#geometry-and-trigonometry | Mathematics | Page 320

mossy valve Oct 28, 2020, 12:53 AM

#

i dont understand why theres an arrow

#

is J 18 degrees?

#

is j also 6x+18 im so confused

upper karma Oct 28, 2020, 12:57 AM

#

@mossy valve the angle J is (6x+18)°

#

and since it's isosceles triangle, then their basis will be equal to each other

#

6x+18 = 8x

mossy valve Oct 28, 2020, 12:58 AM

#

that makes sense since both of the other sides are equal

grizzled lantern Oct 28, 2020, 2:15 AM

#

they are equal

silent plank Oct 28, 2020, 2:15 AM

#

where did you get stucj?

grizzled lantern Oct 28, 2020, 2:15 AM

#

AM and MB are equal

#

so AM+MB = AB

silent plank Oct 28, 2020, 2:17 AM

#

what exactly were you solving?

grizzled lantern Oct 28, 2020, 2:17 AM

#

Should get 16x+32=x^2+7x+10

chrome crater Oct 28, 2020, 2:17 AM

#

OH YEAH

#

BRO

grizzled lantern Oct 28, 2020, 2:17 AM

#

then we subtract the left side and get x^2-9x-22

chrome crater Oct 28, 2020, 2:17 AM

#

thanks

#

so smart 🙂

#

i feel dumb

#

thank you

#

both of you

grizzled lantern Oct 28, 2020, 2:18 AM

#

no problem

magic wasp Oct 28, 2020, 3:00 AM

#

hey guys

#

!av

#

never mind lmao

#

hopin yall are doin good

#

peace out

#

:P

next jackal Oct 28, 2020, 4:29 AM

#

I didn't understand why we had to take a line which passes through 0,1 and r,0 . Why did we take an X intercept in the first place?

📎 Screenshot_328.png

#

📎 Screenshot_329.png

lean citrus Oct 28, 2020, 4:37 AM

#

I didn't understand why we had to take a line which passes through 0,1 and r,0 . Why did we take an X intercept in the first place?
@next jackal I don't understand your question; a straight line with a y intercept of 1 should have a an x-intercept

#

Except for the line y=1 which does not intersect the circle at any other point

#

If your question is "where did this construction come from" I would recommend you read the whole proof

#

The aim is to find a paramteric form of Pythagorean triples

next jackal Oct 28, 2020, 4:43 AM

#

The aim is to find a paramteric form of Pythagorean triples
@lean citrus didn't we already do that by showing (a/c)^2+(b/c)^2=1?

lean citrus Oct 28, 2020, 4:43 AM

#

No see that's not a parametric form

next jackal Oct 28, 2020, 4:44 AM

#

I didn't get what the number r signifies here, is it just an x intercept? or is it like an input for a side of triangle?

lean citrus Oct 28, 2020, 4:44 AM

#

Yes when you construct the line, you consider r to be the x intercept

#

Now the goal is to express both x and y in terms of r

next jackal Oct 28, 2020, 4:45 AM

#

why in terms of r?

lean citrus Oct 28, 2020, 4:45 AM

#

Well that's what we are aiming to do!

#

Define all possible Pythagorean triples

#

What we have done is derived a general form of the Pythagorean triple in terms of a single paramter

#

Lemme give you a simple example

#

"all odd numbers are of the form 2n+1, where n is an integer"

#

So you define all odd numbers in terms of a single parameter n. By considering different values of n we are to get different odd numbers

#

In a similar way, the goal here is to get a form for the Pythagorean triple in the form of a single parameter

#

x^2 + y^2 = 1 is a bit difficult to work with

#

Coz there are 2 variables

#

i feel like I'm not being able to clear your doubt

upper karma Oct 28, 2020, 4:59 AM

#

r is a parameter which can take any real value ,

#

Put value of r

#

You'll find different rational coordinates

lean citrus Oct 28, 2020, 4:59 AM

#

Yeah I wanted to get that through to researcher

upper karma Oct 28, 2020, 5:00 AM

#

I understood that, It was meant for researcher haha

lean citrus Oct 28, 2020, 5:00 AM

#

I think I'm not very good at explaining things

next jackal Oct 28, 2020, 5:00 AM

#

I want to know how we're taking more than the radius of the unit circle

hollow raven Oct 28, 2020, 5:01 AM

#

What is the problem?

next jackal Oct 28, 2020, 5:01 AM

#

like any rational number, more than unit circle, how're we getting a solution for a pythagorean triple?

#

I'll try to go through the whole proof again

lean citrus Oct 28, 2020, 5:02 AM

#

I'll try to go through the whole proof again
@next jackal good idea

upper karma Oct 28, 2020, 5:02 AM

#

I mean. When you square and add those coordinates in terms of r, you get 1

#

no value of x and y will be >1 for any r

next jackal Oct 28, 2020, 5:03 AM

#

it just looks like as r increases, value of y intercepting the circle increases too, but doesn't cross 1, if you take from 1,0 .

#

how do both increasing values add to 1?

lean citrus Oct 28, 2020, 5:04 AM

#

X decreases as r increases

next jackal Oct 28, 2020, 5:04 AM

#

what? x is r right?

lean citrus Oct 28, 2020, 5:04 AM

#

No?

upper karma Oct 28, 2020, 5:04 AM

#

no

lean citrus Oct 28, 2020, 5:04 AM

#

See see see see see

upper karma Oct 28, 2020, 5:04 AM

#

x is (2r)/(r^2 +1)

lean citrus Oct 28, 2020, 5:05 AM

#

r is a term we have assumed so that we can express both x and y in terms of r

next jackal Oct 28, 2020, 5:05 AM

#

but we take x as r in the point r,0?

#

r is a term we have assumed so that we can express both x and y in terms of r
@lean citrus ohhhh

#

but....

upper karma Oct 28, 2020, 5:06 AM

#

you derive them in those terms

lean citrus Oct 28, 2020, 5:06 AM

#

but we take x as r in the point r,0?
@next jackal no... That point is not in question... We are concerned about the point of intersection of the circle, and the line between 0,1 and r,0

#

This particular construction allows us to express both x and y in terms of a single r

next jackal Oct 28, 2020, 5:06 AM

#

ohh okay

#

umm......

upper karma Oct 28, 2020, 5:07 AM

#

x= 2r/(r^2+1) , y= (r^2 -1)/(r^2+1) are coordinates of a point on circle, just expressed in terms of r

#

To repeat what Ragnarok said

next jackal Oct 28, 2020, 5:10 AM

#

okayy

#

thanks

upper karma Oct 28, 2020, 5:39 AM

#

PROVE: $\frac{\left(tanx+secx-1\right)}{tanx-secx+1}=\frac{\left(sinx+1\right)}{cosx}$

somber coyoteBOT Oct 28, 2020, 5:39 AM

#

A pOtaTo:

📎 763343571121733642.png

upper karma Oct 28, 2020, 5:40 AM

#

could somebody help me with this?

silent plank Oct 28, 2020, 5:45 AM

#

what have you tried

lean citrus Oct 28, 2020, 5:47 AM

#

could somebody help me with this?
@upper karma do you need a solution or a hint?

#

I like to prove such identities by simply taking the expression on the right, and multiplying and dividing it to the expression on the left

#

It's a bad method, but usually works

upper karma Oct 28, 2020, 5:49 AM

#

I did a little bit, wait a min. I'll send you

hollow raven Oct 28, 2020, 5:50 AM

#

Yeah see whats on the right, and try to make it fit the left side, even if it doesnt look conventional

#

thats what i personally do too

silent plank Oct 28, 2020, 5:50 AM

#

you can use that idea like this:

#

LHS $= \underbrace{\frac{\tan(x)+\sec(x)-1}{\tan(x)-\sec(x)+1} \cdot \frac{\cos(x)}{\sin(x)+1}}_{\text{problem reduces to showin that this is 1}} \cdot \frac{\sin(x)+1}{\cos(x)}$

somber coyoteBOT Oct 28, 2020, 5:50 AM

#

ramonov:

📎 220517274845577216.png

silent plank Oct 28, 2020, 5:50 AM

#

which would be considered more acceptable than mixing the two sides directly

lean citrus Oct 28, 2020, 5:51 AM

#

Thanks @silent plank for translating t through Texit

upper karma Oct 28, 2020, 5:51 AM

#

📎 image0.jpg

lean citrus Oct 28, 2020, 5:52 AM

#

I mean, it's a good idea to break down stuff into sin and cos

#

Not much helpful here tho

#

At least I can't intuitively rearrange this into the expression required

#

📎 IMG_20201028_111706.jpg

#

@silent plank mentioned this method

upper karma Oct 28, 2020, 5:54 AM

#

ohh i am getting it. Thank you

lean citrus Oct 28, 2020, 5:59 AM

#

I'm sorry my handwriting is extremely bad

hollow raven Oct 28, 2020, 6:00 AM

#

@lean citrus 1000000 times better than mine

lean citrus Oct 28, 2020, 6:00 AM

#

😅

hollow raven Oct 28, 2020, 6:01 AM

#

do you write in one flow?

#

like it seems like cursive without the joining lines

#

its really neat

#

meanwhile my sin looks like sn^2 x when i write it

#

cause i write it so quickly and mercilessly 😅

lean citrus Oct 28, 2020, 7:11 AM

#

lol

next jackal Oct 28, 2020, 7:38 AM

#

I found my mistake. @lean citrus I was assuming all the points on a unit circle were rational coordinates.

dark sparrow Oct 28, 2020, 7:38 AM

#

(1/sqrt(2), 1/sqrt(2)) would like to have a word with you

next jackal Oct 28, 2020, 7:39 AM

#

yes I did something similar to contradict myself

#

but thanks

lean citrus Oct 28, 2020, 7:40 AM

#

I found my mistake. @lean citrus I was assuming all the points on a unit circle were rational coordinates.
@next jackal lol

next jackal Oct 28, 2020, 7:41 AM

#

yeah that's why I didn't get why we were doing this derivation

#

So if one coordinate is rational, does it mean the other has to be rational too?

dark sparrow Oct 28, 2020, 7:48 AM

#

on the unit circle? no

#

counterexample: (1/2, sqrt(3)/2)

next jackal Oct 28, 2020, 7:51 AM

#

okay thanks... I couldn't find one. BTW why does this book say something that seems like it's saying that?

