geometry-and-trigonometry

[upper karma]

Yeah :)

Id love to

mmmm

well put in a request? not rlly sure how these things work but hhhhhHHHHHHHH

Me neither lol

oop i smell radiation, gotta head out

[upper karma]

Lol

[nova sand]

Helloo, if i have a square with 4 half sirkles in. and i need the find the circumference of the square without the 4 half sirkles, cant i just take circumference of the full square - circumference of the 2 sirkles?

[limpid tide]

do you have a picture

[nova sand]

need to find the circumference of the yellow part

[upper karma]

well, if 4 cm is the diameter of the circle, how long is the radius?

[nova sand]

2cm

[upper karma]

good

and do you know the formula for the circumference of a circle?

because this is half that

(since it's half a circle)

[nova sand]

pi*d

[upper karma]

https://www.google.com/search?&q=formula+for+the+circumference+of+a+circle

but yes, I suppose 2r = d

[nova sand]

but can i do circumference of the full square - circumference of the 2 sirkles?

for the circumference of the yellow part

[upper karma]

there are 4 half circles, yes

or you could think of them as 2 full circles

[nova sand]

yeah thats what i did

[upper karma]

then, yes

perimeter of the square (4 * 8) minus the circumference of 2 full circles

(pi*4 * 2)

[nova sand]

32

[upper karma]

minus... ?

[nova sand]

wait no

6,8

thats what i got

but the solution in the back of the book says 41

wich is why i asked here

[upper karma]

"6.867258771281" is closer to 6.9

if you round it off

[nova sand]

yeah

[upper karma]

you sure it's not area that they're asking you to find?

[nova sand]

they are asking for both

[upper karma]

it can't be 6.9....

hmmm

[nova sand]

i got the area correct and then 6,8 on the ones we just talked about

but it says 41...

confused

[silent plank]

perimeter of orange figure = perimeter of square - 4diameters + 4semi-circle arcs

[keen zenith]

hello, guys hi. so i am taking a trig test today and its not timed, but on a worksheet and i'm so lost

[nova sand]

-4 diameters? @silent plank

[keen zenith]

so can i send this test to you guys and someone possibly do it so i can see how they did it? I can use my outside sources so

[silent plank]

those 4cm parts

[nova sand]

all 4 of them?

[silent plank]

yes

[upper karma]

UPDATE: this is wrong 😦 needs to say 12.57/2 because it's half of a circle

8 * 4 = 32

as I was saying... @nova sand 8 * 4 = 32, but you need to subtract the diameter of each half circle from it

[nova sand]

okay

[silent plank]

subtracting 4*4cm from the perimeter of the 8cm square will get you the length of the red parts

[keen zenith]

i dont know if my answers are even correct

[upper karma]

@keen zenith we'll get to you in a moment 🙂

[keen zenith]

@upper karma okay sorry. thank you though~

[silent plank]

we're not allowed to help people in tests

[upper karma]

even if it's a finished test?

[keen zenith]

im allowed to get other people to help me though

[upper karma]

or wait, are you supposed to send this in the mail or something?

because then no

sorry

[keen zenith]

im not supposed to send it in

[nova sand]

ahh okay @silent plank

[keen zenith]

like my instructor said that im allowed to get other people to help me

im not lying either

our entire class is working together

[upper karma]

🤔

[silent plank]

and then you just need to add on those semicircle arcs

not sure why it's even called a test then

[keen zenith]

it was a test before we went online and then everything changed and she is keeping the same sheet for us to do

[nova sand]

this is what i did

nvm u guys cant understand norwegian lmaoo

[silent plank]

so is it actually just homework now? is it getting assessed?

[nova sand]

im not quite understanding what u mean i need to take away

[keen zenith]

its homework, not graded, but we need it to review for our upcoming exam so its helpful if we have the answers and work

[silent plank]

to find the full perimeter, you'd need to add arcs of those 4 semi-circles

which would be equivalent to added the circumference of 2 full circles

[nova sand]

so its not 8+8+8+8

[upper karma]

8+8+8+8 = 8 * 4

which is the formula for the perimeter of a square

[silent plank]

calculations are split into multiple parts

as mentioned earlier do you see how:
4*8 - 4*4 gets you the red stuff?

[upper karma]

to get the measurements of the ones in red, you need to subtract the small line from the big line

[nova sand]

yeah Ramonov i see that

[upper karma]

you do this for every side (meaning you do this 4 times)

and you get the ones in red

then just add the the half circles (either 4 half circles, or 2 full circles)

[silent plank]

and you know how to calculate the length of the arc of a semicircle?

[nova sand]

so thats gonna be 16 on the reds

[silent plank]

yes, the reds will have a total length of 16cm

[nova sand]

pi*r^2 for calculate the length of the arc of a semicircle

?

[silent plank]

no, that is the formula for the area of a circle with radius r

[upper karma]

that's for a full cricle

C = pi * d = 12.57
but since you need half of that...

or you could think of it like 12.57 is for a full cricle, and you need two full circles

[nova sand]

C is the arch? sorry if im not understanding quite, blame it on the language barrier

[upper karma]

C = circumference (of the circle)

[silent plank]

the length of an arc is just: r*theta
and for a semi circle it is: pi*r

[nova sand]

so 12.57*2

[upper karma]

yes

41

does it say to which decimal to round it off to?

