geometry-and-trigonometry

[grizzled zinc]

Yeah i hate the formatting

Id prefer bigideasmath as review but theres slader that people can use to find answers

Oh nvm though my teacher got back to me.

[acoustic jungle]

they add up to 180

it's a cyclic quad

[latent iron]

What would be the diameter of the tire for this question?

[dusky surge]

@latent iron
Hello ryan. What do you think of about the diameter of the tire? What does it relate to?

[latent iron]

I'm not sure.

It's a new type of problem I've never encountered.

[dusky surge]

I see.

So, as you see, the cosine function is an oscillating function.

[latent iron]

Okay.

[dusky surge]

And it acts like a point moving on a circle

,w graph cos(x)

[somber coyote]

Display results online and refine query

Requested by Biscuit#5208

[dusky surge]

So can you guess it? What does it relate to?

[latent iron]

Uh.

[dusky surge]

This nail moving up and down

[latent iron]

yeah, i got that

i just don't get how we would figure out the diameter cause wouldn't it be consistent

so would it be +25 ?

[dusky surge]

So, the distance between the topmost point and the bottommost point will be the diameter

The vertical distance

[latent iron]

OH

omg

im so dumb

So id need to plot the graph right?

[dusky surge]

Yep

It will definitely help you figure it out

[latent iron]

it also asks how many times it rotates in 10 seconds

and then

[dusky surge]

Yea

Is it in degrees or in radians?

[latent iron]

At what time in the first two cycles (starting t=0) will the nail be 30 cm?

degrees

[dusky surge]

Yep, then if you draw it out, it will definitely help ya

[latent iron]

wait so

how would the graph look

cause i tried using desmos but it isn't workin

[dusky surge]

Did you change back to degree?

Default should be in radians

[latent iron]

where do i switch it

[dusky surge]

Settings

Top-right

[latent iron]

would it look

like this ?

[dusky surge]

Yep

Now all left is how good you can make use of desmos

[latent iron]

thank you so much

ill let u know if i need help with more!

i appreciate it a lot

[dusky surge]

You're welcome!

[latent iron]

@dusky surge

[dusky surge]

A negative cosine equation?

[latent iron]

yea

its apart of my hw but idk how to do it

[dusky surge]

That means it's something like -acos (bx+c)+d

Let's start from reading some important information from the graph

Like, the amplitude, the period, the middle point, and the shift of the cosine

[latent iron]

Alright

[upper karma]

doesn't look like a cosine

[dusky surge]

First one, what is the amplitude of this graph

[upper karma]

"negative cose equation"?

[dusky surge]

Ryan, got the amplitude?

[latent iron]

one sec

[dusky surge]

Sure, take your time.

[latent iron]

its 3 right

[dusky surge]

Yep.

What will the amplitude of a -cos(x) be

[latent iron]

i have no idea

LOL

im confused

[dusky surge]

We are comparing

Both that drawn function and a negative cosine function

Do that we can make a "negative cose equation" for the graph

[latent iron]

would it be

like

-3 ?

but amplitude can't be negative right

so

[dusky surge]

Try and use desmos

It always helps

[latent iron]

i dont have

the equation

[dusky surge]

-cos(x)

=y

We are comparing both the draw function and a negative cosine function

[dusky surge]

Ouch XD

Seems that it is not working in degrees.

I check in radians it worked, but we are working on degrees right?

[latent iron]

yes

[dusky surge]

Hmmm..... Do you know how to use geogebra?

[upper karma]

@latent iron https://www.khanacademy.org/math/trigonometry/trig-function-graphs

this teaches you how to identify the amplitude, midline, period and phase shift

[latent iron]

@upper karma i know how to figure them out there isn't a full graph given tho

@dusky surge no idea what that is

[upper karma]

you don't need a full graph

you can determine it by looking at the first half

or even the first quarter

it's obviously twice as long as half

and 4 times longer than 1/4

if 2.5 is half.... how long do you think is the entire thing?

or if a quarter of the graph is 1.5... how long do you think is the entire thing?

[latent iron]

5

[upper karma]

look at the graph

is 2.5 half?

[latent iron]

yeah

looks like it

[upper karma]

how about now?

is it still 2.5?

[latent iron]

o wait no

its 6 my bad

[dusky surge]

I still haven't figured out why desmos' not working in degrees...

[latent iron]

so one period is 6

amplitude is 3

[hearty vault]

What is the answer here?

[dusky surge]

Sorry, @hearty vault this is channel is currently occupied

[hearty vault]

What does that mean?

[dusky surge]

Ryan, try y=cos(360x)

[latent iron]

ok

[hearty vault]

what is the answer? i just need answer

[upper karma]

@hearty vault means move to a questions channel

[hearty vault]

ha ok

[upper karma]

And we don’t hand out answers here

We help you solve them

[upper karma]

For Desmos, you can change it to degrees from the wrench icon

[latent iron]

i got that

[dusky surge]

Yea. Cool

[latent iron]

what from this point?