#

📎 Screenshot_361.png

dark sparrow Oct 28, 2020, 7:51 AM

#

does it?

#

the book simply says the construction starts with one rational point and generates more points which are distinct from the first but are still rational

next jackal Oct 28, 2020, 7:52 AM

#

okayy....

dark sparrow Oct 28, 2020, 7:53 AM

#

it does not say that a point on the unit circle with one rational coordinate automatically has the other coordinate also rational.

upper karma Oct 28, 2020, 7:56 AM

#

'Many more', not 'only'

next jackal Oct 28, 2020, 7:57 AM

#

okay got it

next jackal Oct 28, 2020, 10:01 AM

#

while I understand that statement on the graph, I don't understand how that fits into the equation....

📎 Screenshot_362.png

dark sparrow Oct 28, 2020, 10:11 AM

#

$\frac{2r}{r^2 + 1} = \frac{2}{r + \frac{1}{r}}$

somber coyoteBOT Oct 28, 2020, 10:11 AM

#

Ann:

📎 183668144404037632.png

dark sparrow Oct 28, 2020, 10:12 AM

#

as r goes to positive or negative infinity, the denominator does the same

#

and $\frac{r^2 - 1}{r^2 + 1} = \frac{1 - \frac{1}{r^2}}{1 + \frac{1}{r^2}}$, so as $r$ grows large $\frac{1}{r^2}$ approaches 0 and so that fraction approaches 1

somber coyoteBOT Oct 28, 2020, 10:13 AM

#

Ann:

📎 183668144404037632.png

next jackal Oct 28, 2020, 10:16 AM

#

Makes sense

upper karma Oct 28, 2020, 10:52 AM

#

solve for x

📎 122475653_3363509627097258_942078861704958219_n.jpg

next jackal Oct 28, 2020, 10:54 AM

#

https://www.youtube.com/watch?v=SqaUj1rzHq4@round plume

YouTube

Michael Penn

A relaxing geometry problem.

We look at a solution to a nice and quick geometry problem. Inspired by this:

https://math.stackexchange.com/questions/3360970/how-can-i-find-the-side-length-two-squares-inside-an-equilateral-triangle?rq=1

Please Subscribe: https://www.youtube.com/michaelpennmath?sub_confirm...

▶ Play video

upper karma Oct 28, 2020, 10:56 AM

#

thanks man

#

wait a second...

#

He mentioned "Angle-Angle-Angle theorem"

#

but that is technically impossible

next jackal Oct 28, 2020, 10:59 AM

#

??????

#

what's impossible in that?

upper karma Oct 28, 2020, 11:00 AM

#

i mean an angle does not change if you increase the side length

next jackal Oct 28, 2020, 11:00 AM

#

length of all the sides. not just one.

#

So both those triangles in the corner are similar.

upper karma Oct 28, 2020, 11:01 AM

#

oh

#

yeah

next jackal Oct 28, 2020, 11:01 AM

#

catthumbsup

upper karma Oct 28, 2020, 11:02 AM

#

Pi_thonk

lean citrus Oct 28, 2020, 11:23 AM

#

lol we have 3b1b emotes?

lunar sand Oct 28, 2020, 1:09 PM

#

is this right

📎 unknown.png

mossy plinth Oct 28, 2020, 1:10 PM

#

https://cdn.discordapp.com/attachments/718907419580956674/770991829046788116/unknown.png

untold coyote Oct 28, 2020, 1:11 PM

#

I need help with easy math guys

#

📎 unknown.png

#

what is the mapping rule

tardy fulcrum Oct 28, 2020, 2:27 PM

#

guys i need help

chrome crater Oct 28, 2020, 2:30 PM

#

ok

tardy fulcrum Oct 28, 2020, 2:31 PM

#

with geometry

chrome crater Oct 28, 2020, 2:31 PM

#

alright

#

ok then

tardy fulcrum Oct 28, 2020, 2:45 PM

#

ok well then can someone else help 💀

novel ginkgo Oct 28, 2020, 3:39 PM

#

how would i find the missing angles. im so confused sorry

📎 unknown.png

grizzled lantern Oct 28, 2020, 3:45 PM

#

oh boy

upper karma Oct 28, 2020, 4:28 PM

#

Can anyone help me with these two problems, I've been sick for a week and I have no idea what I'm doing.

📎 ayo.PNG

dire wasp Oct 28, 2020, 4:48 PM

#

I’m confused

📎 image0.png

upper karma Oct 28, 2020, 4:50 PM

#

📎 unknown.png

#

confused on 18 and 20

silent plank Oct 28, 2020, 5:02 PM

#

what are you confused about?

upper karma Oct 28, 2020, 5:19 PM

#

Hi, if I have a point O on acute triangle ABC, such that angles OAB = BCO and ABO = OCA how can I prove that line AO is normal to BC?

glass rover Oct 28, 2020, 5:44 PM

#

yo can someone help me out with this one

#

what is the smallest positive integer value of k such that sin(theta) is negative. (theta is 56 degrees)

#

@silent plank if you understand

grizzled lantern Oct 28, 2020, 5:50 PM

#

@upper karma Gotta know your angle theorems

upper karma Oct 28, 2020, 5:51 PM

#

i figured it out

#

thanks anyway

grizzled lantern Oct 28, 2020, 5:51 PM

#

gj

upper karma Oct 28, 2020, 5:51 PM

#

ty

glass rover Oct 28, 2020, 5:51 PM

#

it would be a linear pair right

#

like 110 for the first one

#

70 = 2y

grizzled lantern Oct 28, 2020, 5:52 PM

#

i think its 2y+70 = 180 @glass rover

glass rover Oct 28, 2020, 5:53 PM

#

or 110 = 180-2y

#

would be the same answer either way no?

#

oh wait no

#

70=180-2y

#

i took this math a year ago u got me confused lmao

grizzled lantern Oct 28, 2020, 5:54 PM

#

wdym?

glass rover Oct 28, 2020, 5:54 PM

#

since its a linear pair it would be 110+2y = 180

grizzled lantern Oct 28, 2020, 5:54 PM

#

its a supplementary angle so two angles = 180

glass rover Oct 28, 2020, 5:54 PM

#

oh thats what i meant

#

but wouldnt 70=2y since opposite angles?

grizzled lantern Oct 28, 2020, 5:56 PM

#

I think you are right

#

I was just going off internal angles and angles whos vertices touch

glass rover Oct 28, 2020, 5:57 PM

#

yeah im confusing mysekf

#

have fun with geometry...

upper karma Oct 28, 2020, 6:04 PM

#

@upper karma by normal do you mean perpendicular

#

ye

#

well i'm not sure how you can prove this but.
triangle ABO is equal to triangle ACO

#

How so?

pseudo pumice Oct 28, 2020, 6:22 PM

#

anyone able to help me essentially walk me through this college trig course? or have the time to do so? Looking for like tutoring kind of thing idk how that works or if it is allowed etc. very behind, have no clue what's going on....class is online...sigh

upper karma Oct 28, 2020, 6:25 PM

#

you need help with sin, cos, tan?

#

@upper karma I'm considering that the statement doesn't necessarily need to be true

#

oh yea and wuf idk how to prove it, i just maed a triagnle with those angles specified

#

Because I was playing around and I got that angles between AO and BC (k and k') are in this relation: k - k' = a2 - b2, where a2 = BAC - BAO and b2 = angle ABC - angle OBA

#

So they are perpendicular if a2 = b2

pseudo pumice Oct 28, 2020, 6:35 PM

#

That's part of it I'm not too sure where that stops and if I get into something different but for the most part I'm seeing a lot of that

balmy egret Oct 28, 2020, 6:41 PM

#

📎 20201028_144126.jpg

#

Can anyone help me with reasoning

upper karma Oct 28, 2020, 7:50 PM

#

@upper karma solved it

#

Just have to do bunch of similarity stuff

#

Then prove a2 = b2

#

And from there prove k = k'

#

And btw ACO and BCO are not similar unfortunately

#

cuz a1 != a2

#

but AB'O and BOA' are

surreal vector Oct 28, 2020, 8:26 PM

#

is there a way of solving for x given just these details?

📎 unknown.png

umbral snow Oct 28, 2020, 8:41 PM

#

y is easy to find, then x is a cosine law away

#

Wait, but that bottom triangle is impossible haha

#

@surreal vector

#

Yeah both those triangles can't happen

surreal vector Oct 28, 2020, 8:44 PM

#

oh my bad

#

I tried to solve it using cosine and sin

#

but I got that x is shorter than 30 which shouldnt be possible

#

why cant they happen tho?

umbral snow Oct 28, 2020, 8:59 PM

#

The bottom one is a 45 degree angle, which means both sides should be the same

#

Similar logic on top @surreal vector

surreal vector Oct 28, 2020, 9:15 PM

#

The bottom one? where do we have a 45 degrees angle here?

#

oh isnt the left side is just 25 - 13?

#

Like this, right?

📎 unknown.png

#

which leaves me with x = 32.310

#

x^2 = 12^2 + 30^2

astral maple Oct 28, 2020, 11:01 PM

#

📎 Screenshot_20201029-055552_Samsung_Notes.jpg

tardy fulcrum Oct 28, 2020, 11:53 PM

#

📎 image0.jpg

#

anyone ?

novel berry Oct 29, 2020, 12:03 AM

#

what is the issue?

#

Do you see any nice property of the triangle?

#

@tardy fulcrum

tardy fulcrum Oct 29, 2020, 12:05 AM

#

nice property? LMAO what is that?

novel berry Oct 29, 2020, 12:05 AM

#

a property is a feature that the triangle would have

#

do you know what it is called when a triangle have two sides of the same length?

tardy fulcrum Oct 29, 2020, 12:06 AM

#

isosceles ?

novel berry Oct 29, 2020, 12:06 AM

#

Yes

#

What do you know about these triangles?

tardy fulcrum Oct 29, 2020, 12:08 AM

#

sum is always 180° , has two equal angles , two equal sides obviously

novel berry Oct 29, 2020, 12:09 AM

#

sum is always 180 degres goes for all triangles! This an important part of solving it yes.

