geometry-and-trigonometry

[weak shoal]

yea

Yea sure

pi = 180

so each angle has measure 60 degrees

or pi/3 radians

? have you not covered this in class?

[ancient vine]

2a

[dark sparrow]

have you made a diagram?

[upper karma]

F

[finite vigil]

Does anyone have time to help me with some Geometry IXL work? Similarity Theorem? If so please @ me thank you in advance! 😄
If I don't get the ping I won't know you responded haha

[upper karma]

[finite vigil]

Should I send the question?

[upper karma]

[finite vigil]

Also I love your legoshi picture

Beastars is bomb! I wish there was another season

[upper karma]

there will be a second season

[finite vigil]

Oh crap wrong ss

also yayyy

this is the one haha my bad

[glacial haven]

just use pythagorean

theorem

[upper karma]

it's been a long time since ive worked with similar triangles, so lemme try to recall how things should work out here

[finite vigil]

The pythagorean theorem doesnt work for this one @glacial haven

also okay np TTerra

[glacial haven]

why doesnt it?

oh

i didnt read it carefully

[upper karma]

well, first you want to figure out some similar triangles in the diagram here, and then look at the ratios of corresponding sides

those ratios will be equal iirc, so then you use the given information to find KM

just a matter of picking the right correct triangles and ratios

which is, for some reason, escaping me

alright, here is my suggestion: see if the triangle i shaded in red is similar to the entire triangle (i am choosing to use this triangle in the diagram because i know two sides of it, and so it's more likely to tell me more information via similarity)

and then because the triangle in red shares a side with the entire triangle, you can use the fact that corresponding sides of similar triangles have equal length ratios to finish the problem

(i actually haven't done this in about 4 years so sorry it took so long to get something out)

@finite vigil

(im assuming the similarity theorem has something to do with justifying that the triangles are indeed similar)

[glacial haven]

wait you do use pythagorean

[upper karma]

maybe you can use pythagorean and some trigonometry

but it's asking (implicitly) for similarity to be used

[meager pendant]

I need some help with this question.

[upper karma]

Draw DA

<COA=2*<CDA

<CDA=<CDE-...

[finite vigil]

ty terra for the help my teach explained we can solve it thru a table

[arctic vortex]

is this good for my angles of elevation project

[upper karma]

@meager pendant still here?

[meager pendant]

yeah

[upper karma]

Solved it?

[meager pendant]

not yet

[upper karma]

Have you taken a look at my advice?

[meager pendant]

not yet

I’ll do it tomorrow morning tho

cause it’s almost 1 am

[upper karma]

Ok

[meager pendant]

thanks for your concern tho , I’ll report back to you

[upper karma]

Ok

[livid sequoia]

I got the answer for this but I want to know how to actually do it step by step can someone explain?

[dark sparrow]

how did you get the answer?

[livid sequoia]

My teacher told me last second and she left, I wanted her to explain how but the class ended

[dark sparrow]

okay...

big picture, there are two steps here

• calculate the area
• calculate the total cost

the area can be calculated in a bunch of ways, but they'll all involve rectangles combined in one way or another

[livid sequoia]

A = L x W, so the L would be 3 and the W would be 3.6?

[dark sparrow]

this is just one of the rectangles in one of the possible area calculation methods

3.6 * 3 is the area of the red rectangle

[livid sequoia]

oh

[upper karma]

[livid sequoia]

I use pi?

[dark sparrow]

...no

there are no circles here

if you subtract off the area of that little white rectangle in the corner, you will get the area of the bedroom.

[livid sequoia]

3.6 minus the 1.2?

[dark sparrow]

what does that give you?

[livid sequoia]

2.4

[dark sparrow]

no

i didn't ask you to calculate it

i asked you what the answer would represent for this problem

[livid sequoia]

that would give you a rectangle

[dark sparrow]

no, it'd give you a number.

what does the number represent?

what does this 2.4 figure mean for the problem?

[livid sequoia]

the width

[dark sparrow]

the width of the white rectangle, great.

so now what's the area of the white rectangle?

[livid sequoia]

2.4 times 0.6= 1.44

[dark sparrow]

great

so do you understand how to continue the problem from here?

