geometry-and-trigonometry

[upper karma]

bc it technically is 7/1

yeah that makes sense

[thorn talon]

yeah

well -7/1

[upper karma]

can you explain what you mean by flipping both sides

[thorn talon]

well take this example:

1/x = 1/2

[upper karma]

yeah

[thorn talon]

what do you think the solution for x should be?

[upper karma]

2

[thorn talon]

yeah

[upper karma]

yeah?

[thorn talon]

One way to achieve this is to take the reciprocal of both sides

Which is also the same as taking both sides to the -1 power

[upper karma]

you mean ^-1

[thorn talon]

Yeah

[upper karma]

i dont understand what you mean by flipping both sides tho

ohhhhh

wait so your saying

1/x=1/2

and the flipping being

x/1=2/1

[thorn talon]

Yeah

Exactly

[upper karma]

and then solving it

so its on the bottom

bc when something is on the bottom you can times it with whats on the other side to solve it

if that makes sense?

[thorn talon]

Hmm

What do you mean?

[clear haven]

cross multiplying??

[upper karma]

i mean like

when its like

2/x=2

you can go 2 ÷ 2

[clear haven]

yeah

that's all cool

[upper karma]

yeah

[brave hill]

https://artofproblemsolving.com/wiki/index.php/2018_AMC_12B_Problems/Problem_21

could someone please explain to me how they're solving this

i can't understand this no matter how hard i tried

i know the formula for incenter and shoelace but I can't visualize this image nor establish proper coordinates for everything

[exotic meteor]

Let's say we have plane

$\frac{x-x_0}{A}+\frac{y-y_0}{B} + \frac{z-z_0}{C} = 1$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

What is the vector of the normal?

n(A,B,C)?

Doesn't seem to be right... :/

Found the solution

Apparently I am supposed to use the algebric form of the plane formula

Which is

$A(x-x_0)+B(y-y_0)+C(z-z_0) = 0$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

And n(A,B,C) is the normal of that plane

[blazing kelp]

sin+cos/sin-cos+sin-cos÷sin+cos=2sec^2÷tan^2-1

Prove

<@&286206848099549185> help

[warm pine]

Can you write it out more clearly

Either LaTeX or on a piece of paper

[blazing kelp]

Actually I only need to know sin2 - cos2 = tan2 - 1

[gritty siren]

$\sin +\frac \cos \sin - \frac \cos \sin + \cos $

[somber coyote]

Tuong:

[gritty siren]

?

[blazing kelp]

$ sin2 - cos2 = tan2 -1

sin to the power 2

sin squared

[gritty siren]

$\sin^2-\cos^2=\tan^2-1$

[somber coyote]

Tuong:

[blazing kelp]

si

yes

[gritty siren]

doesn't look true

[fleet wolf]

[blazing kelp]

damn my text book prints wrong stuff

@gritty siren that is only the denominator part

let me write whole stuff

sin+cos/sin-cos+sin-cos÷sin+cos=2sec^2÷tan^2-1

[gritty siren]

$\sin+\frac\cos\sin-\cos+\sin-\frac\cos\sin+\cos$

[somber coyote]

Tuong:

[blazing kelp]

no no wrong

[gritty siren]

you're writing it ambiguously

Use parenthesises

[blazing kelp]

s and c are sin and cos respectively

@gritty siren what do you think ?

[gritty siren]

1=cos²/cos²

[blazing kelp]

yes

[gritty siren]

use that

[blazing kelp]

oh okey

ty I solved it

[lusty oar]

hey everyone i’m new

trigo and log i think is my absolute worst

[plucky marlin]

[lusty oar]

i love math a lot but i acc cannot stand trigo

i need help

for question vi, we are meant to use the R formula

so we are allowed to pick between sin and cos, as long as we put it correctly

but this bitchass question

1 is wrong and 2 is correct. why is 1 wrong. what mistake did i do

[exotic meteor]

Test

I'm trying to solve this problem

There are two lines, find the parameter lambda so that the two lines intersect

$a... \frac{x-2}{\lambda} = \frac{y-1}{1} = \frac{z-2}{0} \ b... \frac{x-5}{2} = \frac{y-2}{3} = \frac{z-3}{1}$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

Right so the first part is to convert the two lines in parametric form

[dire rampart]

dividing by 0

[exotic meteor]

No

It doesn't mean that

Apparently dividing by zero in that is not really dividing by zero, it's just a way of writing it. it does not have the same meaning

Anyways

$a... \ x = 2 + \lambda t \ y = 1+t \ z = 2$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

$b... \ x = 5+2t \ y = 2 + 3t \ z = 3 + t$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

Alright, that's correct

Now, I am supposed to find lambda

$2+ \lambda t = 5 + 2t \ 1 + t = 2 + 3t \ 2 = 3 + t$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

From this

Okay I found the problem

I chose the last one because it's the simplest to answer but apparently I need to choose one that has t on both sides

From the last one I get that t = -1

And from the middle one I get that t = -1/2

t = -1/2 is the correct answer

Or now...

hmm

Or not...