[nova sand]

nope

[silent plank]

an arc here refers to a part of the circumference

[nova sand]

wdym 41 DSpider?

[upper karma]

16 + (12.57 * 2) = 41.14 = 41

[silent plank]

if it doesn't ask you to round,

you should leave your value exact

(16 + 8pi) cm

[umbral blaze]

Help

[nova sand]

i think i understood now Dspider and Ramonov

[upper karma]

@umbral blaze https://www.youtube.com/watch?v=I1C-fJyMQ1M

sure thing

[nova sand]

thanks kings 👑

[upper karma]

I agree that if they don't mention anything about rounding it off you should leave it as (16 + 8pi) cm

@umbral blaze I also solved it last week here using the law of sines: https://discordapp.com/channels/268882317391429632/326138757474680852/702219100176384105

if you pick a side and give it a value (e.g. "1.0"), you can figure out all angles from there using that law

what's weird is that you posted the exact same picture as that guy did

with the same gradient at the top

🤔

[strange laurel]

Because <DBC is a 90° angle, i can assume b is mdpt right?

[upper karma]

B is a mid point, yes

[strange laurel]

ok ty

[upper karma]

because you can also think of it like this:

ADC is an isosceles triangle

which means BD splits it in half

since these angles are congruent:

[strange laurel]

any1 know how my friends are getting 58 as the measure of angle CBD

[twilit zenith]

It seems to me that angle CBD should be 116° - like the angular magnitude of the arc CD.

Could it be that they were referring to CAD, by any chance?

[strange laurel]

yeah i also thought that it should be 116, but i don't think that they were referring to CAD bc the problem asks to find what BDC is

[twilit zenith]

I see.
B is the center of the circle, by the way, right?

[strange laurel]

it doesn't say in the problem.

i thought you could assume if it looked like it was

[twilit zenith]

I assume that was the intent in the question and they forgot to mention it, but I'm not sure.

Could you please take a picture of the entire question, just to be sure?

[strange laurel]

ok

[acoustic jungle]

you're trying to find BDC?

yes DBC is 116

that's defined.

[strange laurel]

ohk idk why my friends are all getting 58

[acoustic jungle]

they must be wrong then.

[little coral]

how would I draw -390 degrees on a coordinate plane?

[acoustic jungle]

draw -30 degrees

[little coral]

how?

[acoustic jungle]

tan -30 = what

it will be the slope

[little coral]

That’s how I drew 240 degree

[acoustic jungle]

-390 degrees is basically -30 degrees.

because 360 is a full rotation

and -30 is also 330 degrees

from another full rotation

[little coral]

oh ok

[dim gyro]

halp

im clueless as of what to do

i just know to use the pythagorean theorem

[acoustic jungle]

do you know the 30 60 90 triangle.

or at least or to derive that.

You would use the pythagorean theorem to derive that.

try finding the height of the equilateral triangle.

I'm going to tag you

@dim gyro

[dim gyro]

oh sorry

[acoustic jungle]

what are you sorry about lol.

[dim gyro]

not responding

[acoustic jungle]

it's ok.

do you understand what I mean.

[dim gyro]

somewhat

like this

[acoustic jungle]

what do you think the height of the triangle is.

[acoustic jungle]

no.

triangle ABC

you don't know what those degrees are.

[dim gyro]

m

[acoustic jungle]

Find the height of triangle ABC then you can apply the pythagorean theorem to find BE.

[dim gyro]

mm ok

wait so i think im doing something wrong

@acoustic jungle how do i find out the height of ABC

[agile tree]

Would anyone willing to help me?

[weak shoal]

Sure, what do you need help with?

[agile tree]

If the relation above describes y as a function of x.

@weak shoal

[weak shoal]

Yea sure

So what are you struggling with? What do you need to determine if the above is a function of x?

[agile tree]

thanks bud

yea im not really understanding the question

[weak shoal]

Don't multipost, by the way

[agile tree]

oh ok

not sure

[weak shoal]

Right, so let's think about this logically, okay?

[agile tree]

wich one is the right chat

[weak shoal]

The idea here is that every x should be mapped to a unique y

[agile tree]

oh ok so this would be a relation?

as it would be a function right?

[weak shoal]

Yea if it was a function, every x must be mapped to a unique y

In other words, if i pick a specific x, i shouldn't be getting multiple values of y

So, let x = 3, okay?

[agile tree]

so this relation describes a function

[weak shoal]

Then, 3 = y^2 -6 => y^2 = 9

No listen to me

[agile tree]

oh ok my bad

[weak shoal]

I'm literally walking you through that right now

So, anyways, we picked a specific x and we've said that y^2 = 9

But y = 3 or y = -3

In other words, there are two possible y-values that this x-value could be mapped to

So, it's not a function

[agile tree]

wait so

only one function per

relation

usually?

[weak shoal]

You're getting the wrong idea there

A relation is either a function or it is not

That's something you decide once you look at the properties of the relation

[agile tree]

oh ok i see

[weak shoal]

So, we have a clear definition of what a function should be and if a relation satisfies that, it is a function

[agile tree]

for every y there is one specific x right

oh ok i appreciate it im getting confused in my head

[weak shoal]

No

For every x, there must be one specific y

[agile tree]

thanks appreciate you bud

honestly helped me i am getting these confused for some reason

[weak shoal]

It takes time to get used to them

So take your time

[agile tree]

ok ima try some more home work problems like the ones i just asked about

[weak shoal]

Alright 🙂

[outer smelt]

can some one help me with this

[acoustic jungle]

apply the tan function twice

@dim gyro search up 30 60 90 triangle and proof.