[hearty vault]

And we don’t hand out answers here
@upper karma I studied this like 4 years ago. I just need answer.

[dusky surge]

So, now we compare the amplitudes.

[latent iron]

its 4,5

4.5*

oh wait

oops

its 2

[dusky surge]

Before we got the amplitude 3 and this cos(180x) we have amplitude...2

So it's 1.5 times bigger

[upper karma]

No way to verify that. You could be asking for an answer to a test question. Either way, I will not just give you answers.

[dusky surge]

Then we just multiply it by 1.5 and head on to the next step

Which means we are closer to the answer.

We have y=-1.5cos(180x)

Next is the period.

[latent iron]

the period is

6

for the original graph

[upper karma]

@hearty vault google "the hardest simple geometry problem"

[latent iron]

they gave

[upper karma]

https://www.youtube.com/watch?v=S2GA1zMdl-E

[dusky surge]

And comparing our made up graph y=-1.5cos(180x)

What is the period of this graph?

[latent iron]

one

the other one was 6

[dusky surge]

So... What do we do now?

It's 6 times

[latent iron]

uhhh

im not sure lol

[dusky surge]

Okay. For periods, for longer periods, low the number it is.

6 times is longer

So we divide x by 6

Remember to keep track on our made up function in desmos!

[latent iron]

kk

[upper karma]

forget desmos

you should be able to determine the amplitude, midline, period and phase shift of any graph

just by looking at it

[dusky surge]

The new one about the period

We stick the division with x

[upper karma]

so the amplitude is 3, period is 6

what's the midline?

[dusky surge]

So we have y=-1.5cos(180x/6)

[latent iron]

the midline is

-2

[upper karma]

why -2?

[upper karma]

and you think there's a minus sign in front of 2?

[latent iron]

OH MY BAD

2

[upper karma]

up is always positive

and right is always positive

same thing on a unit circle

[latent iron]

mb

[upper karma]

you really should go through those lessons on Khan Academy

https://discordapp.com/channels/268882317391429632/326138757474680852/708353405575495769

the guy explains it far better than we could

[latent iron]

which video

[upper karma]

no... not one video

you can't just cover this stuff with one video

it's multiple videos

[latent iron]

yeah but

ive alr watched some of my teachers videos

[autumn gazelle]

is this true? i dont wanna pull out my copy of elements xd

[upper karma]

no

[autumn gazelle]

damn

ok well the question is

[upper karma]

so what r u looking so answer

express MP in terms of a b and k ?

[autumn gazelle]

yea

[upper karma]

The diagram is wrong

[upper karma]

Can someone help me with a problem?

i can try

[rich wolf]

4/7=x/310u^2

[upper karma]

@rich wolf ejem exam ejem

looks like a test

[glad oak]

how do you use the angles inside circle theorem to solve this

[soft gulch]

15sinA+8cosA=8

Find 8sinA -15cosA

[glad oak]

me?

[soft gulch]

If you can

[upper karma]

it's easy: open up Youtube, search "circle theorems", click the first 2-3 videos

they should cover it pretty well

[soft gulch]

15sinA+8cosA=8, Find 8sinA+15cosA

[upper karma]

@soft gulch what if you replace sinA with, say... x

and cosA with y

?

15x+8y=8

can you solve it then?

[soft gulch]

x=8(1-y)/15

(64(1-y)/15)+y

64-64y+15y/15

Nowhere

No conclusion

[wooden current]

@autumn gazelle do you still need help?

[upper karma]

I'm trying to find a and theta of this triangle using cosine law but im running into a few problems

I was able to use google to correctly calculate everything, so I know the way I've used the formula isn't wrong.

However, when I run it through a code that I've wrote using the above equations I get stuck on the first a = ~ formula.

gs = a.

And it's giving me 22.13 instead of 27.25.

I've checked my code numerous times but I can't figure out what's wrong with it.

is it something got to do with the sine?

cause I got stuck because of sine ambiguity before

I corrected my code and it's returning even more absurd value.

oh wait

ummm

yeah my bad. it turns out i forgot to square the sine in my code. BUT it's giving me on error on other problems.

Btw this isn't homework. I've been out of hs/uni for a while and it's a personal project to code a aerial/nautical navigation computer.

💩

@acoustic jungle

[acoustic jungle]

Find 8sinA -15cosA

15sinA+8cosA=8

I squared both sides and the -240 sinacosa canceled out, giving me +-15

@soft gulch

[upper karma]

uhh I found out that it's not a math problem but a programming one.

it's all good now.