Okay, I found out the problem

I have to use different variables for the perametric form of the lines

Which means that the line b is actually

$b... \ x = 5+2u \ y = 2 + 3u \ z = 3 + u$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

Than I put the equal one to another

$2+ \lambda t = 5 + 2u \ 1 + t = 2 + 3u \ 2 = 3 + u$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

From that system of equations, lambda is -1/2

Which is the correct answer 😃

[proper citrus]

Anyways, say we have a paralellogram ABCD

AB = 8m

AD = 2m

And we know the paralellograms area

Can we find angle DAB?

umm

Anyone?

I formed an equation:
2 * (1/2) * 8m * 2m * sin (DAB) = area

But

That doesn't give a correct answer

Even though it should

No?

[lusty oar]

how is the parallelogram shaped? @proper citrus

where is the A B C D

[proper citrus]

uhh

the A B C D go clockwise and the A is at the top left corner of the paralellogram

and the parallelograms top slants towards the right

[lusty oar]

no other values?

only AB and AD?

[proper citrus]

we have the area

[lusty oar]

oh whats the area, sorry??

[proper citrus]

11.05m^2

[plucky marlin]

,w 11.05/16

[somber coyote]

Display results online and refine query

Requested by soap#7503

[plucky marlin]

,w arcsin(0.690625)

[somber coyote]

Display results online and refine query

Requested by soap#7503

[plucky marlin]

,w 0.762353 convert to degrees

[somber coyote]

Display results online and refine query

Requested by soap#7503

[proper citrus]

that's what I got

[plucky marlin]

so that's apparently wrong ?

[proper citrus]

however the correct answer is 136.4

[lusty oar]

this gave me a headache

[plucky marlin]

,w 180 - 43.68

[somber coyote]

Display results online and refine query

Requested by soap#7503

[plucky marlin]

yea

[proper citrus]

why'd we subtract from 180 now?

[plucky marlin]

you know arcsin's range is restricted

from 90 to -90

[proper citrus]

Sorry for being dumb, would you mind explaining?

[plucky marlin]

arcsin(x) cant spit out a value that's not in between [-90 , 90]

in this case DAB is greater than 90

so we need to subtract the ans from 180

[proper citrus]

are we talking in radians rn?

[plucky marlin]

oh sorry

[proper citrus]

it's 180/pi right?

[plucky marlin]

what's 180/pi ?

[proper citrus]

like it's the ratio used to convert from radians to degrees and vice versa

[plucky marlin]

yea

basically $1^{\circ} = \frac{\pi}{180}^{c}$

[somber coyote]

soap:

[proper citrus]

c?

[plucky marlin]

c for radian

[proper citrus]

ah

[plucky marlin]

like that circle thing used to show that it's in degrees

[proper citrus]

yeah no i realized

but how does arcsin's range being from -90 to 90 mean subtracting the answer from 180?

[plucky marlin]

this gotta do wiht unit circle

[lusty oar]

from the parallelogram itself, you can tell that the angle is more than 180

isn't that like the easier explanation

i mean more than 90

sorry

[proper citrus]

more than 180?

[lusty oar]

90

[proper citrus]

okay

[lusty oar]

and the value you got is an acute value

because when u do trigo and that formula, it gives u in a triangle form, hence always acute

do u follow??

[plucky marlin]

https://prnt.sc/mg1o8p

[proper citrus]

yeah but the question didnt require the value to be obtuse

[plucky marlin]

u can seen that sin of theta and 180 - theta is the same

[lusty oar]

it didn't but the fact that it's a parallelogram tells u that it should not be an acute angle

because logically, DAB is more than 90

[plucky marlin]

but arcsin wont give you that other angle

it'll just give you that boring theta

[proper citrus]

hmm, yeah youre right spider

[lusty oar]

a lot of the times, you'll get an acute angle but the question requires obtuse due to the question giving diagrams etc

i hope it helped?? idk hahah sorry

[proper citrus]

it did

<3

[plucky marlin]

my unit circle drawing was waste

😩

[proper citrus]

heh

I kinda did know that

[plucky marlin]

that's what i was trying to say man

sin(x) = sin(180 - x)

[proper citrus]

that sin(x) = sin (180-x)

ReEEE

[plucky marlin]

[proper citrus]

[dire rampart]

[proper citrus]

wait a minute

,tex sin^-1 (sin (a)) = a

[somber coyote]

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

[proper citrus]

uhh

[upper karma]

Guys whats the difference between a infinite series and an infinite sequence

[proper citrus]

,tex sin^{-1} (sin (a)) = a

[somber coyote]

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

[proper citrus]

There

That's correct, right?