#

And yes https://dr282zn36sxxg.cloudfront.net/datastreams/f-d%3A1c54f3ea2a10f06e79bb2058aa20098632e3e76423653d63c97b9379%2BIMAGE_TINY%2BIMAGE_TINY.1

tardy fulcrum Oct 29, 2020, 12:09 AM

#

i’m just confused on what i’m supposed to do because we literally just learned this

novel berry Oct 29, 2020, 12:10 AM

#

okay

#

so the triangle is isosceles then the two angels at the bottom are the same

#

Can you see the image I sent?

tardy fulcrum Oct 29, 2020, 12:11 AM

#

yes

novel berry Oct 29, 2020, 12:11 AM

#

okay so the third unknown angel what must it be?

tardy fulcrum Oct 29, 2020, 12:14 AM

#

well in the notes we took my teacher takes two angles and puts the like equation and equals it to another one and breaks it down to find x- i feel very dumb rn so sorry if i’m confusing -

novel berry Oct 29, 2020, 12:14 AM

#

don't worry it's fine, but I am asking about the top angel, do you know what it is in terms of x?

#

looking at the image I sent with the isoscele triangle

#

It should be the same as either the bottom left angle or the right angel. Which one of them and why?

tardy fulcrum Oct 29, 2020, 12:17 AM

#

well i thought the left and right angle were equal and the top is different than the other two -

novel berry Oct 29, 2020, 12:17 AM

#

it says that they are different the bottom left is 2x+11 degres and the bottom right is x-2 degrees

#

So the angels opposite of the two sides that have the same length are the angels who are equal

#

look at the image I sent to confirm that you see this

upper karma Oct 29, 2020, 12:34 AM

#

📎 image0.png

pearl lava Oct 29, 2020, 12:42 AM

#

The pentagon to the right has an area of 7 square centimeters. If you enlarge the figure by a scale factor of 3, what is the new area?

📎 0b14f08cabba60f300cb73ca313ac3a0.png

#

is the answer 63?

novel berry Oct 29, 2020, 12:53 AM

#

How did you get to this answer?

#

@pearl lava

pearl lava Oct 29, 2020, 12:56 AM

#

I used a square. If the sides of the square is 2, the area is 4. If I triple the sides, the area is 36, which is 9 times. Then I tried it with a circle. If the radius is 2, the area is 4pi. If I triple the radius, the area is 36pi

#

so I did 9*7

novel berry Oct 29, 2020, 12:57 AM

#

Cool

#

So the way I see is in the new figure all lengths are increased by a factor of 3 like you just said

#

since area is length squared this factor 3 gets squared

#

so 7cm^2 in the first image becomes 7* (3cm)^2=7*9cm^2

#

Nice way of trying some shapes you knew 👍

pearl lava Oct 29, 2020, 12:58 AM

#

thanks

meager pendant Oct 29, 2020, 1:05 AM

#

📎 image0.png

#

I need some help with this question

cinder nacelle Oct 29, 2020, 2:15 AM

#

How do I do proofs good

winged vine Oct 29, 2020, 8:45 AM

#

hey

#

How do I do proofs good
@cinder nacelle by doing more proofs

upper karma Oct 29, 2020, 9:35 AM

#

$prove:\frac{\left(sin\theta :+cos\theta :+1\right)}{sin\theta :+cos\theta :-1}-:\frac{\left(1+sin\theta :-cos\theta \right)}{1-sin\theta :+cos\theta :}=2\left(1+cosec\theta \right)$

somber coyoteBOT Oct 29, 2020, 9:35 AM

#

A pOtaTo:

📎 763343571121733642.png

upper karma Oct 29, 2020, 9:35 AM

#

could somebody help me with this?

#

What have you tried

drifting orchid Oct 29, 2020, 3:32 PM

#

can someone solve b part

📎 unknown.png

#

and explain pls

loud flare Oct 29, 2020, 7:28 PM

#

📎 image0.jpg

long shale Oct 29, 2020, 7:47 PM

#

📎 unknown.png

#

On the triangle ABC, AB=AC. ∠ BAC = 120°. Point D lies on BC and point E lies on AC. ∠ BAD = 80° and ∠ ABE = 20°
Find ∠ ADE

hidden crescent Oct 29, 2020, 7:49 PM

#

ab = ac therefore angle b is 60 degrees

#

note isoscles

long shale Oct 29, 2020, 7:50 PM

#

isnt angle ABC 30?

#

since 180 - 120 = 60, 60 * 1/2 = 30

#

i got all the angles but BED, ADE, DEC, CDE

#

this is what i got so far

#

📎 unknown.png

hidden crescent Oct 29, 2020, 8:01 PM

#

you can do something with triangle ADC

#

to find angle ADC

#

should bring you closer to your answer

long shale Oct 29, 2020, 8:01 PM

#

ADC is 110, BEC is 140

upper karma Oct 29, 2020, 10:24 PM

#

How do I perform rigid transformations on a triangle without a coordinate graph? These transformations are translations, rotations, and reflections. I know how they work, but I don’t know how to construct them. These transformations have to be conducted using a straightedge, compass, and protractor.

pearl lava Oct 30, 2020, 1:24 AM

#

how would I start to do this?

📎 221c2608ff67c9622de872746c31af45.png

silent plank Oct 30, 2020, 1:26 AM

#

image is blurry, are both those angles labelled as theta?

pearl lava Oct 30, 2020, 1:26 AM

#

yes

silent plank Oct 30, 2020, 1:26 AM

#

angle bisector theorem

pearl lava Oct 30, 2020, 1:27 AM

#

thanks

#

is the answer 6?

silent plank Oct 30, 2020, 1:33 AM

#

yes

upper karma Oct 30, 2020, 1:48 AM

#

📎 unknown.png

#

I wrote this problem and solved it but if anyone likes, could you plz also solve it so I can see if I did the problem correctly? Thx

#

(also I put the problem in this channel since the problem's basically just a ton of applied trig)

ripe osprey Oct 30, 2020, 2:40 AM

#

thonkzoom what's the point of making it three blocks instead of two

#

aside from one of the parts

upper karma Oct 30, 2020, 3:04 AM

#

one of the parts

#

for tenshun

quiet mason Oct 30, 2020, 7:34 AM

#

@upper karma send your work and people can review

rocky birch Oct 30, 2020, 8:19 AM

#

is there a way to prove that the max value of a*sin(x)+b*cos(x) is sqrt(a^2+b^2) without using calculus?

dark sparrow Oct 30, 2020, 8:24 AM

#

yeah using the pythagorean formula or the R-formula or whatever else it's called nowadays

#

https://brilliant.org/wiki/trigonometric-r-method/

Trigonometric R method | Brilliant Math & Science Wiki

The trigonometric R method is a method of rewriting a weighted sum of sines and cosines as a single instance of sine (or cosine). This allows for easier analysis in many cases, as a single instance of a basic trigonometric function is often easier to work with than multiple ar...

rocky birch Oct 30, 2020, 8:25 AM

#

hmm, i didn't know about this

#

thanks so much!

tacit cedar Oct 30, 2020, 11:45 AM

#

Hi, guys. I have a question. Given the volume of the pyramid is 500cm^3, solve for the height. Based on the formula V = 1/3bh, is this question even solvable or is the answer just h = 1500/b.

upper karma Oct 30, 2020, 2:14 PM

#

Hi guys! Can help me?

#

The following figure represents the flat pattern of a solid. The volume of this solid is:

a) 20 (Root of three)
b) 75
c) 50
d) 100
e) 100 (Root of Three)

Sorry if there is something wrong written by google

#

📎 1.jpg

novel berry Oct 30, 2020, 2:29 PM

#

@upper karma base area times height! Also don't multi post.

upper karma Oct 30, 2020, 2:29 PM

#

Ok! Thank you!

@upper karma base area times height! Also don't multi post.
@novel berry

#

Thank you

novel berry Oct 30, 2020, 2:30 PM

#

Do you see the shape it forms?

#

if you fold it

dry pulsar Oct 30, 2020, 3:27 PM

#

Could someone help me with this?

📎 image0.png

#

I have no clue how to start it

dusky surge Oct 30, 2020, 3:29 PM

#

First hint: what do the red lines represent?

dry pulsar Oct 30, 2020, 3:30 PM

#

90 degrees

#

@dusky surge

dusky surge Oct 30, 2020, 3:30 PM

#

Second hint: what is the sum of all interior angles of any triangle?

dry pulsar Oct 30, 2020, 3:31 PM

#

180

fast tide Oct 30, 2020, 3:31 PM

#

write it down as an equation = 180

dry pulsar Oct 30, 2020, 3:31 PM

#

I tried but I did something wrong

fast tide Oct 30, 2020, 3:32 PM

#

2x+13+5x+90=180

#

2x+13+5x=90

#

7x+13=90

#

7x=77

hidden crescent Oct 30, 2020, 3:33 PM

#

One angle is a right angle

fast tide Oct 30, 2020, 3:33 PM

#

x=11

hidden crescent Oct 30, 2020, 3:33 PM

#

therefore the other two angles are complementary

#

yes thats correct

fast tide Oct 30, 2020, 3:34 PM

#

addition subtraction multiply and divide

#

don't get scared by the math

dry pulsar Oct 30, 2020, 3:34 PM

#

Thanks

fast tide Oct 30, 2020, 3:34 PM

#

👍

odd blade Oct 30, 2020, 4:26 PM

#

Guys if you use the same reflection twice the object will land where it started right

olive portal Oct 30, 2020, 4:47 PM

#

help?

