to apply a theorem

[wanton lake]

yep

[upper karma]

So the interior angle is...

Idk

Bruh

[wanton lake]

me?

[upper karma]

Yeah i was asking you

[wanton lake]

ok so what is it?

[upper karma]

I mean, did you managed to solve the problem?

i think i gpt it thx

[wanton lake]

no....

[upper karma]

Ok

So

We need to apply a theorem called the complementary angles

Look it up

[wanton lake]

yep, 2 angles that add up to 90 degrees?

[upper karma]

Add up to 180*

[wanton lake]

mb

[upper karma]

A straight line will have always 180

So yeah

supplmentary angles

True

supplementary angles add up to 180

They are called complementary as well i think

[wanton lake]

ok ok

[upper karma]

Anyways, if you have one angle as (x+100) and the other one you dont know it, but you know that they add up to 180. The equation is...

[wanton lake]

give me a sec

[upper karma]

Ok

[wanton lake]

ahhh im not to sure sorry

[upper karma]

Its okay

Ill give you the general equation

a+b=180

Try to apply it, if you still dont know, lmk

[wanton lake]

oh, mb dat equation

so then

lets say 'a' will be (x+100)

[upper karma]

oh, mb dat equation
yeah thats the supplementary theorem

lets say 'a' will be (x+100)
yes

And b the missing one

[wanton lake]

yep

[upper karma]

The equation is ...

[wanton lake]

(x+100) + b = 180

[upper karma]

Yes

So

We need b to apply the polygon sum-angles theorem

Like we did before

So b is

$b=180-(x+100)$

[somber coyote]

Al3dium:

[upper karma]

$b=-x+80$

[somber coyote]

Al3dium:

[upper karma]

Mb

Ty

Got until here? @wanton lake

I only cleared "b"

[wanton lake]

ok wait

yep

makes sense

[upper karma]

Ok

Last step

b is the interior part of the angle

This one

Wait

[wanton lake]

yes

it is and i will wait :)

[wanton lake]

yep

[upper karma]

So now that b=-x+80

[wanton lake]

indeed

[upper karma]

And applying the theorem of before

What equation do we get?

The one we did on the last problem

Called the polygon sum-angles theorem

What equation do we get?
.

[wanton lake]

give me a sec gotta grab sumthin soz

equation is
90+120+ (-x + 80) + (2x + 55) ?

[upper karma]

Thats an expression, not an equation

It lacks of something

[wanton lake]

ahh mb

[upper karma]

... and...?

[wanton lake]

is it

ye idk

[upper karma]

What does an equation need

[wanton lake]

(-x + 80) = b ?

[upper karma]

$a + x = z$ IS an equation.

[somber coyote]

Al3dium:

[upper karma]

$a + x$ IS an expression

[somber coyote]

Al3dium:

[wanton lake]

kk

[upper karma]

Do you see the difference now?

[wanton lake]

i do

[upper karma]

90+120+ (-x + 80) + (2x + 55)
Whats missing here then?

[wanton lake]

the equals sign

[upper karma]

Yes

And that equation, equals to ...

[wanton lake]

90+120+ (-x + 80) + (2x + 55) = 360?

[upper karma]

YES

[wanton lake]

yay

[upper karma]

Now try simplifying

Lmk if you need help still

[wanton lake]

ok

so (-x + 80) + (2x + 55) = 150?

[upper karma]

Yes

Now do away the parenthesis

[wanton lake]

hmm, could you help me a bit with that?

[upper karma]

Yep

So we see that the parethesis are summing

And they are not being multiplied or anything else

[wanton lake]

yep

[upper karma]

So we can just "remove" them

$-x+80+2x+55=150$

[somber coyote]

Al3dium:

[upper karma]

Go on yourself now

[little osprey]

occupied?