[wooden current]

[upper karma]

Lmao

[barren tide]

Can anyone help me on what to do next?

here's a better picture

[upper karma]

that looks right

[acoustic jungle]

how is that right

(a+b)/(a-b) is not b/-b

it should simplify to cos^2-sin^2 over 1

[barren tide]

Could you walk me through how you got that?

[acoustic jungle]

I multiplied the denominator and num. by sin^2x

sin^2(x)

[upper karma]

i said it looks right

not it is right

lol

[barren tide]

oh fuck he's bread now

[barren tide]

I was wondering if someone can walk me through how to simplify this equation

I dunno how to simplify (sqrt(3) -1)/(1+sqrt(3))

[glacial haven]

multiply the top and bottom by 1-sqrt 3

[rich wolf]

,w (a-b)/(b+a)

[somber coyote]

Display results online and refine query

Requested by AMD#4434

[barren tide]

I got -1 but it says that it's not the answer

Am I doing it wrong?

[rich wolf]

Nope

[acoustic jungle]

(a+b)/(a-b) is not b/-b

[barren tide]

I don't know what that means

[rich wolf]

@acoustic jungle that's not the same thing as what the problem says

[merry abyss]

oof

[rich wolf]

$\frac{\sqrt{3}-1}{1+\sqrt{3}}\times\frac{1-\sqrt{3}}{1-\sqrt{3}}$

[somber coyote]

AMD:

[barren tide]

ooooohhh

[rich wolf]

yes

that's what it is

[barren tide]

I went to Mathway and it showed me how to answer the numerator. I've always had trouble understanding this formula

Just to clarify, So the formula is that you multiply the numerator by (-1) and multiple that by the numerator and the denominator.

[gentle pasture]

Anyone need help?

[carmine sundial]

can someone explain the instructions for me? i dont understand the length part.

does it mean if the answer is (example) 45.7892 will turn to 45.78?

[upper karma]

Idk

I’m also confused

[prime jewel]

anyone knows how to get the ordered pair?

[upper karma]

Dont u just plug -4 into both equations ?

[brisk holly]

nvm i got it

pi/10 right?

[rich wolf]

$H(t)=-7\cos(\frac{\pi}{10}(t-2))+7$

Wtf

[brisk holly]

lol

[somber coyote]

AMD:

[rich wolf]

How many times in what interval?

[brisk holly]

A pendulum is swinging back and forth. After t seconds, the horizontal distance from the bob to the place where it was released is given by

thats all it says

[rich wolf]

It has to give an interval

Take a screenshot

[rich wolf]

Oh thats some crucial info right there

I see its referring to frequency

I dont think pi is the right answer here

[brisk holly]

so whats the answer

[rich wolf]

You are trying to find when the function is equal to its own midline

So when H(t)=7

So set the function equal to 7

$7=-7\cos(\frac{\pi}{10}(t-2))+7$

[somber coyote]

AMD:

[rich wolf]

Now find two adjacent solutions

[rich wolf]

Yeah

So what is the distance between each of those numbers

[brisk holly]

10?

so thats the answer?

[rich wolf]

Yeah checks out

[brisk holly]

sick it was right

thanks

[rich wolf]

You should pay me

[brisk holly]

AMD is already screwing us though with them not supporting b450 for no reason

[rich wolf]

I have a family

Please

[brisk holly]

so do we

[rich wolf]

Lmao

[dark notch]

how to do 1 through 8? these are practice questions

i just need explanation for each with answers

[upper karma]

@dark notch https://www.youtube.com/watch?v=MyzGVbCHh5M

this is basically all you need to solve all of those

[dark notch]

i don’t get these can someone answer for me pls

[upper karma]

watch the video

[dark notch]

so the lines sit on the circle

bro wtf

nah sorry

idk

[upper karma]

which one don't you understand?

@dark notch

aaaand you're offline

https://youtu.be/GppOSNTi5OA?t=717

there you go

[upper karma]

i don’t get these can someone answer for me pls

lmao

[carmine sundial]

hello?

triangle reference

[winged hill]

so C is 90 degrees, since all angles in a triangle must add up to 180, 180 = 90+B which is 32*15'(whats this?)+x, solve for x

x should be angle A

wait oops their message was sent a long time ago

[dark notch]

okay i am stuck on one problem can anyone pls answer it for me with explanation

[upper karma]

So first of all, you know that in a quadrilateral opposite angles are equal right?

And adjacent are supplementary

If you make Angel CDB An Inscribed Angle you get 80

So you know that Angle CDB = 80 right

Then you set the 2 opposite angles = to each other

80 = 11x + 16

@dark notch You got it?