📎 werg.JPG

#

i need to prove this

odd blade Oct 30, 2020, 4:48 PM

#

0_0

olive portal Oct 30, 2020, 4:49 PM

#

i can’t do anything no more

#

my head hurts

#

this thing haunts mr

#

me*

earnest echo Oct 30, 2020, 4:52 PM

#

What is 0 choose n??

olive portal Oct 30, 2020, 4:53 PM

#

combinations

#

of n things taken 0 times

earnest echo Oct 30, 2020, 4:54 PM

#

Shouldn't it be n choose 0

olive portal Oct 30, 2020, 4:55 PM

#

oh fuck

#

damn

#

oh my god

#

it is the other way around

odd blade Oct 30, 2020, 4:59 PM

#

📎 image0.jpg

#

Correct me if I am wrong it has 2 eight

#

2 right diagonal and horizontal

#

@olive portal

olive portal Oct 30, 2020, 5:01 PM

#

i don’t study geometry sry

odd blade Oct 30, 2020, 5:03 PM

#

Ah ok

upper karma Oct 30, 2020, 5:23 PM

#

📎 20201030_131615.jpg

raven heart Oct 30, 2020, 5:33 PM

#

can any kind ppl here help me out with this

📎 Screenshot_547.png

agile tinsel Oct 30, 2020, 6:02 PM

#

Hey guys, can anyone help me with a Vector issue, its for a game engine. I want to somehow offset my vector v (indicated on the image), in a way that it's offset to the left and to the right. Kind of like raytracing in 3 directions to check for empty position to move to. But I have no idea how to approach this offset. I have T(x,y) (desired location), vector v and the position P(x,y).

📎 unknown.png

#

I'd like to try and get something like this

#

📎 unknown.png

wise pawn Oct 30, 2020, 6:04 PM

#

are you trying to get it to shoot around that obstacle then?

#

do you know the width of the blocking thing or would you have to find that out

#

like I'm assuming you're like playing minigolf and you're at P trying to hit your ball into the hole at t, but there's a wall in the way, so you are wanting to shoot as close to the wall as possible to get around it

#

is that basically what you're trying to do?

agile tinsel Oct 30, 2020, 6:06 PM

#

Yeah

#

I can get the size of the obstacle

#

But the thing is, I can offset it to the side of the obstacle, but I have no idea how to prone and see how much should I add to avoid it

#

As depending on orientation it can be +-x, +-y

#

(these are units so I am implementing avoidance and going around

#

)

wise pawn Oct 30, 2020, 6:08 PM

#

I'm not sure I fully understand but

#

what you can do is take the difference between a vector pointing to the corner of the obstacle and your position, that gets you a direction vector pointing exactly where you want to shoot

agile tinsel Oct 30, 2020, 6:09 PM

#

But wouldnt that land me exactly at the corner?

wise pawn Oct 30, 2020, 6:09 PM

#

if your gun/ball you're shooting has width, add half the width to the vector at the corner

#

maybe add a little extra just in case that exact amount is absolutely too close

#

that make sense or should I doodle a little pic to show

agile tinsel Oct 30, 2020, 6:10 PM

#

Literally add? Doesnt it make a difference how the target is orriented?

wise pawn Oct 30, 2020, 6:10 PM

#

yeah, of course

agile tinsel Oct 30, 2020, 6:10 PM

#

📎 unknown.png

#

How can I take that into account?

wise pawn Oct 30, 2020, 6:11 PM

#

I don't know what that is

agile tinsel Oct 30, 2020, 6:11 PM

#

That's my biggest issue

wise pawn Oct 30, 2020, 6:11 PM

#

well I can't help you with your problem if you don't know your problem

agile tinsel Oct 30, 2020, 6:11 PM

#

What?

wise pawn Oct 30, 2020, 6:12 PM

#

I don't know what you're drawing

agile tinsel Oct 30, 2020, 6:12 PM

#

Oh

wise pawn Oct 30, 2020, 6:12 PM

#

these are just triangles

agile tinsel Oct 30, 2020, 6:13 PM

#

P is the position we're at, O is the obstacle we're trying to avoid, v's are the vectors

📎 unknown.png

#

We have a vector pointing at our target behind the obstacle(the middle one) and we have 2 vectors pointing exactly at the corners

wise pawn Oct 30, 2020, 6:14 PM

#

📎 ball.png

#

this is how I'm picturing the scenario

agile tinsel Oct 30, 2020, 6:14 PM

#

Yep

wise pawn Oct 30, 2020, 6:14 PM

#

blue vector is pointing to one corner, and orange dot is where you're shooting from

#

you need just enough offset to pass, ok cool

#

📎 green.png

#

so our problem is now figuring out how to get that green vector and that solves the whole problem?

#

just want to make sure we're doing what we want here cause I'm not psychic haha

agile tinsel Oct 30, 2020, 6:16 PM

#

Yeah we're getting the green vector

wise pawn Oct 30, 2020, 6:16 PM

#

ok what do we know about it?

agile tinsel Oct 30, 2020, 6:17 PM

#

We know the pink vector

#

the center of the obstacle

#

wait

#

We can use the center of the obstacle and the corner

wise pawn Oct 30, 2020, 6:17 PM

#

I'd say we don't know the pink vector

agile tinsel Oct 30, 2020, 6:18 PM

#

I mean we don't know it because we don't know which side we're hitting

#

thats correct

wise pawn Oct 30, 2020, 6:18 PM

#

we know the length of the green vector though, and we know the green vector and pink vector are perpendicular

#

well, let's not care about which side we're hitting

#

we can determine that beforehand

agile tinsel Oct 30, 2020, 6:18 PM

#

Alright

wise pawn Oct 30, 2020, 6:19 PM

#

based on where p is relative to the left or right of the center of the obstacle

#

maybe we should do that first since it's easier

#

and logically happens first too

agile tinsel Oct 30, 2020, 6:20 PM

#

I am following

wise pawn Oct 30, 2020, 6:21 PM

#

📎 proj.png

#

so here I've drawn a dark blue line where the projection will end up, which is to the left of the center of the obstacle

#

so it would really end up shooting to the left in our case

#

actually it should be computed as the center along the light blue vector, not the center of the obstacle, that doesn't matter

#

so you tell me, how do you compute something like this?

agile tinsel Oct 30, 2020, 6:23 PM

#

What exactly are we computing?

#

The dark blue?

wise pawn Oct 30, 2020, 6:23 PM

#

you're wanting to decide which side to shoot towards

#

and we determine which side to shoot to depending on whether p is to the left or right of the center

#

and we can look at that by checking the projection along the light blue vector relative to the center

agile tinsel Oct 30, 2020, 6:24 PM

#

Nono, we should shoot towards both sides

#

As another obstacle may be in place there

wise pawn Oct 30, 2020, 6:24 PM

#

you're shooting towards both sides simultaneously?

agile tinsel Oct 30, 2020, 6:24 PM

#

Yeah

#

The issue is just the green vector

wise pawn Oct 30, 2020, 6:25 PM

#

oh ok then

#

I thought you were deciding where to shoot

agile tinsel Oct 30, 2020, 6:25 PM

#

Sry @wise pawn this is a big complication, I'll try and research more on my own

#

And get some notes ready

#

It's a bother to ask anyone in discord this

#

I'll look at stackoverflow

wise pawn Oct 30, 2020, 6:26 PM

#

well I'm complicating it because you're not paying me, I'm teaching you for free lol

#

I have the answer

#

I'm just not giving free answers that's all

agile tinsel Oct 30, 2020, 6:27 PM

#

Yeah that's understandable

#

I am gonna try this if you're interested in reading it https://gamedevelopment.tutsplus.com/tutorials/understanding-steering-behaviors-collision-avoidance--gamedev-7777

Game Development Envato Tuts+

Understanding Steering Behaviors: Collision Avoidance

Decent NPC navigation often requires the ability to avoid obstacles. This tutorial covers the collision avoidance steering behavior, which allows characters to gracefully dodge any number of...

pearl lava Oct 30, 2020, 6:40 PM

#

Where do I start?

📎 image0.png

obtuse tapir Oct 30, 2020, 6:49 PM

#

Similiar shape

spiral hawk Oct 30, 2020, 6:54 PM

#

First, u want to add 60+100+120 together to get 280, then 360-280 to get an angle of 80 for F

pearl lava Oct 30, 2020, 6:54 PM

#

yes

spiral hawk Oct 30, 2020, 6:54 PM

#

Then your gonna take the FECG and split it in half from F to C

pearl lava Oct 30, 2020, 6:55 PM

#

ok

spiral hawk Oct 30, 2020, 6:57 PM

#

Now u should have two trinagles with the top one u have two known sides and and unknown side

pearl lava Oct 30, 2020, 6:58 PM

#

yes

spiral hawk Oct 30, 2020, 7:00 PM

#

Nvm

#

Wait, missread, FECG is just like a shrunk version of ABCD, since the side lengths of BC is double EC, that means u can double the six the same way to get x = 12, 6*2

obtuse tapir Oct 30, 2020, 7:03 PM

#

Thats what i thought

#

Decided not to say anything

#

To not sound dumb

pearl lava Oct 30, 2020, 7:04 PM

#

oh ok thanks

paper marlin Oct 30, 2020, 7:44 PM

#

ABC is a triangle right angled at B.AB=5 and BC=10,what is angle C?

hidden crescent Oct 30, 2020, 7:47 PM

#

use your trigonometric functions

paper marlin Oct 30, 2020, 7:48 PM

#

Yes?

hidden crescent Oct 30, 2020, 7:49 PM

#

we can take the arctan(5/10)

#

to get angle c

paper marlin Oct 30, 2020, 7:49 PM

#

I did

#

But I still did not get angle C

hidden crescent Oct 30, 2020, 7:49 PM

#

degree or radians mode

paper marlin Oct 30, 2020, 7:49 PM

#

Sorry I am new to trignometry

#

Maybe a bit bad as well

#

Could you please show me the solution for this?

hidden crescent Oct 30, 2020, 7:51 PM

#

ok

#

== atand(5/10)

#

omg I forgot theres no mathbot

upper karma Oct 30, 2020, 7:51 PM

#

,w

hidden crescent Oct 30, 2020, 7:51 PM

#

and this is a math server

somber coyoteBOT Oct 30, 2020, 7:51 PM

#

Please submit a valid query! For example, ,ask differentiate x+y^2 with respect to x.