[umbral snow]

Not always, for example a = 100

[proper citrus]

yeah

it returns 80

owo

Fkin sorcery

[umbral snow]

You're working in degrees!

[proper citrus]

!

[umbral snow]

Notice however that sin(100) = sin(80)

[proper citrus]

Degree users rise up!

[umbral snow]

So sin¯¹(sin(x)) will be a value within -90 ≤ x ≤ 90 that has the same sin of x

[proper citrus]

Ok thank

[umbral snow]

Cool cool. Feel free to ask if you have anything else!

[exotic meteor]

What is the correct way of doing dot products of two vectors?

$\vec{p} = \lambda a + 17b \ \vec{q} = 3a - b \ |a| = 2 \ |b| = 5 \ \angle (a,b) = \frac{2\pi}{3}$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

$\vec{p} \cdot\vec{q} = (\lambda a + 17b)(3a-b)$

[somber coyote]

Autistic Hoodie:

[somber coyote]

The 80's:

[exotic meteor]

How do you test if points are coplanar?

For example I have three points

$A(0,0,2)\B(0,1,6)\C(-2-1,4)$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

Alright, my question doesn't make sense

3 points are always coplanar, lol.

But I learned how to test if 4 points are coplanar

From those 4 points you make 3 vectors and calculate their mixed product and if it is 0, they are coplanar

[zinc grail]

hello

I was completely lost on this

what would be the formula for something like this?

[clear haven]

s=r\theta

[plucky marlin]

$S = r \theta$

[somber coyote]

soap:

[clear haven]

rip me

[plucky marlin]

i win

muahhahaha

[clear haven]

lowercase s better tho

S is arclength

r is radius

theta is the angle

[zinc grail]

3pi = 5x?

[clear haven]

yep

[zinc grail]

so 3/5pi = x?

[clear haven]

i think so yeah

[zinc grail]

how do I find an actual value though

[clear haven]

wym this is in radians

[plucky marlin]

that's the actual value

[clear haven]

so convert pi to 180 degrees i guess

[plucky marlin]

yea

$1 rad = \frac{180}{\pi}^{\circ}$

[zinc grail]

oh i thought the formula was like degrees/360 = rad / 2pi

[somber coyote]

soap:

[plucky marlin]

it's the same thing dude

just a little bit manipulated to make ur life easier

[zinc grail]

so 3/5 rad = 180/pi

I dont see how the answer is 108 from this

sorry

[plucky marlin]

$\frac35 rad = \frac35 \frac{180}{\pi}^{\circ}$

[somber coyote]

soap:

[zinc grail]

what would these type of questions be called?

[plucky marlin]

geometry....

[zinc grail]

yeah thats why I posted it here

[plucky marlin]

unit conversion

[upper karma]

hi

rectangle
because what you're doing is
you're bisecting a rectangle diagonally
thats how you get the two right trianges
if you put two equilaterals together you get a diamond
or if you switch which side oyu couple them together
you certainly do not get a rectangle```

https://www.mathsisfun.com/triangle.html

so how does

a = bh/2 work for all triangles?

[bleak barn]

I hope this is the right channel

[shadow anvil]

@bleak barn You here?

[plucky marlin]

cos^2(x) = (1 + sin(x))(1 - sin(x))

[upper karma]

Guys... Can you help me with this problem?

In the figure below, the triangle ACE is equilateral and the angle DBC = DCB = 20. Calculate the measure of the angle "x".

[glad ocean]

ok so

In triangle DBC:
angles DBC + angle DCB = 40

so angle BDC is 140°

angle AEC and angle ACE is 60 because equilateral triangle

I wanna do this problem lol

[upper karma]

How to find the amplitude of this function?

https://cdn.discordapp.com/attachments/263571731791216640/542025412885348352/Screen_Shot_2019-02-04_at_5.51.28_PM.png

9i please

[umbral snow]

y' = 3cos(x) - sec²(x) = 0
3cos(x) = sec²(x)
cos³(x) = 1/3
cos(x) = 1/∛3

[upper karma]

Lovely

[umbral snow]

Oh wait do you know differentiaton?

[winged saddle]

bruh arauto that problem is literally ipossible

[upper karma]

@winged saddle I saw it in my book... The book says that the answer is 50.. But I really don't know how to discover it...

@glad ocean I'm trying to do that question since morning

[glad ocean]

wait

In the book does it give any reason why

if not but you really need the hw grade just show work for a wrong answer

[winged saddle]

uh arauto the problem is angle E and angle D, cuz if u find 1 or the other, u can know ht answer but if u dont find them its impossible

i cant really find a way to find it using traingles, so u should probably set up a system of equations

and try too find x

i mean

try to find

either E or D

[upper karma]

@glad ocean Unfortunately my book gives no reason why for it... And I translated the question to english (because It's a book in another language)

@winged saddle Yep.. I was thinking about it.. 🤔 .. That question looks so easy but It's so confuse

@winged saddle
I was thinking of finding X by the external angle theorem, using D and C. Anyway, it did not work out ..