[urban egret]

@outer smelt what did you start with?

[agile tree]

How bout for a question like this one @weak shoal

[outer smelt]

@urban egret what do you mean by that

[urban egret]

@outer smelt what work have you completed so far

[outer smelt]

in online school we have learned tan sin cosine

[agile tree]

can anyone help me our

out

[urban egret]

@agile tree if someone does not help you I will but im helping @outer smelt now

[agile tree]

ok sounds good thanks bud

[urban egret]

You also requested someone and please stay in one room

[agile tree]

i would appreciate it

i am

learned my lesson

thanks

[urban egret]

@outer smelt what about the angles of the triangle I would start with the triangle on the left to find the length where both triangles meet

[outer smelt]

how would you find that

[urban egret]

SOHCAHTOA have you heard that before

[outer smelt]

yea

[urban egret]

Ok so let us look at where it meets

[outer smelt]

ok

[urban egret]

Your angle is 65 your hypotenuse is 26 and you don’t have lengths where it is opposite or adjacent do you agree?

[outer smelt]

opposite

[urban egret]

I was asking a question to see that we don’t have given lengths of adjacent or opposite?

[outer smelt]

yes

[urban egret]

Ok awesome so we would only need to see what equation we would use out of SOHCAHTOA to find the opposite angle as you said earlier

Sine and tangent is something we can use

Cosine can not help us solve the length of opposite

We would use sin because we have a given length for hypotenuse and we can solve the opposite length

[outer smelt]

ok

[urban egret]

So sin(65)=o/h

[outer smelt]

ok

[agile tree]

top cat

will you be able to help me out?

[urban egret]

So our h is 26 and you can solve for the opposite does that make sense. After you solve for the opposite you can use the angle 47 degree to solve for the x variable

@agile tree I’m still consulting with @outer smelt I need to make sure he understands this I will get to you when we are done

[outer smelt]

yes

[agile tree]

ok thanks

[outer smelt]

wait what is the eqaution for the opposite

[urban egret]

You would use sin(65)=o/26

[outer smelt]

ok

[urban egret]

You can multiply the 26 on each side to solve for the opposite

@agile tree

[agile tree]

wouldnt it be square root of x-3

or +3

not really sure

[urban egret]

Rad(X+3)

[agile tree]

wait so like this

[urban egret]

You go the opposite than what you think it is

When it is left/right

[dim gyro]

@acoustic jungle so like this

[agile tree]

so it would look like this

[urban egret]

It is in the radical

[agile tree]

sure?

[urban egret]

Positive cause your going left/right

[agile tree]

ok thanks

and one question also

what is this asking

not understanding this

rewritten function

[urban egret]

Write it similar to that equation

Look up completing the square

[agile tree]

so x^2(-4x-5)^2+k

thats how?

@urban egret

[urban egret]

No doesn’t look like it

[agile tree]

jeez how im so lost

[urban egret]

When you multiply it out is it equal to what they gave you?

[agile tree]

nope

how wouls i do this?

looked it up

[urban egret]

Completing the square should be looked up

[agile tree]

its saying

divide by 2

[urban egret]

Yes the middle term you divide by 2 then square it to add to both sides of the equation

[agile tree]

so

[urban egret]

Let me find something

[agile tree]

ok

[upper karma]

factor out a 1 from the 1st 2 terms

[agile tree]

ok so

i followed the steps

this is what i got

[urban egret]

Cause they are asking you to put it in vertex form

[agile tree]

plus or minus square root of 13 over 2+ 1/2

now what do i do with that

after i completed the square

like how would i put it in the equation

[urban egret]

You leave everything on the same side

[upper karma]

you plug it into x

[urban egret]

Look up put it in vertex form using completing the square I feel it is too convoluted to explain on here

You have to leave everything on the same side

I put up the wrong picture

[upper karma]

yah i was gonna say that doesn't look correct

[agile tree]

wait

it didnt load

so

i did it worng

[urban egret]

https://www.chilimath.com/lessons/intermediate-algebra/completing-the-square-step-by-step/

I was trying to find a simple screenshot but I couldn’t follow the directions on this link it does it step by step everything needs to be on the same side.

[agile tree]

thats the exact form

now how would i plug it in the equation

@urban egret

after completing the square

[upper karma]

you don't have an x in your equation

[agile tree]

so what would the rewritten function look

thats completing the square is what i got

[upper karma]

Yes

[agile tree]

xi yes what

[upper karma]

i think that's just the point you plug in for x

[agile tree]

yea ur write

then how would i rewrite it for y

[upper karma]

i think you plug it in to the factored equation

then you get an answer

[urban egret]

@agile tree I posted a link above for you to follow my previous post I posted the wrong thing

They want you to rewrite as the vertex equation.