[upper karma]

Yeah

[little osprey]

ok

[wanton lake]

sorry Denton

[upper karma]

But in a few mins i dont it will

As we are almost at end

[wanton lake]

so i will subtract the 80 and 55 from 150...

i think

[upper karma]

Yes

Although thats a weird wording

But yeah

[wanton lake]

ye soz

then

that equals 15

[little osprey]

Dont be sorry

[wanton lake]

so does x = 15?

[upper karma]

that equals 15
@wanton lake yes

so does x = 15?
@wanton lake but no

[wanton lake]

ok ok

[upper karma]

Oh wait

Yeah it does my bad

[wanton lake]

it does? nice

[upper karma]

I didnt saw that the -x+2x = x

So yeah

You are done

@little osprey now it is free

:)

[little osprey]

Ah thanks

y=3.497sin(0.5250(x+1.505))+13.95, C=-(1.505)-1.505
Would
C=1.505 or -1.505
because in the orignal equation it is represented as A*sin(B(x-c))+D

I got very confused tbh

[high zephyr]

c=-1.505 in this case

if u're comparing your eqn with the original one

[little osprey]

AHHHH

[high zephyr]

x+1.505=x--1.505

[little osprey]

Wait why am I so blind

hmm so -

but you said if comparing equation with original one

So then

why is

y=A*sin(Bx+c)+D a thiing

[high zephyr]

what i did just then is i compared the two eqns

i didnt bother looking at the other values

oh

im not too sure but the eqn is for determining transformations for a sine wave

for me im familiar with y=asin(nx-h)+k instead

gtg now

[little osprey]

ah no worries

thanks for the help

I got confused at something that was completely trivial

[high zephyr]

ah alrighty

[royal cipher]

y all quick question ples

how come this circle of a wave function is not 2pi

normally half a circle is pi but here it is half pi

how come is that

[solid swan]

what's the equation though

like maybe
x = cos2t
y = sin2t

[royal cipher]

yeah that s right

i get they do 2pi/2 but i dont get why

[solid swan]

what why

[royal cipher]

like shouldnt every circle be 2pi?

[solid swan]

x=4cos2t - 1
y = 4sin2t

no maybe they like it

or "make it more challenging"

[royal cipher]

it s weird cos the question is calculate the length of the piece above y=2. what they do is calculate the radian of the area between t=0 and C

then they do that times 2. then pi minus then

do u get what i mean

so basically pi - the two lengths between y=0 and y=2

do u get that?

this way they get the length of the piece above y=2

my question is shouldnt they do 0.5pi - the two length, since the top half of the circle is 0.5pi and not pi?

[solid swan]

the length of the piece above y=2

arc length or the chord

to be honest the parametric equations don't really matter that much

you could do 4cos42069t - 1 and 4sin42069t and it would still give a circle

treat it like a regular circle

CM is 4, and CD is 2

so you can find the angle

[royal cipher]

yeah i guess. that s how i accepted the answer lol. It s nothing like a circle but treat it like one

[solid swan]

well it's 2 circles

just do it like normal

[royal cipher]

yeah i have the angle

but isnt it weird they do pi minus the two radians instead 0.5pi?

[solid swan]

no it's not weird

[royal cipher]

like what is have that circle. is it pi or 0.5pi

[solid swan]

because you're dealing with an actual circle, not the equation

[royal cipher]

ah ok

can i ask u about the third step in this question?

in the third step they have the length/rad of the up piece above y=2. That is (2/3)pi. to calculate the length of that piece, they do (2/3)pi * 4

so they basically calculate the perimetr of that piece

using 2pi*r

but isnt the length of that piece just (2/3)pi?

[solid swan]

you mean to t=3pi/4?

[royal cipher]

uhm no

this is what they do

[copper crest]

can someone help me with this?

[royal cipher]

(2/3)pi is the length of that and 4 is the radius

[solid swan]

i think what they mean is like

a circle is 2pi radians

[royal cipher]

i dont know what that means sorry, the question is calculate the length of the piece of path/orbi taht sabove y2

[solid swan]

have you found the angle CME?