[livid sequoia]

you add the 3.6 with 1.44?

[dark sparrow]

why would you add a length to an area

[livid sequoia]

10.8 + 1.44?

times*

waiy

[dark sparrow]

10.8 m^2 is the area of the red rectangle
1.44 m^2 is the area of the white rectangle

what would their sum represent?

[livid sequoia]

the entire rectangle

or bedroom

[dark sparrow]

don't you think the white rectangle, instead of being counted TWICE as you did, shouldn't be counted AT ALL?

[livid sequoia]

I guess not since its not part of the bedroom

can we cut it, do the L x W with 3.6 and 3 and the area for the small piece 0.6 times 1.2?

[dark sparrow]

3.6 * 3 will give you the red rectangle

if you're asking whether or not you can cut it up like this: you can

but the blue rectangle will not be 3.6 by 3

[livid sequoia]

ohh

I subtract 3 minus the 0.6 then use L x W

[dark sparrow]

if you want to phrase it that way, sure...

[livid sequoia]

2.4 times 3.6 = 8.64

[arctic vortex]

is the asnwer to this 1788.30

[livid sequoia]

then I can do 0.6 times 1.2= 0.72+8.64= 9.36

then times that by $9.50

OH

Thank you!

[arctic vortex]

good luck in ur test corosu

[livid sequoia]

I'll get a high score thanks to Ann

[arctic vortex]

yeah ann is a genius at geometry

@dark sparrow sorry for the ping but can u check my asnwer??

for tihs one is the answer 1788.30

[dark sparrow]

did you write this problem yourself

[arctic vortex]

yesss

and its actually a typo

[dark sparrow]

you have conflicting info between the problem statement and the diagram

[arctic vortex]

10000 is actually 1000 and 5000 is actually 500

oh

what do i do need to change?

[dark sparrow]

well ok you've cleared up the altitudes

what's the angle of elevation supposed to be

[arctic vortex]

from the bird to the airplane

[dark sparrow]

i got that much

but is it supposed to be 34 (as per the diagram) or 64 (as per the text)

[arctic vortex]

or is 34

i changed that as wlel

its 34

[dark sparrow]

ok

then the answer becomes 894.1 feet

[arctic vortex]

what

how

is it the distance from the bird to the plane

i used sin34 = 1000/x if that is correct

[dark sparrow]

no it's not

the duck isn't at ground level

it's 500 feet above the ground

so the vertical distance between the duck and the plane is 1000 - 500 = 500

(feet)

[arctic vortex]

ohh

what would happen if i make the bird at ground level

and the plane 500 feet

[dark sparrow]

the calculation would happen to be exactly the same

[arctic vortex]

oh

im confused on how u got 894.1

how did u get that

@dark sparrow

[dark sparrow]

500/sin(34°).

[arctic vortex]

wait

instead of 1000 u did 500?

@dark sparrow ?

[dark sparrow]

the duck isn't at ground level
it's 500 feet above the ground
so the vertical distance between the duck and the plane is 1000 - 500 = 500

[arctic vortex]

OHHHH

wow ur iq mustbe very high

i didnt acknowdledge that

[dark sparrow]

IQ is meaningless

this is stuff from like 5 years ago to me

[arctic vortex]

oh

well i got 894.1 as well

once i did 500 = x(sin34)

[finite vigil]

I need helps

[upper karma]

@finite vigil is it a test?

[finite vigil]

it's IXL

It's a homework

[upper karma]

Okok

[finite vigil]

I'm on this one rn

[upper karma]

Actually its a special right triangle

So do you know the relationship between the sides? @finite vigil

[finite vigil]

no

[upper karma]

@finite vigil srry had to eat, still need help?

[finite vigil]

yes

pls

@upper karma

[upper karma]

Okay so

Hold up

https://images.app.goo.gl/gyzv6xjEvFfrFTL56

Click on it

That's a truth for every 30 60 90 triangle with the angles in those positions

@finite vigil

x can be whatever and the truth will hold as long as it has the conditions i said above

[finite vigil]

I clicked on it

[upper karma]

https://youtu.be/SFL4stapeUs

[finite vigil]

ty

[upper karma]

Here is the proof behind it

Np

[upper karma]

Alternatively, using the law of sine would also work here for a more general approach to these types of question (see also: https://en.wikipedia.org/wiki/Law_of_sines)

[gritty flicker]

Could someone help

[silent plank]

where are you stuck?