[winged saddle]

yeah i tried too

im 2 lazy to set up a system of equation so ill just leave it to u LOL

[upper karma]

😂

[upper karma]

so

you know that one of the angles is 7x and the other is 5x

you know that they both sum up to 90 degrees

solve for that

Do you know why one is 7x and the other is 5x?

[upper karma]

exactly so you know that one is 7x and the other is 5x

and they both add up to 90 degrees

therefore x = 7.5, one of them is 37.5 and the other 52.5

[torn linden]

Ok I have a question about ratios and area

The question: Two similar wood floors have a scale factor of 4/3. It costs $216 to refinish the smaller wood floor. At this rate, how much would it cost to refinish the larger wood floor?

would you do 4/3=216/X OR 3/4=216/X

[lyric wagon]

?

i moved it to #help-2

[torn linden]

can someone help me 😦

[lyric wagon]

sorry i forgot a little bit about scale factors from earlier in the year

so i can't remember which one is right

[torn linden]

well dang

[hardy surge]

I think I have a pretty basic question but I don't seem to be able to google it properly:

I have a point cloud in 3D. I would like to transform these points relative to a given origin. I have the location P and the orthormal basis which I want to use as new.. axes? I just want to construct the translation (mean) and rotation (this one I don't know) matrices for this, if I'm correct?

How do I go about that?

[umbral snow]

@hardy surge
A translation isn't a linear operation, so matricies can't help. You're better off creating a function. What are you thinking of doing?

[hardy surge]

Isn't translation one of the affine transformations?

Anyhow, what I need to do is project the given points (already slected) into a 2d subspace (surface). I want to calculate a 2d convex hull of those points.

The easiest way to do this is probably by translating and rotating to the origin so that the forward dimension can be ignored and for X we use object's local right and for Y we use object's up

[exotic meteor]

$|z-1-i| + |z-3-i| \ge 4$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

I asked my professor and he told me the only way to solve this is using the long claculations

I am trying to do it but it always ends up being incorrect

Or I am missing something...

I got it correctly till this

$6-x \le 2\sqrt{(x-3)^2+(y-1)^2}$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

If I share both sides

That should be like

$(6-x)^2 \le 4((x-3)^2+(y-1)^2)$

[somber coyote]

Autistic Hoodie:

[exotic meteor]

No?

This doesn't make sense

[upper karma]

@hardy surge Look up the affine transformation matrix

It involves augmenting the coordinates with an extra 1, and stripping it off later on.

[hardy surge]

I'm aware of that one, although I am not sure how to apply it.

[upper karma]

Then you can just transform the standard basis, making sure to maintain the 1 on the bottom.

[hardy surge]

ok so I have 3 direction vectors... that span the orthonormal basis relative to my selected point

but I am not sure... what to do with it... how to apply this.

[upper karma]

Lemme explain a bit better

wait that's wrong

Let's just assume that you can find the affine transform of the standard basis.

I believe you actually need 4 vectors, 3 linearly independent, to uniquely determine the affine transformation.

[hardy surge]

yes, that sounds right. Location and... dimensions

Sorry, my mathematical vocabulary is not that great... especially not in English.

[upper karma]

Anyway, you can break the affine transformation into the linear component and the translation component.

The affine part comes from translation.

So, ignoring translation for the moment, you can derive the linear part of the transformation by just transforming the standard basis in your current coordinate system.

Those transformed vectors directly become the columns of the linear part of your transformation matrix, i.e. the A part of the matrix given on Wikipedia.

[hardy surge]

oooh... I remember this, I think

[upper karma]

Then the b is just the translation that you want.

[hardy surge]

I think I get it.

I'll try it out. Thank you!

[glad falcon]

I don't know how to approach this

[astral hornet]

focus on the props of a kite

such as perp diagonals

and one of them perp bisect

and 2 sets of adj congruent sides

[rugged plume]

@everyone can anyone help me with my prove the given identity homework, im really stuck

[upper karma]

the diagram below shows two right angled triangles triangle a is mathematically similar to triangle b the perimiter of A is 12 cm what is the perimiter of B

https://cdn.discordapp.com/attachments/342850939306246145/542499577589137428/6f4f589adf1c9bb00d97f9509262af24.png

is the perimiter 60

[rugged plume]

@upper karma yes

[upper karma]

ok ty

[rugged plume]

np

[glad ocean]

ok I just F---ED UP a test

and I am salty as

fffffffff

[upper karma]

help me pls

how i do di

dis

[astral hornet]

Substitute known values into distance formula and solve for unknown

[upper karma]

yeah but it doesnt work

i cant figure it out

[astral hornet]

What "doesn't work"?