[royal cipher]

yeah

it s 2pi/3

[solid swan]

then you can just use arc length = radius * angle

[copper crest]

can I please have some help?

its the last question on my quiz

[solid swan]

i don't think i'm allowed to help you out on your quiz

[copper crest]

how come

[high zephyr]

read #❓how-to-get-help

[royal cipher]

so is arc length the rad of an angle or is it the circumference of a piece of a circle

judging by what the do it s the latter

i think

[solid swan]

length of a piece of the circumference

[royal cipher]

ah ok

alright

[solid swan]

a piece of a circumference is called an arc

[royal cipher]

handy lol

i thought they were the same

an arc and a rad of the same piece

thanks Conan!

[solid swan]

also has anyone solved this yet

https://discordapp.com/channels/268882317391429632/326138757474680852/703818675748077599

[wanton lake]

does x = 63?

[versed river]

Yes

[wanton lake]

i feel smarter already XD

[gray marten]

@solid swan I tried but failed.

[solid swan]

i also tried but failed

[wanton lake]

now does x = 40 in this one?

[solid swan]

yes

[wanton lake]

ooo nice

im actually getting dem correct for once

not to sure but i think this one is 140 correct?

[versed river]

yeah

good job

[wanton lake]

a streak of 3

now my luck runs out probs XD

is this one 93?

@solid swan soz for ping but u r smart

[solid swan]

what

to do this i'd draw a line through c parallel to AB and DE

then you can split it up into 37 + (180 - 124)

[wanton lake]

yes i did that

[untold flare]

Won't it be 180-93?

[solid swan]

i got 93 too

[wanton lake]

yay

[untold flare]

Isn't E parallel to AB?

DE*

[wanton lake]

dis one is new, since i dont understadn what it means or how to do it

how do i prove that they are the same?

[untold flare]

SAS

Congruence

[wanton lake]

side, angle, side?

would c be the angle?

[untold flare]

Ye?

Yes

Wait tho, is that the question?

[wanton lake]

the questions are on the left

im tryna figure dem out

just having tough tiem getting it a bit

[untold flare]

I mean, the data seems very less

Oh right my bad

[wanton lake]

oh also idk if dis helps but

[untold flare]

Yes

Since they've marked it in the figure

With those dashes (sorry for not using appropriate mathematical terms here..)

[wanton lake]

so, if u dont mind me being dumb
wht would be the answers to the 3 questions?

[untold flare]

Angle ACB = Angle DCE

DE will be 6

If you prove 1st point you can deduce the second point automatically

Since the triangles are "Congruent" they will be having same measurements

[wanton lake]

ahhh yes

i see

so will the answer to the first quesition be SAS?

[untold flare]

Using SAS test yes

That is what i can conclude from the data they've given

[wanton lake]

ahh thx

[untold flare]

I hope you might be having an answer sheet to cross verify it

[wanton lake]

i dont, these are just from a bunch of random stuff online

so waht would the answer for question 2 be?

[untold flare]

Oh, okay.

Errr since those triangles are congruent , which you will be proving in point #1,

You can say that AB = DE

This inturn deduces that DE = 6

May i know which grade math are you trying to solve?

[wanton lake]

year 10 i am in!

[untold flare]

Oh, okay...

[wanton lake]

hm?

something wrong?

[untold flare]

Btw the image which you sent above AC = DC and BC =EC

That was a given right?

[wanton lake]

yes

it was a given

[untold flare]

No, nothings wrong, all cool. It's just that i'm trying to help without knowing much about the question, so was scared if i might misguide you :p

So was verifying things

Do you know why those angles of the triangle can be proved equal? (Angle C)?

[wanton lake]

no, idk. not too good at math sorry

[untold flare]

Okay so, using Vertical Opposite Angles (VOA) you can provr that they are equal

[wanton lake]

ok

ill move on to another thing i having problem with

would i use the same strat here?