upper karma Oct 30, 2020, 7:51 PM

#

Use ,w

hidden crescent Oct 30, 2020, 7:52 PM

#

can someone help me solve arctan of 5/10

upper karma Oct 30, 2020, 7:52 PM

#

wtf is atand

hidden crescent Oct 30, 2020, 7:52 PM

#

in degreess

#

arctan in degress

#

its for Mathbot

#

but I forgot this server doesnt have it

#

i have never use ,w

upper karma Oct 30, 2020, 7:52 PM

#

Just do the same thing with ,w

hidden crescent Oct 30, 2020, 7:53 PM

#

,w atand(5/10)

somber coyoteBOT Oct 30, 2020, 7:53 PM

#

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

upper karma Oct 30, 2020, 7:53 PM

#

If you want an approximation ofc

hidden crescent Oct 30, 2020, 7:53 PM

#

yes 26 degrees

paper marlin Oct 30, 2020, 7:53 PM

#

Thank you

humble pulsar Oct 30, 2020, 8:48 PM

#

what's tand?

umbral snow Oct 30, 2020, 8:57 PM

#

tan in degrees

#

@humble pulsar

#

As opposed to tan which works in radians

humble pulsar Oct 30, 2020, 9:21 PM

#

Oh

upper karma Oct 30, 2020, 9:32 PM

#

draw the height

flat viper Oct 30, 2020, 9:34 PM

#

Ooh i got.it

#

My handwriting 😭

hidden crescent Oct 30, 2020, 9:38 PM

#

$sinx=\frac{3}{\sqrt{10}}$

somber coyoteBOT Oct 30, 2020, 9:38 PM

#

chene12:

📎 730879672958582984.png

hidden crescent Oct 30, 2020, 9:39 PM

#

Therefore opposite = 3 and hyp = $\sqrt{10}$

somber coyoteBOT Oct 30, 2020, 9:39 PM

#

chene12:

📎 730879672958582984.png

hidden crescent Oct 30, 2020, 9:39 PM

#

note these arent exact side lengths, merely ratios

#

therefore adjacent = $\sqrt{ \sqrt{10}^2 - 3^2 }$

somber coyoteBOT Oct 30, 2020, 9:41 PM

#

chene12:

📎 730879672958582984.png

hidden crescent Oct 30, 2020, 9:41 PM

#

adjacent = $1$

somber coyoteBOT Oct 30, 2020, 9:41 PM

#

chene12:

📎 730879672958582984.png

flat viper Oct 30, 2020, 9:41 PM

#

Yeah i solved it

#

Second question is hard

hidden crescent Oct 30, 2020, 9:42 PM

#

image isnt clear enough

flat viper Oct 30, 2020, 9:56 PM

#

📎 20201030_225551.jpg

hidden crescent Oct 30, 2020, 10:42 PM

#

w, rotate

#

,w rotate

upper karma Oct 30, 2020, 11:01 PM

#

,rccw

somber coyoteBOT Oct 30, 2020, 11:01 PM

#

📎 349552552385314818.png

faint gazelle Oct 30, 2020, 11:10 PM

#

i need some help, how would i write an equation to solve for z?

📎 unknown.png

upper karma Oct 30, 2020, 11:10 PM

#

Consider the theorem of the sum of angles on a triangle

faint gazelle Oct 30, 2020, 11:11 PM

#

alright cool, thanks

#

after looking into it a bit more, i'm still a bit confused

upper karma Oct 30, 2020, 11:23 PM

#

Alright

#

Where did you got to

#

If you got to anything

#

The key here is to notice how we can relate the angle 145° to the inside angle in order to apply the theorem mentioned above

faint gazelle Oct 30, 2020, 11:26 PM

#

yeah, i got to that part

#

C = 35 degrees

#

that's all i really have for now

upper karma Oct 30, 2020, 11:27 PM

#

Okay so yeah

#

We want to create an eqn to solve for z

#

And we can use

Consider the theorem of the sum of angles on a triangle

faint gazelle Oct 30, 2020, 11:27 PM

#

angle1 + angle2 + angle3 = 180 degrees

upper karma Oct 30, 2020, 11:28 PM

#

Yes

faint gazelle Oct 30, 2020, 11:28 PM

#

i'm still pretty confused, sorry

upper karma Oct 30, 2020, 11:29 PM

#

We want to apply that idea exactly the same way you'd do with numbers

#

Is the z what confuses you?

faint gazelle Oct 30, 2020, 11:30 PM

#

yeah pretty much

upper karma Oct 30, 2020, 11:30 PM

#

Okay how would you apply it with numbers

#

Imagine you have a triangle with 30 60 and 90 degrees

faint gazelle Oct 30, 2020, 11:30 PM

#

would z equal the 35 degrees?

upper karma Oct 30, 2020, 11:31 PM

#

,calc 180-45

somber coyoteBOT Oct 30, 2020, 11:31 PM

#

Result:

upper karma Oct 30, 2020, 11:31 PM

#

Oops

#

Wait

#

,w calc 180-145

#

No it is not 35

faint gazelle Oct 30, 2020, 11:33 PM

#

i had a feeling

upper karma Oct 30, 2020, 11:34 PM

#

Okay

#

So we said

#

angle1 + angle2 + angle3 = 180 degrees

#

Yes?

faint gazelle Oct 30, 2020, 11:34 PM

#

correct

upper karma Oct 30, 2020, 11:35 PM

#

And we have our triangle, whose angles are 35°, 2z° and (5z-2)°

#

Yes?

faint gazelle Oct 30, 2020, 11:35 PM

#

yes

#

wait

umbral cloud Oct 30, 2020, 11:35 PM

#

So how can you relate those angles to 180 degrees?

faint gazelle Oct 30, 2020, 11:35 PM

#

do you mean 35 instead of 145?

upper karma Oct 30, 2020, 11:35 PM

#

Sid i have the floor please don't interrupt

#

@faint gazelle yes oops

faint gazelle Oct 30, 2020, 11:36 PM

#

OH

#

i understand now

#

so, 35 + 2z + (5z-2) = 180

upper karma Oct 30, 2020, 11:36 PM

#

@umbral cloud i believed it is not hard to understand that i'm helping and i don't like getting interrupted

#

so, 35 + 2z + (5z-2) = 180
Yes

faint gazelle Oct 30, 2020, 11:37 PM

#

okay, thank you

upper karma Oct 30, 2020, 11:37 PM

#

Yw

faint gazelle Oct 30, 2020, 11:40 PM

#

got it all figured out, thank you so much for your help bro @upper karma

upper karma Oct 30, 2020, 11:40 PM

#

Glad to hear!

sharp socket Oct 31, 2020, 12:19 AM

#

I dont know how to do this but I think I've found AC/BC by using the sine rule? My result for AC/BC is sin(arctan(1/2))/sin(arctan(2/3)). Idk how to get AC/AB

📎 Screen_Shot_2020-10-30_at_8.07.34_PM.png

upper karma Oct 31, 2020, 12:50 AM

#

Solve for angle C

unborn bison Oct 31, 2020, 12:50 AM

#

whoops

#

The number of sides of a regular polygon is 18. Find the measures of an interior angle and an exterior angle for each ploygon.

#

i said interior 160 and ext 20

#

potentially

upper karma Oct 31, 2020, 12:51 AM

#

$180(n-2)/n$

somber coyoteBOT Oct 31, 2020, 12:51 AM

#

The Sealman:

📎 544231274030563328.png

upper karma Oct 31, 2020, 12:51 AM

#

n = 18

#

So interior is correct

#

Exterior is 360/18 which is 20

#

Also correct

unborn bison Oct 31, 2020, 12:52 AM

#

yay

#

📎 Screen_Shot_2020-10-30_at_8.55.24_PM.png

#

how does it prove those two are parallel?

#

if they are the same

#

thanks for the help ❤️

upper karma Oct 31, 2020, 12:58 AM

#

Well if the angles are equal then they are parallel

unborn bison Oct 31, 2020, 12:59 AM

#

i mean the lines

#

it says 2 equals 4 proves p and q (transversal above) are parallel

unborn bison Oct 31, 2020, 1:50 AM

#

any ideas?

silent plank Oct 31, 2020, 1:53 AM

#

converse of parallel line theorems

unborn bison Oct 31, 2020, 1:54 AM

#

they didn't put that in the word bank, @silent plank

#

📎 Screen_Shot_2020-10-30_at_9.54.27_PM.png

#

this is all i've got

#

Alternate int, Corresponding, Same Side interior, Alternate Exterior, linear pair, vertical

silent plank Oct 31, 2020, 1:58 AM

#

can you see that 2 and 4 are alternate int?

unborn bison Oct 31, 2020, 1:58 AM

#

i thought 4 would be exterior

#

oh wait im dumb

upper karma Oct 31, 2020, 4:31 AM

#

can someone help me find BDC , AB = AC

📎 nOJpsxq.png

pure cape Oct 31, 2020, 4:52 AM

#

@upper karma what have you tried

rare pollen Oct 31, 2020, 6:06 AM

#

a car with a mass of 1100 kg is travelling at 100km/h and collides with a tree and stops in 0.5 seconds. Calculate the force exerted by the car on the tree

#

someone do it adlays

proven totem Oct 31, 2020, 7:16 AM

#

Can you help me with this problem?

📎 unknown.png

dark sparrow Oct 31, 2020, 7:17 AM

#

can you show what rules you have to choose from?

#

this looks like more of an issue with the answer format than with the underlying math

#

also @proven totem please please PLEASE do not post across multiple channels.

proven totem Oct 31, 2020, 7:19 AM

#

Sorry. My bad, just a newbie

dark sparrow Oct 31, 2020, 7:21 AM

#

yeah again i'd like to see the list of rules

#

there are a bunch of possible ways to do this, i just need to know what your system calls them

devout harbor Oct 31, 2020, 7:23 AM

#

I don't think they will reply, I gave them the answer in #calculus which I shouldn't have done

upper karma Oct 31, 2020, 2:46 PM

#

what is B?