[dark notch]

yes thx

is answer

80

or

i have to solve that equation

[upper karma]

Solve equation

But did you understand what I said

[dark notch]

yes

[upper karma]

👍

Wait

@dark notch wait

Wait

WAIT

HELLO

I MESSED UP

I’m sorry

It’s opposite angles are supplementary

Common mistake

Messed them up

In a cyclic quadrilateral, opposite angles are supplementary. If a pair of angles are supplementary, that means they add up to 180 degrees. So if you have any quadrilateral inscribed in a circle, you can use that to help you figure out the angle measures.

80 + 11x + 16 = 180

What is this.

Oh idk

Idk if I’m correct tbh

[untold lance]

no i think you are right

sum of opposite angles in a cyclic quadrilateral adds up to 180

[upper karma]

He is right.

[upper karma]

No u

Oh wow

[upper karma]

youre also right too

Yay

[acoustic jungle]

80 + 11x + 16 = 180
@upper karma angle D is not 80.

unless 160 is referring to the arc NBC. but it looks like it's referring to arc BC.

I guess it has to be referring to arc NBC otherwise I am not sure how to solve it.

[upper karma]

everybody right

[acoustic jungle]

what is this

[merry abyss]

that guys tag

[upper karma]

Why is angle d not 80

If you consider angle d as inscribed angle CDB it would be 80

Ya idk

@dark notch Did it come out as correct?

[acoustic jungle]

No. Angle NDC is 80 if 160 is referring to arc NBC. Angle BDC is 80 if 160 is referring to arc BC. But BDC is useless and doesn't provide any information.

[upper karma]

Is there anyone that can help me with my geometry hw?

[winged hill]

what is it about 😔✊✊

if it’s some triangle stuff i’m pretty sure i can be of help

[upper karma]

https://media.discordapp.net/attachments/326138757474680852/708818614186410084/image0.png <@&286206848099549185>

someone help

I am also interested in solving this

And I’m curious what the answer actually is

The other guy didn’t get a correct answer anyway either yet

[upper karma]

16 what

this image doesnt make sense

11x + 16 degrees would make sense

you just have 16 units

so frustrating

@upper karma what

@upper karma it just has 11x + 16

what does "16" mean

16 what

11x + 16 is referring to the angle

Degrees

?

How do u find what D is

My man that’s what I’ve been saying awhile ago in chat

@acoustic jungle doesn’t agree

Angle D = 80 right

Ik

But my boi @acoustic jungle Disagrees

@upper karma Shall we prove him wrong?

It’s 80 right

It’s not my work

And then u set them = to 180

11x + 16 + 80

= 180

This isn’t my problem

I’m just curious

It’s someone else’s problem

But is that the answer

It’s not my homework

I was just curious to see the actual answer

But it turns out I was correct

Is the first place

But @acoustic jungle Doesn’t agree

Lol what

It’s this dudes work

@upper karma

?

Can u help

?

No I’m just curious of what fishraider means

Why isn’t it 80

?

@upper karma

@acoustic jungle what do u mean

BDC isn’t useless it is 80

Opposite angles of a cyclic quadrilateral = 180

Set them equal and find X

[vale nimbus]

@upper karma D isnt 80

i dont know what it is, but it isnt 80

[upper karma]

@vale nimbus How so?

Why isn’t it 80

I’m using the inscribed angle theorem

Does this theorem not apply to it for some reason?

Am I missing something?

[remote heart]

that isn't the entire arc

160 is part of it

[upper karma]

@remote heart Arc NBC?

That’s 160?

I don’t really understand

[remote heart]

no, it is not 160

[upper karma]

What r u saying

What is 160 then

[remote heart]

160 is only part of nbc

[upper karma]

It’s arc bc then

Right?

[remote heart]

arc bc is 160 degrees

[upper karma]

And then use inscribed angle theorem and you get angle D = 80

Right?

[remote heart]

no because angle d does not intersect BC

[upper karma]

Ya but how else does angle d correlate to arc Bc

You basically have to just make it a inscribed angle

How else does those 2 connect

@remote heart

Hint says find angle d using 160

Also his homework in the past was focusing on inscribed angles

This was part of his homework of inscribed angles

Wait I’ll show u

[remote heart]

yes, it uses inscribed angles

[upper karma]

Ya so what’s the answer?

[remote heart]

I haven't worked it out yet, but it is not 80 as 160 is reffering to BC, not NBC which D intersects

[upper karma]

Oh

Ya I’m starting to understand now, thanks dude

[remote heart]

sure, let me know if you need anymore help 🙂

[upper karma]

Ya it’s not even my homework tho, I’m just curious

[remote heart]

mhm

[upper karma]

This @dark notch homework

[remote heart]

ik

[upper karma]

Cool

[winter beacon]

area is 38.5 cm^2

and i have to find angle p

is it 58.8?