[untold flare]

SSS test

side side side test

[wanton lake]

yep got it

your pretty helpful

i think i got a test on dis in 1 and a bit weeks

goota study

[untold flare]

Since AC is common it will be equal in both triangles

Paralle sides WILL be equal in a parallelogram

Hence 3 sides are equal

And the next point can be deduced using the congruence proved

[wanton lake]

i see

how should i word that for question 2?

[untold flare]

Hold up lemme write it down

[wanton lake]

kk

u r helpful X)

not to rush u but u there? @untold flare

how do i find x and y?

[untold flare]

I'll tell you in some time, eatin breakfast :x

[wanton lake]

all g

[untold flare]

@wanton lake is Angle ADC 90 degrees?

[wanton lake]

not sure, doesnt say

[untold flare]

Angle ADC And FHE

[wanton lake]

ahhh cool thx!

[untold flare]

?

No i am asking xD

Is there any data given regarding those 2 angles :x ?

[wanton lake]

no

[untold flare]

I am not sure about the x and y problem you gave me.

But the above parallelogram problem for the second statement you can say ONCE the triangles are proved congruent which means the sides are equal, that means the Angles projected by those sides will always be equal

[white cradle]

@vital sleet

it is quite literally plugging in numbers into a formula

do you know the volume of a cylinder with radius r, and height h?

then go review your lesson again

recheck notes/textbook and come back

thats not what this server is for

that is also against the rules .-.

[dark sparrow]

<@&268886789983436800>

[upper karma]

[white cradle]

he just shifted to #help-0

[upper karma]

@white cradle I shifted him outside the server dw 😉

[white cradle]

lmao

kk

[night cape]

a point in the right direction is appreciated. ```
Calculate the area if H is 12cm and the base edge 8squareroot3

[dark sparrow]

the surface area?

[night cape]

yeah

[dark sparrow]

well, don't you have the area already? you just need to plug in $a = 8\sqrt{3}$ and $H = 12$

[somber coyote]

Ann:

[night cape]

what confuses me is $P=2*\frac{a^2\sqrt{3}}{4} + 3aH$
since theres already a sqrt3 there, not sure whether to write it again?

[somber coyote]

ddm4313:

[dark sparrow]

$a = 8\sqrt{3}$

[somber coyote]

Ann:

[dark sparrow]

$P = 2 \cdot \frac{ (8\sqrt{3})^2 \cdot \sqrt{3} }{4} + 3(8\sqrt{3})(12)$

[somber coyote]

Ann:

[dark sparrow]

why would you write 8 and not 8sqrt(3) for a when a is not 8?

[night cape]

just wanst sure

[dark sparrow]

you were overthinking it

[night cape]

alright, now i have $\frac{384\sqrt{3}}{4} + 3(8\sqrt{3})(12)$ unless i did something wrong

[somber coyote]

ddm4313:

[dark sparrow]

you've got a lot of simplification to do there

[night cape]

yeah, ended up going too far and got a crazy number lol

how is this exactly called in english?

[dark sparrow]

what

[stray hearth]

can someone help me ?

[night cape]

this

[solid swan]

triangular prism?

[night cape]

yeah, but like this formula

[silent plank]

formula for surface area of an equilateral triangular prism?

[night cape]

thanks

[stoic timber]

Hi can someone help me out with this, I know how to do these types of equations when the domain in radians but not sure with this one, thanks.

[silent plank]

the domain for this question is in radians

[stoic timber]

I guess I'm just lost, my thought process was to input the 0, and 3 to find the endpoints but im not getting an answer for it

[upper karma]

2pi is a full rotation

so 2pi * x (where x = 3) would be 3 full rotations

you'd end up right where you started

but since it's divided by 3...

[dark sparrow]

y'all overthinking it...

[stoic timber]

I see

i dont know what happened to my brain

[dark sparrow]

i mean you can substitute θ = 2πx/3 and then θ will range over [0, 2π] and your equation will be sin(θ) = -sqrt(3)/2

if that makes it easier for you to comprehend

[stoic timber]

ye I have no idea why I just started overcomplicating it, I had 2pi = root 3/2 and i was having mental block

Thanks guys

[upper karma]

sine of what equals "-√3/2" ?