📎 unknown.png

dark sparrow Oct 31, 2020, 3:07 PM

#

the answer to B is something you should know from theory

#

graphically, going between the graphs of f and of f^-1 is reflection along a line that makes 45 degree angles with both axes

raven heart Oct 31, 2020, 3:14 PM

#

can some1 help me out with this

📎 Screenshot_547.png

zenith garnet Oct 31, 2020, 5:54 PM

#

📎 Screen_Shot_2020-10-31_at_10.39.23_AM.png

#

im not sure how to go about finding this

#

could someone help me?

dark sparrow Oct 31, 2020, 5:56 PM

#

think about the sum of the interior angles of the triangle, & how they relate to x, y and z

zenith garnet Oct 31, 2020, 5:58 PM

#

i do know the interior angles should add up to 180

#

180-(x y or z) = its respective interior angle

dark sparrow Oct 31, 2020, 5:59 PM

#

yup

#

so (180-x) + (180-y) + (180-z) = 180

zenith garnet Oct 31, 2020, 7:26 PM

#

ohhh ok ok

#

nr sure where to go from there, sorry

dark sparrow Oct 31, 2020, 7:27 PM

#

rearrange the left hand side

#

540 - (x+y+z) = 180

zenith garnet Oct 31, 2020, 7:28 PM

#

ahhh oh so u take out the 180*3 and then add 540 to each side

#

tysm that makes sense

slow bluff Oct 31, 2020, 9:14 PM

#

hi

#

can someone explain to me why this is true

#

📎 unknown.png

#

Why does AE = EG

#

and GF = FC

#

what property is this

upper karma Oct 31, 2020, 9:17 PM

#

tangency

#

triangles DAE and DEG are congruent

slow bluff Oct 31, 2020, 9:23 PM

#

oh can't believe i missed that

hushed spear Oct 31, 2020, 10:20 PM

#

Year 9 coordinate geometry

lines with the equation 3x + ky = 11 and 5x + y = 8 are perpendicular, find the value of k.
so i have isolated the gradients and i know that -3/k * -5 = -1 (i think)
because they are perpendicular

hollow raven Oct 31, 2020, 11:49 PM

#

Change from standard into y intercept form and find the value k for the slope in the 1st equation so thats is the negative reciprocal to the slope in the 2nd equation. m_1 => - 1/m_1

#

for example if the first slope was something like k/2 and the second was -1/4, then you know that k/2 = 4 so k = 8.

teal mica Oct 31, 2020, 11:51 PM

#

Can anyone explain how you determine if a fraction is greater than 2pi?

hollow raven Oct 31, 2020, 11:51 PM

#

Wdym?

teal mica Oct 31, 2020, 11:52 PM

#

📎 20201031_165210.jpg

#

I thought the answer was just 1/6 and 5/6

#

📎 unknown.png

#

But 13/6 and 17/6 were also answers

#

But 17pi/6 and 13pi/6 are both greater than 2pi

hollow raven Oct 31, 2020, 11:54 PM

#

Do they give some sort of explanation on why that is so?

teal mica Oct 31, 2020, 11:55 PM

#

📎 unknown.png

#

📎 unknown.png

#

Its vague

#

I don't understand

hollow raven Oct 31, 2020, 11:57 PM

#

Oh I see

#

@teal mica the pi in the sin function is not part of theta's value

#

so it seems really confusing at first

teal mica Nov 1, 2020, 12:00 AM

#

Yeah it's independent from it

hollow raven Nov 1, 2020, 12:01 AM

#

yeah, so as long as theta is greater than two pi, you can just keep on incrementing 2pi as it results in coterminal angles

#

as for calculating it, I guess you just gotta check if 2pi/theta > 1

teal mica Nov 1, 2020, 12:02 AM

#

Sorry kinda got lost there

hollow raven Nov 1, 2020, 12:02 AM

#

no worries

teal mica Nov 1, 2020, 12:03 AM

#

I substituted pi(theta) with x

#

then solved for theta

#

Which got me theta = x/pi

#

Then solved for sin which I got 1/2

#

Then referenced it to the unit circle which got me the pi/6 and 5pi/6

hollow raven Nov 1, 2020, 12:03 AM

#

right

teal mica Nov 1, 2020, 12:04 AM

#

Then I re-subbed my result with x and then solved

#

which got me to regular fractions

#

But i'm unsure how to determine if those fractions are > 2pi

#

Like pi/6 * 1/pi = 1/6, but how should I go about trying to determine if 1/6 is > 2pi

hollow raven Nov 1, 2020, 12:05 AM

#

Yeah, I can't give you a truly simple way as pi is a tricky number to handle

#

but the best you can do is divide 2pi/theta

#

if its greater than 2pi, then the result would be less than 1

#

or no

#

you can just check if theta/2 < pi

teal mica Nov 1, 2020, 12:07 AM

#

If I divide 2pi/theta when theta is pi/6 that would get me 12, but not sure what that means too

hollow raven Nov 1, 2020, 12:07 AM

#

theta would be 1/6

#

so theta/2 = 1/12 < pi

#

same applies for all other theta values

#

lets say i had something like theta = 17/2

#

theta/2 would equal 17/4, which is 4 with a remainder 1 or 4.25

#

Since thats greater than ~ 3.14

#

we know it doesnt satisfy the range

#

Sorry, my wording is sort of confusing

#

But there isn't really a clear cut way to figure out something like that without testing and checking

teal mica Nov 1, 2020, 12:12 AM

#

Hmm

hollow raven Nov 1, 2020, 12:14 AM

#

If anyone would like to contribute please do^^^

teal mica Nov 1, 2020, 12:14 AM

#

Sorry, just trying to digest it

teal mica Nov 1, 2020, 1:54 AM

#

📎 unknown.png

#

So to check I just plug it into the calculator

#

And it equals to 1

next jackal Nov 1, 2020, 6:43 AM

#

there are no fixed points in a translation right?

dark sparrow Nov 1, 2020, 6:53 AM

#

yes

next jackal Nov 1, 2020, 7:03 AM

#

thanks

main mortar Nov 1, 2020, 11:38 AM

#

pls help me

📎 unknown.png

upper karma Nov 1, 2020, 11:40 AM

#

Then what

#

Did you cut off a part of the problem?

main mortar Nov 1, 2020, 11:49 AM

#

yes lol

#

📎 unknown.png

next jackal Nov 1, 2020, 11:51 AM

#

I didn't understand how this conclusion came up

📎 Screenshot_403.png

next jackal Nov 1, 2020, 3:09 PM

#

the answer is P+ rQ where r is the required fraction right?

📎 Screenshot_404.png

dark sparrow Nov 1, 2020, 3:39 PM

#

no

#

it's P + r(Q-P)

hybrid pendant Nov 1, 2020, 4:09 PM

#

Looks complicated

kindred rapids Nov 1, 2020, 4:54 PM

#

Hi guys....can someone help me prove this identity please?

📎 20201101_125358.jpg

#

I've tried so many times but I get stuck and i can't prove it...help

earnest echo Nov 1, 2020, 4:56 PM

#

Show your work

kindred rapids Nov 1, 2020, 4:57 PM

#

My work..? What do you mean?

earnest echo Nov 1, 2020, 4:57 PM

#

What have you tried

kindred rapids Nov 1, 2020, 4:57 PM

#

Okay

#

So I changed tanØ to sinØ/cosØ...then I expanded the brackets and then I added the fractions since they all had the same denominator....
After I saw there was a quadratic in sin so I factorised and then I changed cos^2 Ø to 1 - sin^2 Ø....then I'm stuck

📎 20201101_125853.jpg

upper karma Nov 1, 2020, 5:01 PM

#

,rccw

somber coyoteBOT Nov 1, 2020, 5:01 PM

#

📎 349552552385314818.png

earnest echo Nov 1, 2020, 5:03 PM

#

If you notice\

$\sin²(\theta)-2\sin(\theta)+1=(1-\sin(\theta))²$

#

Use difference of squares in the denominator

somber coyoteBOT Nov 1, 2020, 5:05 PM

#

The Godfather:

📎 302032999120109568.png

earnest echo Nov 1, 2020, 5:05 PM

#

Use difference of squares in the denominator

kindred rapids Nov 1, 2020, 5:05 PM

#

Okay
It would be (1+sinØ) (1-sinØ) ?

#

For the denominator...?

earnest echo Nov 1, 2020, 5:05 PM

#

Yes

#

Now just simplify

kindred rapids Nov 1, 2020, 5:06 PM

#

Okay thank you so much....I'll go do that now

#

But guys....
It has to equal - sinØ- 1 / sin Ø + 1 ...like I sent in the picture of the question
But when I simplify...I'm getting 1 - sinØ / 1 + sinØ

earnest echo Nov 1, 2020, 5:13 PM

#

(1-sin(x))²=(sin(x)-1)²

kindred rapids Nov 1, 2020, 5:20 PM

#

Yes I understand that

#

But the final answer....I'm not getting it

earnest echo Nov 1, 2020, 5:23 PM

#

Show me just the last step

kindred rapids Nov 1, 2020, 5:27 PM

#

📎 Screenshot_20201101-132711_Photo_Editor.jpg

#

Did I do that wrong or..?

earnest echo Nov 1, 2020, 5:28 PM

#

No

#

$(1-\sin(\theta))=-(\sin(\theta)-1)$

somber coyoteBOT Nov 1, 2020, 5:29 PM

#

The Godfather:

📎 302032999120109568.png

kindred rapids Nov 1, 2020, 5:30 PM

#

Oh...I see

#

Let me do that again

#

So what will sinØ - 1 be equal to..?

earnest echo Nov 1, 2020, 5:32 PM

#

Huh?

kindred rapids Nov 1, 2020, 5:34 PM

#

That up there was 1 - sinØ = -(sinØ- 1)
So what is sinØ - 1 equal to?

#

The way they taught us to factorise instead of 1 - sinØ....we would have (sinØ - 1) (sinØ - 1)

#

Which is what I have in the first pic of work that I sent.....