[silent plank]

area is 38.5 cm
cm**^2**

it would be 58.2 degrees if you were asked to round to 1dp

[winter beacon]

yeah my bad

it says

find the measure of angle p to 1dp

[silent plank]

capital P, but yeh. write the ° and it should be fine

[winter beacon]

alright thanks

what about this?

find BC to 3 significant figures

[dark sparrow]

yeah, what about this

[winter beacon]

they asked to find BC to 3 s.f

is it 10.9?

[dark sparrow]

i don't know, is it?

[winter beacon]

idk i'm asking

i think it is

[dark sparrow]

what's making you doubt that it is

[winter beacon]

idk im not that good at math

[upper karma]

Just use law of cosines

[dark sparrow]

well how did you get 10.9

[winter beacon]

cosine rule

[dark sparrow]

if you didn't fuck up your arithmetic, and you wrote down the law of cosines correctly,

then surely it stands to reason that your answer should be correct

[winter beacon]

oh ok

[upper karma]

also, if you stick your hand in water your hand becomes wet

[winter beacon]

alright thanks

[wooden current]

If I stick my hand in water I will disappear into oblivion.

[winter beacon]

What if you dive in then?

[regal shell]

I’d you dive in them, you won’t be wet

[winter beacon]

Hmm

[upper karma]

I'm confused, dive in a pool of hands?

[winter beacon]

Yes

The hands will pull you down slowly

as you try to escape

[wooden current]

kinky

[winter beacon]

You could make a religion out of this

[regal shell]

Y don’t you replace it with quicksand

The hands pull down the quicksand as it tries to escape

[winter beacon]

Wouldn't the quicksand pull down the hands at the same time?

so the hands and the quicksand will keep on pulling each other down

forming a never ending loop

[regal shell]

Cyclical falling

[peak elm]

Hello

Someone help

Sup

Ye

Yurrrr

[harsh pier]

Im not sure but id say just imagine the unit circle and the triangle in it. From there you should be able to get one of the sides to the additional hypotenuse that should be 1 (if you dont have a idea how to get information about one of the sides remember what sin is

From there you should be able to use the method hinted at at the top of the exercise

[peak elm]

Head

Ass

[winged hill]

use the pythagorean theorem to find the missing side, c(hypotenuse),a(side),b(side) , plug in the values you have into this, c^2 = a^2 + b^2 , and solve for the missing value

[peak elm]

The fuck

[winged hill]

so here, they have given u the hypotenuse and the opposite side

to angle theta

using the pythagorean theorem you have to find the adjacent side

the hyp is 20, the opposite side is 17

20^2 = 17^2 + X^2

solve for x

cosine theta should be x/20

[peak elm]

So what is x?

[winged hill]

lemme plug that into cymath im running on no sleep rn

[peak elm]

Lol

Uglyass

[winged hill]

🥴

https://www.cymath.com/answer?q=20^2 %3D 17^2 %2B X^2

heres a link so you can view the process

[peak elm]

Oh aight

Yo Morin

[winged hill]

?

[peak elm]

Is that the final answer

[winged hill]

10.5

^

rounded

[peak elm]

Alright

[winged hill]

to the first decimal place

so the answer would be 10.5/20

[peak elm]

Oh ok

[winged hill]

OR

if it allows u to write roots

then write the root

cause it said "exact answer"

[peak elm]

Oh alright

Thanks

[winged hill]

youre welcome

[upper karma]

<@&286206848099549185> I need help

😭

I need help with my circles homework

¯_(ツ)_/¯

😭

[glacial haven]

post

[upper karma]

Instead of that decimal.

Just write sqrt space 10.

It'll automatically put in a square root sign if you put it in.

[rich wolf]

well you know that 2sqrt10=sqrt(40)

[upper karma]

Oh

Fuck

Damn it

I didn’t put a space when I did sqrt the first time

I’m retarded

[rich wolf]

and you know the circle equation doesn't have r in it

but r^2

[upper karma]

Is the equation right?

[rich wolf]

for which one?

[upper karma]

First one

[silent plank]

no

[upper karma]

What do I change it to then

[glacial haven]

is it 10?

[upper karma]

Maybe

[silent plank]

...=r**^2**

[upper karma]

So 10

Is this right?

[harsh urchin]

https://i.imgur.com/SaT5eFr.png yo for question "iv", am i supposed to get area and length of the smaller sector or the bigger sector? because for the rest of the questions its for the smaller sector. The answer in the book for the "iv" question is for the bigger sector

[upper karma]

@upper karma Yep correct 👍

Lesss gooo

Are circles supposed to be this hard?

No

It’s pretty easy

Oh

I just don’t get the point of circles, and I can’t see it from their side, it seems they don’t have any.