[stoic timber]

pi/3

[solid swan]

what

that's the reference angle

[dark sparrow]

no, sin(π/3) = sqrt(3)/2, not -sqrt(3)/2

[solid swan]

think unit circle

[upper karma]

-pi/3

(which corresponds to -√3/2)

@slate lagoon need to find x "fast"?

is this a test?

[slate lagoon]

yes

[silent plank]

#❓how-to-get-help 6

[upper karma]

we're not allowed to answer tests

[slate lagoon]

oh ok

[feral sail]

not sure if this is the right place to post, but i need to find the x, and im rly confused.

[versed river]

How even

Its the same test

[upper karma]

@feral sail same question as before :)) haha

we're not allowed to help with tests

[feral sail]

not testg he retard

i mean not u

but prismatic

[upper karma]

it just "happens" to be the exact same question that @slate lagoon asked 30 seconds before you?

[feral sail]

he have mental issue

he need help

[upper karma]

@tender sphinx

aaaand you're offline

nice

[cerulean tundra]

its freaking thales lmao

its some elementary stuff

just remove the /3 and you skip 3 steps xD

[upper karma]

I don't get it, he changed his name from "Person" to "Human"

and the "@" id's no longer match

@silent fulcrum#0341

the hell...?

@silent fulcrum#0341

I right click his name, choose "Mention" and it ends up pinging some other random dude

with a different avatar

and different id number

"human"

https://tenor.com/view/baby-monkey-unbothered-whatever-long-day-gif-13606740

[mint olive]

https://gyazo.com/e1daab317bcd37dab541a7876b21f06d

Need to find y

help

plz

[upper karma]

do they give you any angles?

[mint olive]

nope

[silent plank]

consider properties of parallel lines and similar triangles

[mint olive]

https://gyazo.com/72294c104340487e903f850c4eb5d919 which answer is right? ignore the blue outline

[upper karma]

is this a test?

[silent plank]

did you give the problem another attempt after seeing my suggestion?

not going to blatantly give out answers

[mint olive]

Im not sure what you said, because I don't understand english very well

is this a test?
@upper karma its exercise

[silent plank]

which language? someone else might be able assist

[mint olive]

Estonia

[silent plank]

1 sec orientation of that was bad

[mint olive]

okay

[inland kayak]

Is the 5 correct answer?

I think it is 5

[mint olive]

Agreed

@silent plank Thanks for helping!

[upper karma]

why is it 5?

[inland kayak]

BEcause it is logical answer

[upper karma]

what is the logic in that?

prove it

[silent plank]

apply the properties of similar triangles and solve the resultant equation

[upper karma]

right?

[silent plank]

write better 4s. it looks like a lightning bolt symbol

but yes

[inland kayak]

LIke Y must be bigger than 4 and lower than 6.5 thats logic right?

[silent plank]

if you want to find y, don't worry about the middle part involving x

[inland kayak]

and i had 5 in my answers option

IT pretty hard to learn math in home

[silent plank]

LIke Y must be bigger than 4 and lower than 6.5 thats logic right?
where's that coming from?

[night cape]

calculate area of a quadrilateral prism which base is 36cm^2 & diagonal of side 10cm https://cdn.discordapp.com/attachments/501175742504239106/704251884621529169/Capture.PNG

not sure what to do here

[upper karma]

what is P?

[night cape]

p is area

[upper karma]

and B? and M?

H is the ... height?

[night cape]

h = height, b = base

[silent plank]

is that a translation?

[night cape]

yeah

[silent plank]

do you have a diagram to go with it

[upper karma]

P for area is weird

P usually stands for perimeter

or point P

[night cape]

yeah, but everything is upside down in my language

[silent plank]

looks like its supposed to be a square prism

which diagonal are they referring to?

d_bs sounds the most appropriate

[night cape]

yeah

thats it

[silent plank]

given B = a^2 = 36cm^2
were you able to determine a?