#

According to how I learnt.....this is how I'm supposed to write it and from here I need to work

📎 Screenshot_20201101-133726_Photo_Editor.jpg

earnest echo Nov 1, 2020, 5:38 PM

#

Well you see
When you use difference of squares of you get

1-sin²(∅)=(1-sin(∅))(1+sin(∅))

#

Now to cancel the term in the denominator you have to factor out -1

kindred rapids Nov 1, 2020, 5:39 PM

#

Factor out -1 from where exactly?

earnest echo Nov 1, 2020, 5:40 PM

#

From any one of the term in the numerator

kindred rapids Nov 1, 2020, 5:40 PM

#

I got through

#

Lol I'm sorry to put you through that....
Thank you so much, I'm grateful

earnest echo Nov 1, 2020, 5:41 PM

#

Yw

kindred rapids Nov 1, 2020, 5:42 PM

#

:)

upper karma Nov 1, 2020, 5:52 PM

#

📎 Screenshot_2020-11-01_at_9.51.53_AM.png

#

I just need to know what w x and z are

#

Thats all I need help with

#

<@&286206848099549185>

spiral hawk Nov 1, 2020, 5:56 PM

#

Since the trapizods are similar, z is equal to 80, y = 105

upper karma Nov 1, 2020, 5:56 PM

#

I see

spiral hawk Nov 1, 2020, 5:58 PM

#

Then, since one trapizoid is scaled down from the other, u can take (25/10)*7.2=w

upper karma Nov 1, 2020, 5:58 PM

#

Mmhmm

#

Oh that makes sense now

spiral hawk Nov 1, 2020, 5:59 PM

#

Then, its the same idea for x, except now u can make and solve an equation

upper karma Nov 1, 2020, 5:59 PM

#

I get it now

#

Thank you

spiral hawk Nov 1, 2020, 6:00 PM

#

(10/25)*15=x-2

upper karma Nov 1, 2020, 6:00 PM

#

Thank you for helping me out with this

teal mica Nov 1, 2020, 6:52 PM

#

Is anyone able to explain how to solve this by hand? I get how to solve via calculator... cos^-1(-0.123)... but what steps does the calculator perform to come to the solution?

📎 unknown.png

mild cargo Nov 1, 2020, 6:56 PM

#

Calculator uses Taylor series expansion

cerulean dew Nov 1, 2020, 10:35 PM

#

📎 unknown.png

upper karma Nov 2, 2020, 12:32 AM

#

📎 Screenshot_2020-11-01_at_4.31.40_PM.png

#

📎 Screenshot_2020-11-01_at_4.33.43_PM.png

#

📎 Screenshot_2020-11-01_at_4.33.36_PM.png

cosmic pebbleBOT Nov 2, 2020, 12:36 AM

#

From

#❓how-to-get-help

Rule 4

If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once. Please do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

upper karma Nov 2, 2020, 12:37 AM

#

I still need help

#

<@&286206848099549185>

#

Bruh

#

H e l p m e

velvet heart Nov 2, 2020, 1:08 AM

#

I am scare

📎 unknown.png

upper karma Nov 2, 2020, 1:14 AM

#

Help

📎 Screenshot_2020-11-01_at_5.11.08_PM.png

tiny flax Nov 2, 2020, 2:33 AM

#

📎 unknown.png

#

Does anyone know how I can do this?

#

I can't prove angle A is congruent to angle C

median gull Nov 2, 2020, 4:53 AM

#

anyone know what this symbol means?

📎 unknown.png

silent plank Nov 2, 2020, 4:54 AM

#

that is the greek letter theta, usually used as a variable for angles

median gull Nov 2, 2020, 4:55 AM

#

thanks i thought it was beta cause i kept hearing my teacher say it

silent plank Nov 2, 2020, 4:56 AM

#

beta: $\beta$

somber coyoteBOT Nov 2, 2020, 4:56 AM

#

ramonov:

📎 220517274845577216.png

upper karma Nov 2, 2020, 5:22 AM

#

@silent plank

📎 unknown.png

#

how is this not -8

#

shid wrong channel

night snow Nov 2, 2020, 8:24 AM

#

@upper karma
Because you cannot cancel ANYTHING if there is any + or - in the fraction (that are not within brackets)

upper karma Nov 2, 2020, 8:41 AM

#

@night snow shit thanks so much

analog pebble Nov 2, 2020, 8:47 AM

#

Does anyone know how I can do this?
@Mahjestic#9700 Since there are two parallel sides, (AB parallel CD, AD parallel BC), it is given that ABCD is a parallelogram. By the properties of a parallelogram, angle A = angle C

#

then it’s just simple from then on

next jackal Nov 2, 2020, 9:16 AM

#

Am I supposed to find out the values of s and t by substitution?

📎 Screenshot_408.png

next jackal Nov 2, 2020, 10:40 AM

#

📎 Screenshot_409.png

#

📎 20201102_161005.jpg

#

How do I show that $2(pta+qtb) ≥0$? Or less than or equal to all the terms left side of it?

somber coyoteBOT Nov 2, 2020, 10:41 AM

#

lazypawtato:

📎 507948627260473345.png

next jackal Nov 2, 2020, 10:45 AM

#

dead channel

#

ted help me pls

junior light Nov 2, 2020, 10:46 AM

#

Can't promise to be helpful but I'll try.

#

For starters I don't know shit about parametric form of a line.

#

But nevertheless I guess you can translate an answer in regular form to parametric form?

next jackal Nov 2, 2020, 10:48 AM

#

yes

junior light Nov 2, 2020, 10:48 AM

#

So consider that the two lines intersect in two points, (x1,y1) and (x2,y2). Then show that x1=x2 and y1=y2, and hence you prove that the point of intersection, if it exists, must be unique.

next jackal Nov 2, 2020, 10:49 AM

#

no I got that

junior light Nov 2, 2020, 10:49 AM

#

Oh, alright. You're asking about that circle problem?

next jackal Nov 2, 2020, 10:49 AM

#

yes

junior light Nov 2, 2020, 10:50 AM

#

Have you proven that any line passing through (p,q) must pass through the circle?

next jackal Nov 2, 2020, 10:51 AM

#

No that's what I'm trying to do

#

I tried to follow that hint

#

I understood that you can make points on line with distance to centre as the radius

junior light Nov 2, 2020, 10:52 AM

#

I'm seriously oblivious of the parametric form of a line. I'll take a look at it quickly and see if I can add something valuable.

next jackal Nov 2, 2020, 10:53 AM

#

Okay thanks

junior light Nov 2, 2020, 10:53 AM

#

catthumbsup

upper karma Nov 2, 2020, 1:20 PM

#

📎 104203962_901291193720743_2337338904023341810_o.jpg

rain vigil Nov 2, 2020, 1:30 PM

#

Is anyone free to help me here

west basin Nov 2, 2020, 1:30 PM

#

just post your question someone will answer eventually

rain vigil Nov 2, 2020, 1:31 PM

#

📎 20201102_213110.jpg

#

May i know how do i answer 4(b)

#

Can someone give me a hint

#

;-;

devout harbor Nov 2, 2020, 1:38 PM

#

@rain vigil use double angle identity for sin and cos

rain vigil Nov 2, 2020, 1:45 PM

#

Double angle identity? @devout harbor

#

But thats 2sin theta

#

Not sin 2theta

stark mural Nov 2, 2020, 1:52 PM

#

can anyone help me with this assignment?

📎 ds123.png

#

And I have this type of clue

📎 unknown.png

strange quartz Nov 2, 2020, 2:01 PM

#

Hi guys. I have a question in spherical trigonometry. Given two points on earth A(\lambda_A, \phi_A) and B(\lambda_B, \phi_B), and 2 distances a, b. I have to determine all the points X(\lambda_X, \phi_X) on earth such that, the distance between A and X is b and the distance between X and B is a

humble pulsar Nov 2, 2020, 2:05 PM

#

Given two points on earth $A(\lambda_A, \phi_A)$ and $B(\lambda_B, \phi_B)$, and 2 distances a, b. I have to determine all the points $X(\lambda_X, \phi_X)$ on earth such that, the distance between A and X is b and the distance between X and B is a

somber coyoteBOT Nov 2, 2020, 2:05 PM

#

moshill1:

📎 186109587366215691.png

strange quartz Nov 2, 2020, 2:06 PM

#

thanks @humble pulsar

humble pulsar Nov 2, 2020, 2:07 PM

#

👍

devout harbor Nov 2, 2020, 2:07 PM

#

@rain vigil use it to go from sin(theta) to sin(theta/2)

#

For sin(theta/2), sin(theta) is it's double angle

rain vigil Nov 2, 2020, 2:16 PM

#

I have figured it out, needa use triangle form on first quadrant

cosmic timber Nov 2, 2020, 2:17 PM

#

Hello!
Idk if this is the right channel but it has the geometry in the name so yeah, my question is (this might be to easy for you guys but we just got started on geometry last week) how do i draw and label this

📎 IMG20201102221513.jpg

upper karma Nov 2, 2020, 7:05 PM

#

i just finished ch4 test and I am very confused on why one of my questions was wrong

#

my brain is hurting so low iq rn but

#

📎 unknown.png

#

i said sas

#

because perpendicular form right angles

#

right angles are equal

#

pqr = psr

#

qr = sr is given, and pr = pr because reflexive

#

apparently the answer was not possible

#

i have no idea

#

help

silk patio Nov 2, 2020, 9:09 PM

#

📎 image0.png

inland canyon Nov 2, 2020, 10:17 PM

#

📎 unknown.png

#

helppls

#

ik some rules but like i dont think any of them deal with this

#

📎 4ZJUR9C.png

#

and i know like all of the angles have to equal up to 360

#

like x would be 7 right 😔

#

wait nvm i got it lmao

#

it was like 100 + 100 + 80 + 80 and x was 7

#

14 x 7 + 2

#

look at me doing stuff

indigo chasm Nov 2, 2020, 10:42 PM

#

anyone know how to do this

📎 unknown.png

grizzled lantern Nov 2, 2020, 11:05 PM

#

@indigo chasm mbos look up the equation for the volume of whatever trapezoidal prism that is

hidden crescent Nov 2, 2020, 11:13 PM

#

the volume of any prism is given by this formula:

#

$area = base_area * height$

#

$area = base_{area} * height$

somber coyoteBOT Nov 2, 2020, 11:15 PM

#

chene12:

📎 730879672958582984.png

novel berry Nov 2, 2020, 11:23 PM

#

Volume =base_area* height?

sour jacinth Nov 3, 2020, 12:55 AM

#

Yeah

modern pond Nov 3, 2020, 1:39 AM

#

📎 image0.png

#

May anyone please help me with this homework? I’ve no idea what to do

woven walrus Nov 3, 2020, 3:17 AM

#

📎 help.PNG

#

how would one find z

hollow raven Nov 3, 2020, 3:24 AM

#

Find this angle using trig ratios

📎 unknown.png

#

use angle a to find angle b (remember that they both add to 90 degrees)

📎 unknown.png

#

Then simply use trig ratios to solve for z

#

there are other ways to do it too, see if you can find any alternate solutions!