[twin ingot]

Im so lost its not even funny. Worked 2sin^2(x)-sin(x)-1=0 answers in range of [0,2pi). I got 3pi/2 and pi/2. Answer sheet says pi/2, 7pi/6 and 11pi/6 I have no idea where I went wrong

wait... typing it I see where I screwed up

[upper karma]

uh can someone help me

like asap pleas

[plucky marlin]

sure fam

[fossil fern]

sup

[glad falcon]

Is there a formula to order of rotational symmetry for polygons??

[vivid ridge]

@glad falcon Pretty sure that's just the cyclic group of n vertices so the order is n

[rapid hatch]

exercise: The point P' is the refelction of the point P(-1,10) on the line a:x+5y=10 . What's P

like im trying to get the distance of l Pa l with this : lax1+by1+cl / Sqr a2+b2

i end up getting 6 and on geogebra it shows that i should get 7,6....

[rapid hatch]

<@&286206848099549185>

[jovial sierra]

je ne comprends très bien la question. il me semble que le point P est donné, (-1, 10), mais il faut trouver P?

[rapid hatch]

la reflection de P' est P(-1,10) il faut trouver P'

distance de lPal est 7.65

il faut trouver P' , les coordinates

[jovial sierra]

hmmm, moi je commence toujours par dessiner

[rapid hatch]

je sais que P est (-1,10)

je sais pas P

[jovial sierra]

si vous savez B, vous pouvez trouver la difference entre P et B

[rapid hatch]

je sais pas les coordinates de B

c'est dans a:x+5y=10

[jovial sierra]

oui mais...

dsl ça fait des années depuis que je fais les problèmes de ce style

[rapid hatch]

:/

[jovial sierra]

(et le français n'est pas ma langue natale... ayez patience svp)

[rapid hatch]

quoi est ton langue natale?

[jovial sierra]

l'anglais

[rapid hatch]

we can speak english if thats easier

[jovial sierra]

mais je préfère pratiquer mon français toutefois

[rapid hatch]

c'est difficile pour moi

[jovial sierra]

tiens, la ligne PB est la ligne perpendiculaire à la ligne donné, et qui contient P

donc si on sait le formule de la ligne donnée, on peut trouver le "slope" (pente?) d'une ligne perpendiculaire

et puis trouver la ligne PB

qui nous donne le point B car c'est l'intersection

et puis on ajoute [la difference entre P et B] à B

vous avez fait l'addition des vecteurs auparavant?

[rapid hatch]

its without vectors

we havent seen vectors with these kind of exercises yet

[jovial sierra]

je crois qu'il y ait une méthode très simple que je ne parviens pas à trouver

[rapid hatch]

hmmmm

what is a direction vector

[jovial sierra]

it's a vector pointing in the direction that the line is going i guess

okay but if we can find B, then we know that to get from P to B you add or subtract some amount from the x-coordinate and a different amount from the y-coordinate

and then you can just do the same thing to get to P'

[rapid hatch]

how do u get B

[jovial sierra]

by finding an equation for the line PB and then finding the intersection of PB and the original line

[rapid hatch]

im like 100% sure theres an easier way

[jovial sierra]

me too

i do know that my method will work but there's probably a trick you can do that shortens the work considerably

[rapid hatch]

watching a vid on ur mtethod

im giving up lol

i have three more of those kind to do

its 11 30 pm

[worthy nimbus]

Does anyone know of a video

That explains how to graph circles

[silk sequoia]

Hey guys

Im@new to this sever, but can you any of you guys help me with this geometry question

[dapper patio]

maybe

[silk sequoia]

Q#6

[winged saddle]

bruh

ok well

since abcd is a square

then ah = dg = cf =eb

ok well thats simple

letss call ah = a

and ae = b

since all of these triangles all congruent, then all of them are ab triangles, therefore, they all have the same hypotnuse, c

so, efgh is a square, because all of its sides are the same length

i dont know what paragraph proof is

sorry

well

for b), since efgh is a square,

a square just has 90 degree angles

so...

not sure y they asked that thas simple ;-;

@silk sequoia

[supple haven]

Im trying to simplify and im lost now

[ripe sigil]

1-cos^2 is sin^2 right?

and csc^2 is 1/sin^2

[supple haven]

Gmsudjeusis

Thanks

Final answer is 1 right?

Also sorta lost on this one, it wants me to factor but im unsure what i ought to factor out

[umbral snow]

Difference of squares on the numerator

@supple haven

[supple haven]

Difference of squares?

Im blanking

Ohh

[umbral snow]

It's immediate once you do that

[supple haven]

Im left with sec^2 x - tan^2 x

That should be one of the identities i thini

Final answer is 1 right?