Sorry had to get my circle pins out of the way

Puns*

Lol

[silent plank]

bigger sector. (since the bigger arc is highlighted, the marked angle is also of the bigger sector)

[upper karma]

@harsh urchin yes

@harsh urchin yes

[harsh urchin]

oh thank you i didnt see the highlight lol

[upper karma]

@harsh urchin do you see the little curved arrow in the center?

Yes

that's your angle, which represents the arc

it's even colored in blue (green?) if you look close

Aight alright thanks

[languid finch]

can anyone help with this one? thx

<@&286206848099549185>

[upper karma]

Damn i wish the letter was smaller

[upper karma]

Damn it feels good to be a gangster

Lol

[upper karma]

My textbook is calling this a parallelogram

I looked up this definition and it says a parallelogram has 2 pairs of parallel sides

Anyways... is the height of this the line on the origin from 1 to 4 ?

So I can calculate it's area

nvm its 2, -1

lol

if it was a parallelogram the correct point would be (4,4), not (4,2)

[upper karma]

Weird

[stable magnet]

This is for a friend

never learned bearings in my life

[upper karma]

https://www.youtube.com/watch?v=jVSwRDO0cq0

anil kumar 👍

honestly it was the first video recommendation when I googled "how to find the bearing [autocompleted with "...between two points"]"

so I didn't even need to type the whole thing

technology is awesome

that should get you started @stable magnet

or rather, your "friend"

@languid finch do you still need help with that?

eh, you're offline

never mind then

[stable magnet]

thanks

[calm kite]

anyone know this

[umbral snow]

Just the slope of the secant line between the two points

[barren tide]

Can someone explain the process how how it was factored?

[upper karma]

they factored what was common in both terms

both terms had a 6tan(theta)

when u factor out a 6tan(theta) you get whats left of each term

@barren tide

[barren tide]

thanks bro

[upper karma]

np

[barren tide]

How do they get that answer?

[rich wolf]

Turn of scientific notation

Do you see where it says E-6?

Next to the answer on the calculator?

Thats because it's a really small number

[dark sparrow]

you mean E-4

[rich wolf]

Yeah couldnt see it lol

[barren tide]

No not that, how come I get 3.00864.... on my calculator, but the question shows that the answer is 0.00030...

oh is it cuz they moved it?

[dark sparrow]

3.00864E-4 is your calculator's way of writing 0.00030864

in scientific notation

[barren tide]

i hate math

thnx

[dark sparrow]

\verb|3.00864E-4| stands for $3.00864 \times 10^{-4}$

[somber coyote]

Ann:

[barren tide]

This shit hurts my soul. I've been trying to figure out this question for like an hour

You guys are so very nice for helping my little ape brain understand numbers

And I thought I could be a mathematician 😭 😭 😭

[upper karma]

you can be anything you want

❤️

[barren tide]

I want to able to pass my class with an A

[rich wolf]

Bro this is what's called a "fixed mindset"

And you need to get out of it

You are not "good" or "bad" at math by nature. There are merely concepts you have and have not had sufficient exposure and practice with

[upper karma]

UwU

@rich wolf You’re a smart boi

@rich wolf You have growth mindset? 👌 👍

I wish you a good life my friend

Hope you do well in school and everything 👍

[barren tide]

Oh I just mean when I was in highschool, all math was super easy, but now that I'm in college, it seems like a foreign language, and I've never really struggled until now.

[summer agate]

Hey guys

[barren tide]

At least I have the internet tho lol.

"Always surround yourself with people smarter than you." -Snoop Dog

Welcome

Plus like no one is teaching me so I guess that's like the biggest issue I'm facing.

[rich wolf]

That means you have some sort of gap in your knowledge you need to address

Also yeah not having a teacher kind of sucks

[barren tide]

I got off my ADHD medication in highschool, so like it's not as easy to focus, but I won't let that stop me

Wait are there tutors on here?

[rotund rain]

https://cdn.discordapp.com/attachments/496784958430380033/709300221766664202/QVrvbkAaTk7ibily4bS5djYZLPI7wvN7gpPsd0wzDfSCGRoY08m5RXc8FTVNgEW8cU8fSznYof0Vu3Y4XU6AOAMWmxOqX7reo23f.png how do i do this

[silent plank]

what's the angle sum on a line?

[rotund rain]

https://cdn.discordapp.com/attachments/620562644826324993/709303251396329552/ERhmk8vq0d7khiYHZwxgRJq8bA_5lYvtmcvLvENzZ77jDSE02puXcdqJEAimJW0DAR6pzpZPgvnGvU8z3nEKJHvDuuzeMAOnheo7.png

What about dis

[silent plank]

what's the angle sum on a line?

same idea

[upper karma]

I already told him in #help-1 ...