[agile tree]

yea i figured it out i had to do another similar one earlier but it asked for the vertex form not this one was confused

thanks tho @urban egret appreciate u bud

and also thanks to you @upper karma

[upper karma]

np

anytime

[urban egret]

@upper karma somebody thanked you at least 😑😑😑

[upper karma]

yah ik

nova be hating on me

you were the main help tho @urban egret

[urban egret]

Yes but I don’t know what happened and you tried as well that is all that matters @upper karma

Clarification does help I just don’t know what happened

[upper karma]

yah, but you solved the question while i was blocked

lol

[urban egret]

Shoot so you were really blocked I thought they were just giving an idle threat

[upper karma]

She seemed upset and she said I'm blocked so I assumed she wasn't messing around

[dull remnant]

if it's 4sinx=2sinx+1
just making sure, is that the same as 2sinx=1

like subtract the 2sinx from the 4sinx

that works?

[versed river]

yes that works

[dull remnant]

ty

ty

[novel flax]

yes

so 2sinx = 1

sinx = 1/2

so x = 30

30 degrees

or pi/6 radians

[dull remnant]

yes

ty

@novel flax can u answer one more question rq

so ik how to do 2sin^2x-cosx-1=0 because i can use the pythagorean identity but i can't do sqrt3cosxtanx-cosx=0

because there isn't a pythagorean identity for tan

[versed river]

write it all in terms of cosx and cos^2x using the pythagorean identity

i.e. sin^2(x)=1-cos^2(x)

then come back if you still don't know what to do after that

[dull remnant]

but

ik how to do it when it's sin and cos

but i cant do it when it's cos and tan

[versed river]

o

[dull remnant]

do i need to use pythagorean identity of 1+tan^2x=sec^2x

[versed river]

sqrt(3cosxtanx-cosx)=0?

[dull remnant]

no

sqrt(3)cos^2(x)*tanx-cosx=0

[versed river]

write tan as sinx/cosx for a start

[dull remnant]

ok i did

i have it written as the sinx/cosx way and the sqrt1-cos^2x way

do i square everything in the sinx/cosx way

[versed river]

idk what you mean but you should have something like this?

sqrt(3) sin(x) cos(x) - cos(x) = 0

[dull remnant]

there is a cosx before the sin as well

[versed river]

.

shouldnt be

sqrt(3)cos^2(x)*tanx-cosx=0

this is the orignal equation?

[dull remnant]

lemme take a picture

wait i dont ahve my camera

[versed river]

sqrt(3)cos^2(x)*(sinx/cosx)-cosx=0

[dull remnant]

yes

then square everything?

[versed river]

then the cos's cancel

[dull remnant]

o no i wrote it wrong then

[versed river]

is this

the problem

[dull remnant]

yes

o yea

they do cancel

im just realy dumb

sry

[versed river]

dw

so then you have sqrt(3) sin(x) cos(x) - cos(x) = 0

from which you can factor cosx

[dull remnant]

so it's cosx(sqrt3 sinx-1)=0

[pallid cloud]

2sinx=1
sinx=1/2
x=30° or x=150°
Meaning x=π/6 rd or x= 5π/6 rd
The solution is not unique
There are two principal measure and infinty of solutions congruent to them modulo 2π rd

@novel flax @dull remnant

[dull remnant]

ik how to do that one

but the second one is the problenm

[versed river]

so it's cosx(sqrt3 sinx-1)=0
@dull remnant

so then you can say

either cosx=0 or sqrt(3)sinx-1=0

[pallid cloud]

Yes, but I was referring to up there when you and orange got only one solution

[dull remnant]

o yes

either cosx=0 or sqrt(3)sinx-1=0
@versed river
and then the final one for cosx=0 is x=+ or - pi/2 +2pik

but i can't get the sinx one

[versed river]

pi*k-pi/2 suffices for cosx

so sqrt(3)sinx-1=0

sqrt(3)sinx=1

sinx=1/(sqrt(3))

[dull remnant]

but on the table

sinx=sqrt3/3

doesnt correspond

wait why does pi*k-pi/2 suffice?

[versed river]

because that covers every possible value that gives you cosx=0. i.e. the values at the top and bottom of the unit circle

at k=1, it equals pi/2

k=0, it equals -pi/2

k=-1, it equals -3pi/2

k=2, it equals 3pi/2

etc

[dull remnant]

i understand

but why is it pik-

instead of pi/2+

[versed river]

pi/2+ what?

[dull remnant]

i got plus or minus

pi/2+2pik

because the original formula was

x=plus or minus arccos(0)+2pik

arccos(0) is pi/2

so it becomes x=plus or minus pi/2+2pik

[versed river]

i think i know what you mean, but pi*k-pi/2 is a lot simpler and still covers all the solutions

[dull remnant]

ok

for the second solution

sinx=sqrt3/3

it doesnt correspond on the table though

[versed river]

i have literally no idea what table you are talking about

i am not omniscent

[dull remnant]

sorry

[versed river]

its fine

[dull remnant]

radian 0 is sin 0, radian pi/6 is sin 1/2, radian pi/4 is sin sqrt2/2 and radian pi/3 is sqrt3/2

how would i make a general solution out of sinx=sqrt3/3

[versed river]

2* pi*k+sin^-1(1/sqrt(3)) and 2 π k + π - sin^-1(1/sqrt(3)),

[dull remnant]

tysm

sorry for being dumb

[versed river]

dwdw

[dull remnant]

but why did u leave it in simplest form

for the sqrt

[versed river]

do you mean why didnt i change it to sqrt(3)/3?

[dull remnant]

yes

[versed river]

because i had no reason too. if i put it into a calculator its gonna give me the same answer. maybe some teachers would prefer you rationalise the denominator, but i can't see a situation where its gonna make anything easier or harder to do anything to do with it

[dull remnant]

ok ty!