[night cape]

what do you mean

[silent plank]

in your question, what would be the side length of the square base?

[night cape]

36^2

[silent plank]

36cm^2 is the area of the base

also 36^2 is 1296

[night cape]

i need to get height i believe to solve this

[silent plank]

i'm leading up to that

please try to follow these instructions

solve: $ a^2 = 36cm^2$ for $a>0 \
a = \ ?$

[somber coyote]

ramonov:

[upper karma]

B = a^2
B = 36
a^2 = 36
a = sqrt(36)
a = ?

[night cape]

6

[silent plank]

so a = 6cm right?

[night cape]

yeah

[silent plank]

and you are also given that d_bs =10cm right?

[night cape]

yep

[silent plank]

now consider $\triangle NPP_1$ in your diagram and apply pythagoras to determine the height (H)

[somber coyote]

ramonov:

[night cape]

thanks, d_bs is the hypotenuse, yeah?

[silent plank]

yes

[night cape]

thanks

ok i got 16

[silent plank]

16 for what?

[night cape]

nevermind, i got 4 as b

[silent plank]

what's b?

[night cape]

height

[silent plank]

that is incorrect. show your work.

[night cape]

$b=\sqrt{10^2-6^2}$

[somber coyote]

ddm4313:

[night cape]

am i wrong?

[silent plank]

not sure why you introduced another variable b, when you could've just used h,
but that seems ok so far

[night cape]

my bad

i thought it wouldnt be a perfect square

but it seems to be

2

[silent plank]

2?

[night cape]

well yeah, 10^2-6^2 is 4^2 and then 2?

[silent plank]

no how are you getting "10^2-6^2= 4^2"

[night cape]

didnt square it

[silent plank]

$a^2 - b^2 \neq (a-b)^2$

[somber coyote]

ramonov:

[night cape]

ok i got 8 now

[silent plank]

thats better

[night cape]

alright, so height is 8, now we just plug in the values if im correct

[upper karma]

which one is "diagonal of side 10cm" ?

D?

[night cape]

d bs

okay, so i got now

$M = 4ah $ after that i did $M = 4 * 6 * 8$ and $M = 192$

[somber coyote]

ddm4313:

[upper karma]

what is "M" supposed to be?

you know a = 6

you know H = 8

you know the formula for the area of a square prism

[night cape]

yes, i know, im just not sure if its correct which is why im asking

[upper karma]

plug the values in

[silent plank]

in the context of those formulas M is supposed to be the sum of the areas of those rectangular faces

[upper karma]

Area = 2a^2 + 4aH

[night cape]

$P = 2 * 36 + 192$

[somber coyote]

ddm4313:

[silent plank]

that looks okay

[upper karma]

yep

[silent plank]

(include units when writing it on paper)

[night cape]

thanks

really appreciate the help

[upper karma]

264 cm2

just as long as we're absolutely sure that "diagonal of side 10cm" means d_bs and not D

because then the resulting area would be incorrect (since the wrong diagonal was used)

[night cape]

yeah, it is d_bs

[upper karma]

I had some free time

math checks out

but I don't understand something from yesterday

b = 180 - (x + 100)
b = -x + 80

how?

[silent plank]

distributive property

[upper karma]

@silent plank

[silent plank]

um, how are you getting from the first line to the second line?

that isn't how distrubution works

[upper karma]

That is just wrong...?

$b=180-x-100$

If you are looking for that, im out of context here.

Oh its an x lol

[somber coyote]

Al3dium:

[silent plank]

you added an extra 180 for some reason

[upper karma]

^

I don't get it

distributive property is with multiplication

not addition

But what the hell did you do from the 1st to the 2nd

[silent plank]

-(x + 100) is the product of (-1) and (x+100)

[upper karma]

@upper karma
UPDATE: this is incorrect

I did this

Wow

[silent plank]

wow that's even more fuked up then

180 isn't being multiplied to those parentheses

which is what those red arrows seem to suggest...