Well check the time @upper karma

yep

He possibly didn’t understand what Ramonav was saying

So he went there

And then u answered it

[rich wolf]

@upper karma think about it

Whats the definition of tangent

[misty nimbus]

Hello, I'm using geogebra, do you recommand anything else or is it the good tool? (am using the online tool)

I'm trying to get the radius of a circle but I can't get it

[upper karma]

@misty nimbus post the question

do they give you the circumference, area, what?

[misty nimbus]

Sorry so I've finally found out

I just have to create a radius segment. Then I enter the Length function gives me the length of the segment

[upper karma]

so you're good then?

you found your answer

[upper karma]

i'm not sure how to do this

Doesnt load

[blissful gorge]

@upper karma I believe this is in regard to the quadrants of the coordinate plane. Starting from top right and moving counter clockwise they are I, II, III, and IV. tan = sin/cos. So we want tan to be positive. (x,y) is respective to (sin, cos). From this we know that tan will be positive if sin & cos are both positive or both negative. That should narrow you down to 2 quadrants. Then think of what csc is the reciprocal of and you should be on your way.

If that is wrong then I am an idiot and please don't mind me....

[upper karma]

I'm pretty sure it comes out to Quadrant 1 right?

[blissful gorge]

so quadrant I is where both sin&cos are positive and quadrant III is where they are both negative. csc is the reciprocal of sin, so we want sin (our y value) to be positive

that said, yep, you are correct

[upper karma]

i'm pretty sure i have the answers

i'd like to check them with you first though

sin=40/41 of course

cos=9/41

tan=40/9

cot=9/40

sec=41/9

Wth are you saying

csc=41/40

huh?

this is how ur supposed to put in the answers

What are these trigs?

they're based off of the problem i sent

[blissful gorge]

checking

[upper karma]

@upper karma how did you got cos tho

Show work

i worked on with a friend over discord

I mean, how did you got cos value

i kinda just did what he said and put it into a calculator lmfao

...

Im out

i'm just trying to make sure that it came out right

[blissful gorge]

the numbers look right, just be cautious of what quadrant you are in

[calm kite]

how u write this again

[blissful gorge]

@calm kite what is x^2 + 4x + 4 in factored form?

[silent plank]

factor theorem

[upper karma]

@blissful gorge

this is the last problem on the homework assignment

@upper karma use the sine definition

ok it is sin

so it would be sin(32)=14/c?

@upper karma

wait no

so it would be sin(32)=14/c?
@upper karma yeah lol why not

because then it asks something else

it wont let me snip the rest because it scrolls down every time i try to snip

You have to look at the question lol

It says: "Which of the following ratios can be used to find c?"

We used sine's definition/ratio to get c, so it would be the F. option

Its just a question apart

To check if you did it randomly or with trig lol

alright thank you

Np

[ionic bluff]

,calc sin(32)/14

[somber coyote]

Result:

0.039387620088692

[ionic bluff]

,calc 14/sin(32)

[somber coyote]

Result:

25.388688063615

[ionic bluff]

@upper karma

[upper karma]

i got it

[ionic bluff]

ur welcome

[rich wolf]

[upper karma]

Someone help pls

use pythagoreas theorem

Use trig functions too

Find all three angles

[nocturne musk]

hi

Can someone help me understand how to do thia?

This*

<@&286206848099549185>

[glacial haven]

2pi r h

[nocturne musk]

Thank you!

So it would be 2 pi 2 (16)? @glacial haven

Alright thanks!

[glacial haven]

yes

[nimble agate]

???

[regal shell]

Law of sin to find AC

C=180-31-18

Then either law of sine or law of cosine to find AB

[nimble agate]

Yeah

what is m<R?

[regal shell]

Use law of sines

[vital bough]

can someone please explain this to me?

i feel like im overthinking it

[next epoch]

C I think..

[vital bough]

Could you explain how you got to that?

[next epoch]

Bh is the area for a rectangle

So if you tripped each of those numbers

Tripled

Itd be 9 and 6

[vital bough]

Ah okay thank you @next epoch

[regal shell]

C

[next epoch]

Np

[regal shell]

Lol

[next epoch]

Aha any one okay with a dumb question real quick

[regal shell]

Sure

[next epoch]

Let me get a ss brb

im probably overthinking this smh

not an exam btw, it's just a stupid edgenuity question

[regal shell]

I might be dumb cuz I suck at geometry but I got the a, b, c and I don’t remember what lateral faces are

[next epoch]

Alr

[upper karma]

@next epoch What

[next epoch]

@upper karma hm

[full ruin]

help

[upper karma]

help how?

[full ruin]

i don’t want the answer just need the radius of the semi circles

[shut granite]

It's so vague.