[olive scarab]

any idea to solve this problem ?
give ellipse (E) $ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a \textgreater b) $. point O(0,0). 2 points A,B are mobile on (E) that $\widehat{AOE} = 90° $. H is projection of O on AB. Prove that the length of OH is constant.

[somber coyote]

.,.:

[olive scarab]

sorry for bad english

the picture should look like this

[pallid cloud]

First thing I think of is writr the equation of the ellipse parametrically, i.e., x=acosθ;y=bsinθ

Then find xA, yA, xB, yB

Knowing that AOB=90°

So if you plug θ for B, you'll plug θ+π/2 for A

You have this metric relation for the height in a right triangle(some call it inverse Pythagorean theorem, and I hate that)

$\frac{1}{OH^2}=\frac{1}{OA^2}+\frac{1}{OB^2}$

[somber coyote]

Patrick Salhany:

[pallid cloud]

You have to do a lot of simplifications, since there gonna be a lot of trigonometric lines involved

If you get this expression to be constant

Then OH is constant and q.e.d.

@olive scarab

You can also plug θ-π/2 for A, depending on the position of A and B on (E)

[olive scarab]

yes I have thought of $\frac{1}{OA^{2}}+\frac{1}{OB^{2}}$
but I couldnt prove it's constant

[somber coyote]

.,.:

[pallid cloud]

Hmm

[olive scarab]

I tried to calcute OA and OB depend on θ and π/2 -θ too but still cant get to a constant result

[pallid cloud]

So if it didn't work, we have to do it in another way

[olive scarab]

I thought of HB.HC= OH^2 too but couldnt prive it's constant

[pallid cloud]

We have:
$ \ OH \times AB = OA \times OB \ OH^{2} = HB \times HA \ OB^{2} = BH \times BA$

[somber coyote]

Patrick Salhany:

[pallid cloud]

So if anything, we can use the first two

[olive scarab]

hmm... but none of them is a constant unit, I think to prove OH is constant by only that is impossible though

[pallid cloud]

The first one will lead to what we tried earlier

[olive scarab]

yes..?

[pallid cloud]

I tried many things other than 1/OH^2, but they lead eventually to it

So there must be a way to prove that it's a constant

[olive scarab]

I tried many thing and some lead to OH= OH so...

[pallid cloud]

Lol

You were posting for the last week or so some interesting problems about conics

Where do you get them?

[olive scarab]

it's my homework but in vietnamese

[pallid cloud]

Ah

[olive scarab]

cant find them online though...

[pallid cloud]

In what class are you?

[olive scarab]

i'm first year senior high school

[pallid cloud]

Like grade 10?

[olive scarab]

yes

didnt think that foreign school have some education system like us so

[pallid cloud]

Our education system is other than yours
We learn about conics in grade 12

And the classes aren't divided to statistics, calculus, geometry, algebra...

It's just maths, and the teacher goes through all the lessons

They can alternate between the subjects

[olive scarab]

we do that here also

[pallid cloud]

That's good

Because this way I think it's better and more correlation between the subjects, not like isolation between geometry algebra trigonometry calculus...

[olive scarab]

we learn it bit by bit, but some of them are pretty hard :( like these conics assignments

[pallid cloud]

Yeah these are not direct application for the properties

The hardest problems in conics we used to do were like those

Using no coordinates

Just geometry and geometric properties of the conic

To find locus of a point, or to show that a vector or distance is constant

And so..

[olive scarab]

yes, I often do it with coordinates but this is damn hard to think about

[pallid cloud]

In what language is maths taught in vietnam?

@olive scarab

[olive scarab]

erm...Vietnamese?

[pallid cloud]

Ah ok lol

I thought in english

[olive scarab]

if we study in international school, it would be english

English skill of our teacher improve so much through out the time, but it's still not enough for teaching something so complicate like maths in second language

[pallid cloud]

Lmao 99% of schools in my country teach maths in 2nd or 3rd language

Less than 1% teach maths in native language

[olive scarab]

where are you from ?

[pallid cloud]

Because that way it'll be complicated

I'm from lebanon

Native language: lebanese arabic

[olive scarab]

oh, you language look so complicated

[pallid cloud]

70% of schools teach maths in french
29% in english
1% in arabic

Yes it is

[olive scarab]

I like french, but it's not popular in my country so I can only learn it online or in some big city

[pallid cloud]

Yeah french is my favorite language out of arabic english and french

I like to write and read in french

[upper karma]

help

[dark sparrow]

where are you stuck?

[upper karma]

how

[dark sparrow]

...

okay so

[upper karma]

do you know the formula for the area of a rectangle?

[dark sparrow]

the first thing

the first thing you should do

IS STOP BOMBARDING THIS CHAT WITH PROBLEMS

without so much as saying whether you're done with the previous one or not

[upper karma]

😆

[dark sparrow]

do you need help with the first problem you posted? the one with the blue diagram?

[upper karma]

No i did that

I know the formula

[dark sparrow]

if you are done with a problem then please say so

so that we know that we don't need to help you with it

[upper karma]

oh ok

For the second problem

is the square foot 725?

that's what I got

[dark sparrow]

what?

why are you giving an answer in square feet

when all measurements are given in yards and the area is asked for in square yards

[upper karma]

ok

i meant yars

I got 725

is that correct

[dark sparrow]

you got 725 square yards for the area of the lawn?

no, that doesn't look right

can you show how you got 725 so i can look for where you messed up?