[gritty flicker]

entire question

actually

nevermind

i got it

this one

is the one i dont get

[silent plank]

have you tried drawibg a diagram?

[gritty flicker]

i did this

but idk where 1.6 comes in

[silent plank]

his eyes are at the bottom left corner

right?

[upper karma]

but idk where 1.6 comes in
@gritty flicker you know that the eyes are not on the ground right?

[silent plank]

so indicate that that position is 1.6m above the ground

[gritty flicker]

oh

should i do

tan 51 = 26/x

then subtract by 1.6

[glacial haven]

no

redraw the diagram

[quick gyro]

You might want to include the tree, ground, and thomas' eyes

[silent plank]

(use ruler/line tool)

also tan isn't being applied properly

[upper karma]

Think about angles

👏 👏

@upper karma note that the triangle XYZ is isosceles...

Actually equilateral...

[arctic vortex]

can someone check this

[silent plank]

can't you fix your diagram or text for the details to match?

diagram labels could be clearer, that 1000 could be misinterpreted as the distance of that edge instead of plane altitude

also 10000ft actually seems like a more reasonable flight height. i'd be concerned if a plane was flying horizontally (not landing/taking off) at 1000ft

[upper karma]

Actually equilateral...
@upper karma holy fuck why did i not realised lol

@upper karma hello?

[upper karma]

👻 👻

[meager pendant]

i don't know how to find <ADE

[dark sparrow]

look at angles ADE and ABE

notice anything?

[upper karma]

yes its awesome

[meager pendant]

they equal

thank you

[vapid vector]

How do we know that they are equal?

[silent plank]

inscribed angle theorems

[vapid vector]

Oh ok

[meager pendant]

this is the proof i used

[dark sparrow]

this is not a proof

[meager pendant]

oh

ok

[dark sparrow]

it's the statement of a theorem

[arctic vortex]

can anyone check my homework answers for like 10 questions?

and i might need help with some questions

[dark sparrow]

if possible, could you post the questions all at once and your list of answers, so that we don't have to go through them one by one?

i for one find that extremely annoying

[arctic vortex]

oh ok

i will do that

1 second

[upper karma]

👻

[upper karma]

I didn't know 1 second lasted 1h

[arctic vortex]

no wait 1 second

@dark sparrow ann i am back can u still help me sorry

[dark sparrow]

if possible, could you post the questions all at once and your list of answers, so that we don't have to go through them one by one?

[arctic vortex]

oh ok

5 minutes pls

@dark sparrow which one would this be

i think its the pythagerona theroen

[dark sparrow]

so much for not going through them one by one huh

[arctic vortex]

sorry

ok i will do that

i like to do things where after i finish 1 i go to the next

@dark sparrow wait so is it the pythageoran theorem

[dark sparrow]

all at once, please.

[arctic vortex]

oh

ok wait

there are 24 questions in total

[dark sparrow]

all at once.

[arctic vortex]

ok

i understand most of them there are just a few that i dont understand

@dark sparrow

[dark sparrow]

okay, i see the questions. can i have your answers for them all?

no, i am not going to verify them one by one.

[arctic vortex]

ohhhh

ok

but i dont understand this question at all

can u help me with this

[dark sparrow]

what don't you understand about this?

do you know what it means to fold something in half?

[arctic vortex]

i dont know how to get x

[dark sparrow]

that neither answers my question nor makes it clearer to me where your confusion lies.

[arctic vortex]

i dont understand how to get x

ohhhh

[stray pelican]

wait i made a mistake

ignore that

[arctic vortex]

oh

ok

[stray pelican]

ill reupload

[arctic vortex]

whats a side length

is it the hypotenuse

[arctic vortex]

oh

[stray pelican]

make sense?

[arctic vortex]

isnt it suppposed to be tan

tangent

because opposite and adjacent

and yeah it makes sense now

[stray pelican]

I'm just using law of sines

[arctic vortex]

oh ok

[stray pelican]

you get it now?