Is this correct?

Im sorry for asking so many questions, ive got a test coming up and ive been having a ton of trouble with these

[umbral snow]

Looks good to me

[supple haven]

Rad thanks

[umbral snow]

Trig identities are really algebra exercises in disguise. Always try to:
Factor (Difference of squares)
Add fractions together

Those are the common two lol

[supple haven]

Thanks :D

[umbral snow]

I swear there was more than that, I'm drawing a blank

Well, good luck! Feel free to ask if you have anything else

[supple haven]

Thanks fam, ill peobably be asking at lwast another 10 times haha

[umbral snow]

Great! I can't wait

[supple haven]

Would #5 be correct? The question is asking what values theta can be while still having the equation be true

Theta is also in between 0 and 2pi

[supple haven]

What did i do wrong here?

Susiiss i wanna @ the helpers but i gotta wait 10 mins

[fossil fern]

dude fuck I dont wanna do my trig hw this shit gives me aids. all thos identities and shit ahhhhhh

[supple haven]

Honestly same

I was doing great up until a week and a half ago

[fossil fern]

yeah dude second time taking this class and I know I can do decent but like fuck this stuff is so frustrating when u cant have access to a teacher to ask a sumple question. we’re luck we have resources to go to but still fuck me

I dont even wanna think about calculus

or anything above trig haha

[supple haven]

I dont have internet access

And i was pretty much surviving off of Khan Academy but cant use that now aha

[fossil fern]

What u mean bo internet access? u on it now

[supple haven]

Data

Using my phone

[fossil fern]

No wifi at home?

[supple haven]

Dickhead father took it away

Grades have plumetted since then but doesnt matter to him at all

[fossil fern]

Ah I see thats annoying as fuck. tell him u need it to help with homework. im sure u have but he should realize if he’s not dumb.

Beat bis ass

[supple haven]

Hes very dumb

Completely incompatable with reason

[fossil fern]

Well goodluck w ya homework Im gonna do mine tomorrow morning I’ll look for ya on here and show u what I got goin.

im graphin as well as the identities lol woooo

[supple haven]

Rad

Graphing is kind of a pain in the ass but its pretty much the same as the ordinary shifting rules

[silk sequoia]

@winged saddle all thank you !

[winged saddle]

your wecloem

[upper karma]

i require some assistance with geo homework

i literally have no idea whats going on

[tropic island]

What is it?

[upper karma]

https://cdn.discordapp.com/attachments/496892902769688576/542896188005941258/20190206_213526.jpg

you'll havet o expand it to actually make out the numbers

anyone got any ideas?

[winged saddle]

wat

wat r u suposed to do?

write equation of circle?

[upper karma]

i don't know

thats the problem

[winged saddle]

bruh

[west moat]

What

U dont even have a wuestion u just have equation

[upper karma]

my friend is a dick and only sent the problems instead of the equations

[winged saddle]

yeah it doesnt even say antyihng

[west moat]

I think ur supposed to graph it

[winged saddle]

bruh

[west moat]

wait

no

[upper karma]

i'll send the other side, it actually has the question on it

[west moat]

Wtf

[winged saddle]

ok

[upper karma]

https://cdn.discordapp.com/attachments/496892902769688576/542896188928819211/20190206_213518.jpg

[winged saddle]

wow u should have told us earlier

CENTER THE RADIUS

AND GRAPH IT

BRUH

i mena

find the center

and the radius

[upper karma]

why are you yelling at me

[winged saddle]

ok

[upper karma]

don't hit me

[winged saddle]

im nat

im

bruh

ok

i dont-

oK sryy i will actually do it

[west moat]

No he has to write the equation

[winged saddle]

uh

[west moat]

not graph

[winged saddle]

just write the equationa nd then graph it

[west moat]

do u know equation of a circle?

[upper karma]

no

[winged saddle]

ok

[upper karma]

she started talking about circles and i got scared and turned my ears off

[winged saddle]

its in the form (x-a)^2 +(y-b)^2 = c

let me explain

c is the radius squared

so if u want to find the raidus

u take sqrt of c

[west moat]

a and b represent the centre of the circle

[winged saddle]

yes

so if lets say a = 5

then (x-5^2)

oops

i mean

(x-5)^2

x=5

so that it willmake it 0

and u can find y and sketch a point

same thing for y

find hwat valu eof y will make (y-b)^2 == to 0

[upper karma]

so

[winged saddle]

and u can find x

and sketch a point

[upper karma]

lets take the first problem as an example

[winged saddle]

ok

[upper karma]

on the second page

[winged saddle]

ok

so

for this example

u sketched it wrong

[upper karma]

well

[winged saddle]

x = 1/2

[upper karma]

i didn't

that's my friends work

[winged saddle]