[upper karma]

k

So I did 12 * 38 to get the area of the rectangle which was 456

then I did 10 * 20 to get the area of the vertical rectangle which was 200

[dark sparrow]

what rectangle

why 12 * 38

is that meant to be the area of the rectangle to the left of the dotted line?

[upper karma]

OH

THE 8 ISN'T FOR THE horizontal

I see

so it's 360

which makes the answer 629?

[dark sparrow]

where did you get 629 from now

[upper karma]

12 x 30 from this

10 x 20 from this

1/2(10 x 12) from here

[dark sparrow]

and how did that give you 629 in total

[upper karma]

360+200+69

[dark sparrow]

69?

where did 69 come from?

[upper karma]

what

it's 60

so

620

[dark sparrow]

anyway, 620 is the correct answer

[upper karma]

ty

so 31 dollars?

[dark sparrow]

i don't know, you didn't give me pete's rate per sq yd of mowed lawn

[upper karma]

5 cents per square feet

[dark sparrow]

per square foot or per square yard?

[upper karma]

per square yard

[dark sparrow]

,calc 620*0.05

[somber coyote]

Result:

31

[dark sparrow]

yup

[upper karma]

double check calculations multiple times

reread question before answering

does anyone know how to find the area of the square inside

[dark sparrow]

@upper karma this channel is occupied, please move to a free #❓how-to-get-help channel!

@upper karma ok so do you have an answer i could check or a doubt i could clear?

[upper karma]

How do I do the problem

OH

IT SAYS IGNORE THE LEGS

so it's 60 * 36 * 15?

[dark sparrow]

no, why would it be that?

[upper karma]

@upper karma did you even try to solve these?

yes

[dark sparrow]

you're not soaking the table through with paint

[upper karma]

o

[dark sparrow]

you're only painting its top and lateral surfaces

you're not interested in the tabletop's VOLUME

[upper karma]

so what are we looking for?

[acoustic jungle]

Surface area.

[upper karma]

oh

[dark sparrow]

surface area, not counting the bottom face as instructed

[dark sparrow]

@upper karma you are late

[upper karma]

so 7200?

[dark sparrow]

we already got done with these

that doesn't match what i got

can you show your work for how you got 7200 cm^2 as the area to be painted

[upper karma]

what did u get

that's my work

lmao

[dark sparrow]

ew

[acoustic jungle]

please read the question

[dark sparrow]

^

don't rely on online calculators

because you don't need them

[upper karma]

Well Idk the formula

then use your insight

?

calculate area of all sides in question for surface area if you don't know formula.

it's a longer way but it works

[dark sparrow]

you don't need any formulas!!!!!!!!

other than the area of a rectangle

[upper karma]

and that formula doesn't discard the bottom side.

Well if I knew what to do why would I be here lmao

[dark sparrow]

which should be burned in your head by nwo

[upper karma]

So I do Surface area - area in order to discard the bottom?

look
question asks how many things of paint you'll need to get the top and lateral surfaces of the table, discarding the bottom side obviously.

So how do I discard the bottom

[dark sparrow]

calculate its area, then subtract it away

couldn't be easier

[upper karma]

That's what I said

SA - A

[dark sparrow]

total area minus bottom area, perhaps.

[upper karma]

alrighty then.

that's your surface area, discarding the bottom.

I'm so confused at what yall want me to do

was i right in saying SA - A?

yes

cool

if A is area of bottom figure of course.

5040

if that is the correct answer to the SA-A question, hurray. but you are not done yet.

reread the question and then tell me the last step.

I do that /720

[dark sparrow]

5040 cm^2 is the area to be painted, yes

[upper karma]

so the answer

is 7

[dark sparrow]

yes

[upper karma]

you are correct

congrats

the next question has a prism inside another prism

[acoustic jungle]

Is this an assignment

[upper karma]

It is

[acoustic jungle]

or a test.

[upper karma]

An assignment

just ask for help when u need it m8.

I got 888

is 888 correct?

[acoustic jungle]

Good job.

[upper karma]

it's correct?

[dark sparrow]

yes 888 cm^3 is the correct answer

[upper karma]

how do scale factors work

bruh

what is the formula for the volume of a pyramid?

lwh/3

[dark sparrow]

how do scale factors work
volumes scale with the cube of the scale factor

areas scale with the square

[upper karma]

can u explain it without using the word scale

[acoustic jungle]

suppose the width, height, and length is w,h and l

scaling it by 4

means 4w, 4h, and 4l

[dark sparrow]

no, since i'd have to refer to the scale factor

and that uses the word "scale"

so you're kind of tying my hands behind my back by forbidding me from using the word "scale"

[upper karma]

oh

ty fish

[dark sparrow]

but if you scale a shape by a factor of k, its volume is multiplied by k^3

and its surface area by k^2.

yw

[upper karma]

So the volume is 3* 2 * 6

so it would be 12 * 8 * 24

[acoustic jungle]

that's not the volume of a pyramid

[dark sparrow]

the volume of the original pyramid is not 3 * 2 * 6

[acoustic jungle]

that's a rectangular prism

[upper karma]

(3 * 2 * 6)/3

[dark sparrow]

there we go yes that's the volume of the original pyramid

[upper karma]

so (12 * 8 * 24)/3

[dark sparrow]

there we go

[upper karma]

so

800

wait

no

768

[dark sparrow]

yup sounds right

[upper karma]

i don't understand the wording

am i doing volume?

volume, yes

what is the volume of a sphere that has a radius of 0.5 cm?

figure out how many balls you can make with that kind of molten steel

0.52

I don't think you've tried any of these

actually trying to solve them yourself

@upper karma

what

Is the answer 10?