[upper karma]

It should lead up to
$b=180-x-100$

[somber coyote]

Al3dium:

[upper karma]

Bc the minus is acting as an inversion of signs or what Ramonov said, as a product of (-1)

[silent plank]

would you agree that: $180 - (x + 100)$ \
is equivalent to: $-(x+100) + 180$?

[somber coyote]

ramonov:

[upper karma]

yes

[silent plank]

and how would you have simplified the latter

[upper karma]

but how do you get rid of that parenthesis?

Bc the minus is acting as an inversion of signs or what Ramonov said, as a product of (-1)
.

[silent plank]

distributive property

$-(x+100) \equiv -1(x+100)$

[somber coyote]

ramonov:

[silent plank]

if you want

[upper karma]

Multiply -1 to x and to +100

[silent plank]

but that 1 is ass, ugly, unnecessary so we don't write it

[upper karma]

-(x+100) + 180 everything * (-1)
(x+100) - 180

so the parenthesis then?

how do you get rid of it?

What

Multiply -1 to x and to +100
.

[silent plank]

distributive property
distributive property
distributive property
distributive property

[upper karma]

Lmao

distributive property is called "the distributive property of multiplication over addition"

you're saying you're doing this with the "invisible" 1 there?

Multiply -1 to x and to +100
. intesifies

You can do either of both

Which are the same lol

[silent plank]

maybe if i wrote it as...

$(-1)\cdot(x+100) + 180$

[somber coyote]

ramonov:

[upper karma]

b = 180 - (x + 100)
b = -(x+100) + 180 (reversed, same thing)
b = ...
get rid of the parenthesis how?

Bruh

[silent plank]

i'm done

[upper karma]

Five times already

Same

[silent plank]

all my messages seem to be written with invisible ink

[upper karma]

(-1) * (x+100) + 180
-x - 100 + 180
-x + 80

ok, yeah

b = -x + 80

I get it now

distributive property makes sense now

-(x+100) + 180
is the same thing as
(-1) * (x+100) + 180

and when you distribute it you end up with "-x-100+180", which is "-x+80"

I get it

[acoustic jungle]

good job

[high wren]

how would i start this?

[dark sparrow]

if the equation was $\csc(z) = \sqrt{2}$, would you be able to solve for $z$?

[somber coyote]

Ann:

[high wren]

wouldnt it just be

pi/4

and

5pi/4

[dark sparrow]

one of these is a solution of the equation, the other is not, and you're missing infinitely many solutions anyway if you are to solve this on the whole number line.

in other words, no.

[high wren]

pi/4 + 2piN, 5pi/4 + 2piN

?

[dark sparrow]

z = 5π/4 does not satisfy csc(z) = sqrt(2).

[high wren]

oh wait

[upper karma]

@high wren do you understand how to get "z" from "csc(z) = sqrt(2)"?

[high wren]

isnt it

just

like

sin(x) = sqrt2/2

[dark sparrow]

z, not x. but yes

and also, use parentheses with square roots in plaintext.

[high wren]

x = pi/4 and 3pi/4

[dark sparrow]

z

[upper karma]

arccsc

or csc^(-1)

[dark sparrow]

dspider no

[upper karma]

no?

[dark sparrow]

you are not going to tell this person to blindly "arccsc both sides"

i was almost there with my explanation

[high wren]

[dark sparrow]

@high wren yes, csc(z) = sqrt(2), if z is only considered on [0, 2π], has solutions z = π/4 and z = 3π/4. so you've fixed your previous mishap.

and i hope it's not that big of a leap from that to saying that the solutions of csc(z) = sqrt(2) on the whole number line are given by z = π/4 + 2πn and z = 3π/4 + 2πn, where n ranges over the integers

does that make sense to you

[high wren]

ok