bruh

ok well he did it wrong

[upper karma]

dog i wasn't here to get this worksheet

[winged saddle]

x is 1/2 in this case lol

XD

[upper karma]

so i asked him to send my a photo

[winged saddle]

his y is good

[upper karma]

yeah

[winged saddle]

ye his radius is 3

so

[upper karma]

x is 1/2 cause it has to equal 0

[west moat]

um i think all the answers are wrong

[winged saddle]

he did it right for ther adius

[west moat]

lol

[winged saddle]

bruh

[upper karma]

so

[winged saddle]

im just explaining

shh

[upper karma]

x = 1/2 and y = 5.2

5/2

[winged saddle]

yes

that is corerct

so

[upper karma]

and the square root of 9 is 3

[winged saddle]

for the questions on the first page

yes

so thts radius

ok

anyway

[upper karma]

now i need graph paper

lemme grab it

[winged saddle]

ok

for the questions on the first page

u can put the equations into the form (x-a)^2+(y-b)^2=r

and then ssolve for radius and center

etc

there u go

gtg b4 teacher sees me

[upper karma]

ok

thank you

brother

[west moat]

U can show me r working out

I think i can help if u need

[upper karma]

so i'm confused

i get that you can calculate x and y

[west moat]

show me what uve done

[upper karma]

but why is x - 1/2 squared?

on my paper

[west moat]

Um what question is this

[upper karma]

i've written (x-1/2)^2 + (y-5/2)^2=9

number one second page

[west moat]

question 7?

[upper karma]

questin 1

https://cdn.discordapp.com/attachments/496892902769688576/542896188928819211/20190206_213518.jpg

[west moat]

oh oh my bad

[upper karma]

is fine

[west moat]

ok so (x-1/2)^2 means the centre point is at 1/2

For the x coordinate

[upper karma]

alright

i get that

[west moat]

same applies for the y

so what would y be

[upper karma]

5/2

is that right?

[west moat]

Yep good so the centre is at (0.5,2.5)

mark that on ur graph

[winged saddle]

shay

[west moat]

ye

[winged saddle]

an intuitive sense

is that

[west moat]

And the 9 is the radius squared

[winged saddle]

becasue there is 2x values for every 1

on the circle right

like if y is 3

then x can be something and something

[upper karma]

i can't find any fucking graphing paper and that's an oof

[winged saddle]

but they are not the same values

so u have to take the square root

to get 2 values

[west moat]

Just draw it

[winged saddle]

so u can plot a circle

or else u will just get 1 value for 1 y

[west moat]

and label the coordinates that are important

[winged saddle]

1 value of x for 1 y

sry im bad at explain

[west moat]

Ok so we found the centre of the circle, now we need to find the radius which is the root of 9

[upper karma]

yep

3

[winged saddle]

shay

[west moat]

Yep so from the centre

[winged saddle]

for example

lets say y = 0

then there is 2 x values possible

if u only have linear equation

u can only have 1 x value

per 1 y value

if its squared

u can take the square root

u get a positive and a negative answer

so there is 2 answres

and so 2 x values

[upper karma]

alright

[winged saddle]

i dont think i make u understand

[upper karma]

you make me understand

but i have to move on, i need to get this done fast

so the radius is

[winged saddle]

K

[upper karma]

3

meaning the diameter is 6

[west moat]

So from the centre

u draw 3 out

so the centre is 0.5,2.5

I think u can do that

fuck im shit at explaining

[upper karma]

i did it

how do you find the height of the circle?

[west moat]

what do u mean the height?

[upper karma]

aren't i trying to try and draw out the entire circle?

[west moat]

Ye

[winged saddle]

yes

[upper karma]

so i wrote out the diameter

but now it's just a line

[winged saddle]

the height is the diameter

ok

well

so now u know the center of the circle right

and u just take a compass

and draw it

[west moat]

Just add 3 to the left right top bottom from the centre

and then connect it

[winged saddle]

ye

[west moat]

and u got ur circle

[winged saddle]

[upper karma]

so for the one that goes down

is it 0.5, -0.5

[west moat]

Yes

And then to the left

which should be -2.5,2.5 i think

[upper karma]

yep

[west moat]

and then u do top and right

and u should have 4 points and just connect it

[upper karma]

yep

[west moat]

Ok send me a pic of it

[upper karma]

i don't have my phone

[west moat]

Oh ok

i think u got it anyways

[upper karma]

yep i think i did

[west moat]

Im gonna sleep its late and i gotta wake up early

gn buddy

[upper karma]

night

[tropic island]

Look at my name!

[winged saddle]

wow!

am i supposed to call u 3.1215926.... or pi

[upper karma]

https://cdn.discordapp.com/attachments/496892902769688576/542896188928819211/20190206_213518.jpg

how do i solve 3 or 4?