[upper karma]

Prove that for any triangle ABC: h<=sqrt(p(p-a)), where h is the altitude dropped from A, a is the length of BC and p is half of the perimeter of the triangle.

[sacred pendant]

Yo

[merry abyss]

For adv. functions, do we have to memorize these? Or just the sum and difference formulas?

[sacred pendant]

What do I do for this one

[merry abyss]

Picture isn’t loading for me lol gimme a sec

[sacred pendant]

Same, this app sucks

[merry abyss]

Might just be discord then

[sacred pendant]

Am I stupid or something because I literally have all good grades yet I have a low F in geometry and no matter how much tutoring or help I get I can never get it

[drifting parrot]

You're not stupid, MyMathLab is a pain in the butt to understand

[sacred pendant]

Like I’ve always struggled in math since elementary

[drifting parrot]

i'm in the same boat. Is there a specific math subject that you don't get or is it math overall?

[sacred pendant]

Like literally everything lmso

Lmao

[drifting parrot]

What math are you taking?

[sacred pendant]

Geometry

[drifting parrot]

Idk if this will help you, but there are videos on youtube that I have used.

What are you learning in Geometry rn?

[sacred pendant]

I’m just accepting the fact that I’m gonna have to take it this summer and there’s no point to keep wasting my time trying to do it

[upper karma]

i finished my assignment

thanls to those that helped

[drifting parrot]

btw, I have a math problem that I need help with

[sacred pendant]

We’re doing “volume of prism and cylinder”

[upper karma]

Ok

[drifting parrot]

ok @sacred pendant

[upper karma]

@drifting parrot Send

[drifting parrot]

@sacred pendant https://youtu.be/EDIbRdg9ozg

[upper karma]

Find area of circle and multiply it by height

Same thing with prism

[drifting parrot]

@upper karma

[sacred pendant]

Lmao wtf is that guy wearing

[drifting parrot]

he does that a lot lol

man, this picture won't load

[upper karma]

Yep

It ain’t loading for me either

[drifting parrot]

hopefully that helps you @sacred pendant

ok it's finally up

@upper karma

[upper karma]

No

It doesnt load for me lol

[drifting parrot]

You can’t see it?

[upper karma]

Nope

Ok now

[drifting parrot]

Ok. Good

What am I doing wrong in the equation?

[gentle bear]

https://gyazo.com/50d9a75a988a63ac557da1c8d3d45aa9

Having a little trouble with how to start this

[soft dagger]

do you know what is means by x and y coordinates of the vector?

[gentle bear]

adj / opp ?

[soft dagger]

yes and no, you use those concepts to find their values

but the coordinates themselves are just the magnitudes in the x and y direction that the vector points

[gentle bear]

it looks familar

i feel like i just multiply 84 cos 73 for x

but that seems way to easy

[soft dagger]

no you're absolutely correct in your thinking!

however, notice that the angle is below the xaxis

for cosine that doesn't matter because cos(-x) = cos(x) but to get the y component you have to make sure it is sin(-73)

[gentle bear]

yeah

so -73

[soft dagger]

yes

[gentle bear]

x = 24.56

y= -80.33

[soft dagger]

yup

[gentle bear]

ugh

@soft dagger

question

nvm sorry

[drifting parrot]

Can someone help me with my problem?

[soft dagger]

sure

[gentle bear]

https://gyazo.com/1ffe92b5de5a3c26106a4bd6b130963f

Am i just subtracting ( 6i - i ) , (-4j +2j)

[dark sparrow]

no, you're not finding $\bd{u} + \bd{v}$

[somber coyote]

Ann:

[dark sparrow]

nor $\bd{u} - \bd{v}$ for that matter, but what you wrote doesn't even represent that

[somber coyote]

Ann:

[gentle bear]

what is this called if you can point me in that direction so i can research it @dark sparrow

[dark sparrow]

dot product. or scalar product.

[gentle bear]

ty

i got -6-8

[upper karma]

Don't you just multiply the terms with i wirh eachother and then multiply terms with j wirh eachother and then add them?

[gentle bear]

-14

[upper karma]

Are I and j imaginary numbers in this question ?

[gentle bear]

no

from the video i watched they just put it into component form

and just multiplied like you said

then solve

[soft dagger]

i and j are the unit vectors for the xy plane

u • v = u1v1 + u2v2 + ... + unvn

[drifting parrot]

@elfin radish do you want me to send it again?

[devout kraken]

Hii

can anyone help me, we’re starting trig but I’m so confused

[soft dagger]

on?

[devout kraken]

This

[gentle bear]

https://gyazo.com/e50398fbdb99c377e56e0ea2717b2763

Never seen this before but I found a resource which looks like

I need to use dot product

and then this formula

https://gyazo.com/739052859542c3d24d82ae5e5640632e

[soft dagger]

@devout kraken do you see how it says sin(angle) = oppisite / hypotenuse?