[upper karma]

^

[full ruin]

wdym?

i understand it

[upper karma]

theres like no info to solve this question

[full ruin]

wait no i don’t

[upper karma]

lel

[full ruin]

i’m thinking the diameter is the like diameter of the circle

and the height is the 10ft shown

but idc

idk*

[upper karma]

what

oh, well if thats the case then the radius is half of the diameter, so 5

[full ruin]

yeah

but that’s a weird question

[upper karma]

it thats not the case, then you cant answer the question

Who wrote this question? I’m going to put them into a propane tank and ask them to find the volume, because that’s Just about the same difficulty here.

[full ruin]

LMAOO

[shut granite]

Right

[upper karma]

two hemispheres = one sphere

so you have one sphere, with a diameter of 10 (radius 5)

and one cylinder, also with a diameter of 10 (radius 5)

google the formula for volume of cylinder and volume of sphere

add them

and there you go

Noice

yes but the diameter of the cylinder isn’t necessarily the diameter of the sphere, from the looks of it... or am I wrong?

Nah it is

Hmm

if you cut a sphere in half...

Becomes 2 hemispheres

and stick that half to a cylinder with the same diameter

Oh I see

yes, the radius of the cylinder is the radius of the hemisphere(s)

ahh

Makes sense now

Thx

the diameter of the half circles ain't even looking close to 10 ft compared to the height of the cylinder

dafuq

Deal with it

Nothing is drawn to scale

sometimes they do that on purpose

just to throw you off

deal with me xi

[full ruin]

please don’t judge me LMAOO

but is it pythagorean?

[upper karma]

this is a judge free zone

[full ruin]

i’m pretty sure it’s pythagorean theorem

but i’m being slow

[upper karma]

pythagorean theorem seems like the most correct answer here

Help

Find X

[full ruin]

i-

have fun

[upper karma]

soft tingz

[full ruin]

i’m gonna cry i hate my teacher

[upper karma]

Here's a hint.

Look at the 1st and 4th line.

Then 2nd and 5th line.

Then 3rd and 6th line.

This ordering sucks ass tbh.

[full ruin]

i-

i’m already lost but i’ll try

[upper karma]

Also, A n B and A n C suggests that this is something that BOTH A and C has.

Thus, everything in the first two lines is something that A contains.

Which, then, you rule out the first two answers.

[full ruin]

oh

thanks

so it’s the last one

cause a also contains 4

[upper karma]

How do you know that?

Look at the 4th-6th line.

[full ruin]

oh ur right

hmm

[upper karma]

4th and 6th mb*

If something is in

A u B but not in B u C, then that "something" is also part of A.

[full ruin]

so the answer is the 3rd one

LMAO?

[acoustic jungle]

A doesn't have 4 because B and C have 4 and A n B doesn't include 4

[upper karma]

A = {1, 2}
B = {3, 4}
C = {5}

A U B = {1, 2, 3, 4}
B U C = {3, 4, 5}

Everything in A U B that isn't in B U C is part of A.

Yeah 3rd.

[full ruin]

you guys should be my teachers

like .

[upper karma]

I'm sophomore in HS. Far from being any teacher.

[full ruin]

oh LMAO

[acoustic jungle]

@upper karma I don't know how to do it. Where did you find it online?

[upper karma]

Random image online I found

[upper karma]

bruh

[winged hill]

but why is it so low quality

and doesn’t it say x is a right angle

i googled up the picture

and sorry i’m sure that isn’t a right angle it’s just low quality

it’s pretty tricky tho

[thorn grotto]

Show that the lines ac and bc are perpendicular

Aka slope ac = -1/(slope of bc)

Aka (slope ac)(slope bc) = -1

@thin pendant

[brazen roost]

Just Google circle theorems

Here

What are ZT and XY

They're chords

Google theorems relating to chords

[rich wolf]

@upper karma V is the center right?

What if I told you

That YX and ZT are the same length

[brazen roost]

Hey

[rich wolf]

https://www.onlinemathlearning.com/chords-circle.html

[brazen roost]

If you can prove it, you can name the theorem

[rich wolf]

Pretty sure its called equidistant chords theorem

[brazen roost]

Psh

That's boring

Call it the Golden Legs theorem

[upper karma]

Call it the Golden Legs theorem
@brazen roost 🤣

[upper karma]

https://discordapp.com/channels/268882317391429632/326138757474680852/709596326270337054
where does sin(x+20) come from?

[upper karma]

Law of sines lol

OC/sin(x) = CD/sin(x+20)

where's the x+20 coming from?

it should be CD/sin(angle COD)

Hm

O+x+20=180 maybe

But no

Hm yeah its incorrect i think too @upper karma

Or idk what he did

[winged pagoda]

So i know the answer 9, but can someone tell me why this is wrong 10(x+6)=144?

shouldn't this also give me the answer?

[dark sparrow]

10(x+6) is not the area of this shape.