[winged saddle]

u wanna

put it in form

(x-a)^2+(y-b)^2=r

where r is radius squared

so

its kinda mixed up rn

u probably

waanna

group the x terms together

so

u have x^2

[upper karma]

mhm

[winged saddle]

8x

and nothing

so u wanna complete the square

[upper karma]

i tihnk it's 8y

[winged saddle]

to make it a peffect square

[upper karma]

it's just hard to see

[winged saddle]

and make it (x-a)^2

ok

anyway

it will become

wait

shay

[upper karma]

yes

[winged saddle]

did u learn how to complete the square

in quadratics

[upper karma]

a long time ago

[winged saddle]

ok

so u know?

or u forgot?

[upper karma]

no

forgot

[winged saddle]

ok

so

lets say

u have

x^2+8x=0

and u wanna make it

(x+4)^2

=something

u wanna make a perfect square

like in this situation

for the circle

so

wat u wanna do

is make a constant term

understand?

so

what u do is

u divide 8x by 2

u get 4x

and u take the coefficient

which is 4

u square it

it becomes 16

and that is ur constant term

so

look this is what it becomes

x^2+8x+16-16=0

its an equation remember so if u add 16 u gotta substract it

[upper karma]

mhm

[winged saddle]

and u can make x^2+8x+16 equal to (x+4)^2

right

so u just havve

[upper karma]

yes

[winged saddle]

x+4^2

= 15

i mean

=16

i just moved the 16 to the other side

ok ?

so lets dot hta

for the circle

so

i already did for x

so

now it become

(x+4)^2+y^2+8y+28-16

right

so now u wanna put the ys togehter

[upper karma]

yes

[winged saddle]

same thing

u take the coefficient on the y^1 term

which is 8y

u divide it by 2

4y

u square it

16

and then u have ur constant term for the perfect square

so now it bcome

[upper karma]

so it's the same thing for the x that it is for the y

[winged saddle]

(x+4)^2 +(y+4)^2 +28 -16 -16

well

in this case yes

becuz it is all 1second power term

and 8 for the first power term

but other wise it would be differnt

and then

[upper karma]

alright

[winged saddle]

u have ur equation

so u just simplify for radius

it become

(x+4)^2+(y+4)^2=4

so radius is 2

and center is at -4, -4

there u go

do u have question

[upper karma]

no

i got it

[winged saddle]

ok

u understand?

if no i can explain the next 1 aswell

[upper karma]

mhm

alright

[winged saddle]

ok

[upper karma]

if i don't get it i'll @ you

[winged saddle]

alrigh

[vapid condor]

Can someone explain me Betwenness in definition?

[winged saddle]

no comprendo sire tthats grraph theory

im bada

bad

[vapid condor]

What about vertical angle theorem?

[winged saddle]

?

wdym

vertical angles

bruh

[vapid condor]

Lol. That is our lesson for today. ;-;

Vertical Angles Theorem

Wait.

[winged saddle]

vertical agnles

https://simple.wikipedia.org/wiki/Vertical_angles

are u in middle school?

or something

[vapid condor]

Highschool

[winged saddle]

bruh

u dont learn vertical angles in high

[vapid condor]

Grade 8

[winged saddle]

but thats not highschool

[vapid condor]

Wutt.

[winged saddle]

flkwfl

are u not in us

US

oops

[vapid condor]

Nooo

[winged saddle]

???

no comprendo sire

[vapid condor]

Idk what you are talking about. But I'm not living in US

[winged saddle]

ok

[vapid condor]

It's our lesson in 8th Grade

[winged saddle]

o ok

[vapid condor]

ABout theorems ;-;

[winged saddle]

i learned that in 7th grade laowle

its basically

[vapid condor]

Oh

[winged saddle]

this

the mesure of angle 1 = angle 3

and the mesure of angel 2 = angel 54

i mean 4

not 54

[vapid condor]

Wut is that?

[winged saddle]

it works for any of these lines

like say u have 2 lines

that intersect

like they cross eachotehr

at any point

ok and it will make 4 angles like this:

there

yeah

it is always true

that angle 2 = angle 4

and angle 1 = angle 3

[vapid condor]

Oh okayy.

Can it be 1 and 2?

Nah
?

[winged saddle]

no

those equal 180

[vapid condor]

Oh okay

[winged saddle]

if u add them

[vapid condor]

If you add 1 and 2 its 180?

[winged saddle]

yes

u know that?

[vapid condor]

Yes

[winged saddle]

suplementary angles?

[vapid condor]

Yesss

[winged saddle]

oh ok

yes make sense?

[vapid condor]

How did you know that its 180 tho

[winged saddle]

ok

u have a line

[vapid condor]

Oh

Okay

I already get it

lolll

[winged saddle]

u split it

its still 180 if u add them

a +b still 180