[devout kraken]

yes

[soft dagger]

and you're given an angle and a side length

see if you can plug those in to that equation

[devout kraken]

okieokie

so then it would be. sin(53)/1 = 14/x

?

[soft dagger]

yes

[devout kraken]

so then would you just

OH

okay that makes sense

thank you

[soft dagger]

np

@gentle bear you do the second equation because the geometric definition of a dot product is |u||v|cos(theta)

[devout kraken]

okay so i think i messed it up

where do i go from sin(53) = 14/x

[gentle bear]

im confused @soft dagger

what second equation

[soft dagger]

@devout kraken if a = b/c then c = b/a, does that help?

[devout kraken]

not really fjfj

[soft dagger]

@gentle bear the theta = arccos(u•v/|u||v|)

[gentle bear]

but i need to find the dot product

first no ?

[soft dagger]

@devout kraken you're trying to find x, so you need to get x on it's own

@gentle bear yes

because in euclidean space u•v means two things

u1v1+u2v2 AND |u|||v|cos(theta)

therefore you can use u•v to find theta without already knowing theta

[gentle bear]

dot product i got is -28

then i need to find

magnitude

[devout kraken]

So would i cross multiply and divide .80 by 14?

[gentle bear]

of -63 and 35

no

yeah i got that

far

i thought now i take magnitude

[soft dagger]

i forgot the arccos my bad

yes

[gentle bear]

and do sqrt(-9)^2 + (7)^2

[soft dagger]

yes

[gentle bear]

got sqrt(130)

[soft dagger]

@devout kraken

[devout kraken]

oh my god

[soft dagger]

@gentle bear now multiply that with |t|

[gentle bear]

wait

that's right

sqrt(130)

[soft dagger]

yeah

[gentle bear]

2 sqrt(2405)

divide that by -28

[soft dagger]

what

[gentle bear]

💀

i did

sqrt(130) * sqrt(74)

[soft dagger]

oh okay

but it's -28/2sqrt(2405)

not the other way around

[gentle bear]

i got -0.2854

now i gotta do the arccos

[soft dagger]

yes

[gentle bear]

since the dot product wasn't 0 that means these angles aren't parallel

it's neither

i believe

[soft dagger]

no a 0 dot product would be orthogonal

but it is indeed neither

[gentle bear]

this second one

is orthogonal tho 🙂

https://gyazo.com/6a4c521a24e30d4d2858f97b9b70d9f9

i just did -2/4 and compared to 4/-8

and it matches

wait

it's parallel

i got it mixed up

[soft dagger]

yes

2 vectors are parallel in R2 iff they're scalar multiples of each other

w = -2u so they're parallel

can you think of why a dot product of 0 would imply they're orthogonal?

[gentle bear]

😦

no

i don't even know what orthogonal means

[soft dagger]

oh

[gentle bear]

online math is hell

[soft dagger]

orthogonal means they're perpendicular

and remember that u•v = |u||v|cos(theta)

and if the magnitude of both vectors is >0

how could u•v=0?

[pulsar merlin]

Im in algebra 1

and my teacher is giving us geometery

i have no clue how to do this

help........plz

[gentle bear]

i got 180 degrees

which makes sense

[upper karma]

Do you know what the y intercept is in this question?

[soft dagger]

@gentle bear 180° is correct

[pulsar merlin]

no...\

[devout kraken]

How would I start the equation for the inverse trig side

[upper karma]

So what u do is: Pick two points from the line and subtract the x coordinates with eachother and subtract the y coordinates with eachother

Then divide the differnece in y wirh the differnece in x

That's your slope

[pulsar merlin]

OHHHH

[upper karma]

Then, to find the y intercept, you look at the line where x=0

[soft dagger]

@devout kraken if sin(x) = opp/hyp then arcsin(opp/hyp) = x

[upper karma]

In this case ×=0 at -1000

[soft dagger]

inverse trig is just the inverse function so it's basically how you reverse a trig function

[upper karma]

So your y intercept is -1000

[devout kraken]

So would it be

90 degrees = sin-1 (24/40)

[upper karma]

So the equation would be, in slope intercept form, y=x-1000

[pulsar merlin]

ohhhhh yo thank u

yall the goats

especiallyu in times like these

[soft dagger]

@devout kraken no, 90° = sin-1(a/b) only if a=b

you're trying to find the angle so you have x° rn

[devout kraken]

my head hurts

I still don’t get it

[soft dagger]

okay let's start with something simple

what is sin-1(sin(20))

[devout kraken]

-.77

[gentle bear]

https://gyazo.com/bf81a2de077e6c4fe0edf5fd93e1efff Is there a name for this so i can look up what I need to know

[devout kraken]

?

[soft dagger]

no

sin-1(sin(20)) = 20

[devout kraken]

it just cancels out?

[soft dagger]

yes, can you think of why?

[devout kraken]

Negative and positive cancelled out since they’re the same value ?

[soft dagger]

i meant arcssin by sin-1 btw leaf

is that more clear?

[devout kraken]

kind of yeah

[soft dagger]

@gentle bear you solved a problem like that a while ago

[gentle bear]

have I

cause I am lost than a m'f

[soft dagger]

the gave you an angle and a magnitude, draw the diagram

it's exactly like this question

@devout kraken arcsin(sin(x)) = x because arcssin is defined to be the inverse of sin

so if you have sin(x) = opp/hyp

then applying arcsin to both sides will get you what you want

[devout kraken]

Got it

Thank you