#

@woven walrus

woven walrus Nov 3, 2020, 3:26 AM

#

ty

#

so 9/2?

hollow raven Nov 3, 2020, 3:27 AM

#

im not sure, i havent solved it pandaThink

woven walrus Nov 3, 2020, 3:27 AM

#

well thanks anyways 👍

hollow raven Nov 3, 2020, 3:28 AM

#

my pleasure

violet totem Nov 3, 2020, 4:58 AM

#

https://cdn.discordapp.com/attachments/269688654170030080/773048050729091142/image0.jpg

#

No idea what I do for 1.

silent plank Nov 3, 2020, 5:16 AM

#

your 4th statement isn't given

#

what does: $\overline{BD}$ bisects $\angle ABC$ tell you?

somber coyoteBOT Nov 3, 2020, 5:17 AM

#

ramonov:

📎 220517274845577216.png

violet totem Nov 3, 2020, 5:24 AM

#

Given?

silent plank Nov 3, 2020, 5:57 AM

#

no idea why you skipped line 1,
that's what's given, but what does it tell you about the angles involved?

hollow tendon Nov 3, 2020, 9:00 AM

#

geometry is my favorite

hushed spear Nov 3, 2020, 9:28 AM

#

how

#

hello everybody i need help with bearings

#

i have no idea what is going on

#

9th grade

upper karma Nov 3, 2020, 9:54 AM

#

Have you sorted the possibility of posting the problem you are stuck with

#

@hushed spear

hushed spear Nov 3, 2020, 9:54 AM

#

yes

#

well honestly

#

i have no questions

#

but i have this one

#

Alana hiked 10km in the direction N70E. Bree set out from the same point and hiked 24km in the direction 340T.

what is the difference between the two,

what is the true bearing of Alana from Bree,

and how far west of Alana is Bree, correct to the nearest km?

#

@upper karma

upper karma Nov 3, 2020, 10:06 AM

#

i have no questions
@hushed spear
I am new here, could you elaborate on what 340T is

#

it can easily be solved by resolving them into vectors if T represents a direction

hushed spear Nov 3, 2020, 10:16 AM

#

340t is true bearing

#

and what is a vector

coral dagger Nov 3, 2020, 12:36 PM

#

📎 image0.jpg

#

My teacher said that the green line is tangent

#

I’m not sure why

#

is it because when Sine=1 and cosine = 1

#

1/1 = tangent?

stiff bough Nov 3, 2020, 12:50 PM

#

📎 unknown.png

#

<@&286206848099549185>

vapid lodge Nov 3, 2020, 4:35 PM

#

<@&286206848099549185> Is the slant height of a pyramid the same as the lateral edge height

teal mica Nov 3, 2020, 5:02 PM

#

@woven walrus Did you get help on your problem?

#

Use the Law of Sines

main pagoda Nov 3, 2020, 5:14 PM

#

Is this right

📎 image0.png

dark sparrow Nov 3, 2020, 5:17 PM

#

no

bright hollow Nov 3, 2020, 5:51 PM

#

<@&286206848099549185>

📎 image0.jpg

#

anyone know proofs?

upper merlin Nov 3, 2020, 5:54 PM

#

i cant see what u have to prove

#

what angles are those

#

nvm thats 6 and 2

bright hollow Nov 3, 2020, 5:55 PM

#

i need to fill in each empty box

upper merlin Nov 3, 2020, 5:57 PM

#

ah i see

#

ok so i would do

#

1 is congruent it 7 (given)

#

6 is cong to 7 (vertical are cong)

#

1 is cong to 6 (transitive property????????)

#

measure of 1=measure of 6 (def of congruent angles)

#

measure of 1+measure of 2= 180 (linear pair)

#

measure of 6+measure of 2=180 (substitute m<6 for m<1)

bright hollow Nov 3, 2020, 5:59 PM

#

🐐

#

appreciate it

upper merlin Nov 3, 2020, 5:59 PM

#

yeah did u understand all the steps

bright hollow Nov 3, 2020, 6:00 PM

#

yeah

#

ik proof i just dk where to put every theorem and things like that

#

proofs*

upper merlin Nov 3, 2020, 6:00 PM

#

yeah it can be confusing because there are a lot of ways to prove this

bright hollow Nov 3, 2020, 6:00 PM

#

yeah

upper merlin Nov 3, 2020, 6:00 PM

#

np

bright hollow Nov 3, 2020, 6:00 PM

#

thanks man

glad oak Nov 3, 2020, 7:18 PM

#

Factor the expression and use the fundamental identities to simplify

📎 unknown.png

#

<@&286206848099549185>

sharp plume Nov 3, 2020, 7:23 PM

#

What's csc^3x and so on write them out

glad oak Nov 3, 2020, 7:27 PM

#

what should I do after I split up the exponents?

sharp plume Nov 3, 2020, 7:36 PM

#

Show me your work

glad oak Nov 3, 2020, 7:44 PM

#

📎 unknown.png

#

I thought about it and I just used foil

📎 unknown.png

#

and then I can get (cot^2x)(1-sinx)
(I don't know if the way I got (1-sinx) is correct I did 1/sinx - sinx/sinx)

sleek beacon Nov 4, 2020, 12:15 AM

#

I've been on this for 2 hours I hope someone can help me 😥

📎 20201103_191425.jpg

upper karma Nov 4, 2020, 1:05 AM

#

can someone help me with math

#

in mathematics voice

tidal ember Nov 4, 2020, 1:08 AM

#

I’ve been stuck on this question for a little while and I haven’t figured out the steps in doing it, If someone could dm me the steps in how to do it I would appreciate it (we’re only using sine/cosine law and sohcahtoa in this section of my math course)

📎 image0.png

sullen nimbus Nov 4, 2020, 1:09 AM

#

There's a rectangle that's 4 vertices by 5 vertices and you need to find the amount of rectangles in it someone asked me that and I said 60 but apparently it's wrong

#

I don't get how it's not

upper karma Nov 4, 2020, 1:09 AM

#

If someone can help me with geometry and call and talk please dm me

weak gyro Nov 4, 2020, 1:29 AM

#

Can someone tell me why is 20 being divided by 3?

📎 unknown.png

#

The base is 30in and the height is 20in

still mica Nov 4, 2020, 1:44 AM

#

is the term "parent function" referring to what the shape of the function will look like when graphed?

stuck crypt Nov 4, 2020, 1:55 AM

#

📎 Screen_Shot_2020-11-03_at_8.54.13_PM.png

#

📎 Screen_Shot_2020-11-03_at_8.54.19_PM.png

#

is anyone able to help me? ik its probs not that hard but im a bit confused bc im rushing this lesson

stuck crypt Nov 4, 2020, 2:12 AM

#

<@&286206848099549185>

lament shell Nov 4, 2020, 4:06 AM

#

https://gyazo.com/484069d84ae2f4ef3784c8c96974a774

Gyazo

#

need help

grizzled lantern Nov 4, 2020, 4:18 AM

#

their slopes should be equal in order for them to never intersect

#

find the pairs that make cd's slope equal to ab's slope @lament shell

azure reef Nov 4, 2020, 4:19 AM

#

or you can just draw and see

#

i mean

#

you are not required to make proofs etc

lament shell Nov 4, 2020, 4:21 AM

#

wait so what numbers would i use, just random ones?

grizzled lantern Nov 4, 2020, 4:24 AM

#

Test them

stuck crypt Nov 4, 2020, 4:25 AM

#

📎 Screen_Shot_2020-11-03_at_11.25.34_PM.png

#

can someone help me understand what to do

sullen nimbus Nov 4, 2020, 5:00 AM

#

what pair of angles have to be equal for both triangles to be the same @stuck crypt

#

(Shouldn't be too hard as long as you know the AAS rule)

stuck crypt Nov 4, 2020, 5:06 AM

#

yea i figured it out i should’ve read more carefully thank you!

oak stag Nov 4, 2020, 7:51 AM

#

Hi hi! So I'm struggling with hyperbolas (I'm only in high school dfhds don't judge me too hard lmao). I don't have a specific question, I'm just finding it hard to understand the concept in general?? If anyone is willing to help, I can send some examples of questions at the level I need to understand?

#

(I'm going to be on another tab, so if anyone gets to this can you ping me pretty please <3)

oak stag Nov 4, 2020, 8:14 AM

#

<@&286206848099549185> (I think I'm allowed to ping here? It's been 15 minutes. If I'm not supposed to, my bad bahaha)

next jackal Nov 4, 2020, 9:22 AM

#

@oak stag hey, I just came here for help too now, I'd suggest you to ask in other questions channel with greek letters beside it, because I'm doing that too now. Sorry for guiding you to an inactive channel

oak stag Nov 4, 2020, 9:29 AM

#

Ok, thanks ❤️

bright hollow Nov 4, 2020, 11:44 AM

#

<@&286206848099549185>

📎 image0.jpg

next jackal Nov 4, 2020, 12:37 PM

#

@bright hollow please, only call helpers after 15 mins of submitting your question and no one responding. And if this channel seems to be inactive, ask in one of the questions channel

bright hollow Nov 4, 2020, 12:40 PM

#

@next jackal where should i ask then

next jackal Nov 4, 2020, 12:41 PM

#

see the channels under category "math help"?

#

ask in any of them which seem to be free

obsidian thicket Nov 4, 2020, 2:19 PM

#

Can I get help with geo

#

📎 Screenshot_20201104-091808_Canvas_Student.jpg