[arctic vortex]

yesss

[stray pelican]

good

[arctic vortex]

can u help wioth this one now??

the other one makes sense now

[dark sparrow]

to be honest i'm not terribly inclined to help at this point

[arctic vortex]

oh

[stray pelican]

im not gonna tell you the answer i want you to figure it out but ill give u a hint

it looks like you should use the pythagorean theorem

[arctic vortex]

oh ok

i will use that

by the way how do u get honorable rank

@stray pelican can u check this if it is correct

[arctic vortex]

guess what guys

i got a 23/24 on my homework

now i need to study for the test

yesss im quite a genius

[royal edge]

AED is a straight line. I am completely stumped on this question

It’s not a parallelo gram

[upper karma]

@royal edge consider sine law

On the triangle CEB

Then cosine law

[upper karma]

👻

??

[weak shoal]

@upper karma What is the issue you're having?

[upper karma]

i dont rly understand

it at all

...

[idle bobcat]

Hello, so I know that <QPR + RPS = 180, and thus u can solve for x, but can't you do the same thing by doing 180-<QPR = <RPS?

because if you subtract one angle from 180, cant you find the other angle

[stone adder]

Yeah, you can do that

since 180 degree - <QPR = <RPS is derived from <QPR + <RPS = 180 degree

It's the same thing, since in the end you're asked to find the value of x, which <QPR and <RPS has x on them

[idle bobcat]

ok just making sure, because when i did it this way i keep getting the wrong answer

so im doing 180-2x+122 = 2x+22

[stone adder]

Can you show your work?

Oh

$$180-(2x+122)=180-2x-122 \neq 180-2x+122$$

[somber coyote]

Wilston Lynx:

[stone adder]

Recall the distributive properties

when you work with parentheses

[idle bobcat]

ahhhhh

yeah that makes sense now

tyvm!!

[stone adder]

You're welcome!

[silent plank]

what have you thought about?

[upper karma]

Everything

The meaning of life

[tidal river]

Can someone help me with the geometry question

[royal edge]

Quantum try drawing it out first

Let’s start with that, and once you do that, tell me

[upper karma]

F

[royal edge]

Xi can you send me that profile picture?

[upper karma]

Hmmmm

[royal edge]

Pretty please?

[upper karma]

@tidal river #6241

@royal edge

Help this guy

[upper karma]

Anyone know what the foci for this is

[stone adder]

What's foci?

[upper karma]

idk but it's asking for it

focus

oh

how do you get it

yes

what

I believe there's a formula

yep

Google it

k ty

[sleek thistle]

how do i find y in this?

[upper karma]

@sleek thistle (4y+4) and 60 are supplementary angles

Do you know what that means?

[sleek thistle]

ohh you’re right omg

180

[upper karma]

Yeah they add up to 180°

Set up an eqn and solve for y

[sleek thistle]

okay thanks lmao

[upper karma]

Np

[finite vigil]

Could anyone help me with a trig IXL? I need to get up to an 80 if so please @ me ty in advance

[deep stump]

can you help with these problems ?

[upper karma]

seems pretty hard

[dark sparrow]

@deep stump do you still need help with these

and if so, which ones

ok, now that you've (presumably on accident) DM'd me saying you do need help on these

which ones do you need help on

@deep stump are you gonna respond here or are you gonna insist on DMing me even after i've made it clear i would prefer you didn't?

[deep stump]

@dark sparrow It's just clear to me now, thanks

[dark sparrow]

okay so are you gonna answer my question

which of these problems do you need help with

so that we can go through them in order

[deep stump]

@dark sparrow They clear to me now, when I need help with other ones, I ll ask again, thanks

[dark sparrow]

ok

[sleek thistle]

anybody know of a rule or just how to get this length?

don’t even know how to start

[dark sparrow]

consider that ABCD is a rectangle, as the problem literally tells you

what do you know about the diagonals of a rectangle

[sleek thistle]

oh

well the diagonals are congruent

[dark sparrow]

yup

and what can you say about BC, the other diagonal of the rectangle?

[sleek thistle]

that it’s equal

[deep stump]

@dark sparrow available now?

I have this problem to find the eccentricity of an ellipse if the sides of the square inscribed in it pass through the foci of the ellipse. Can you help?

[sleek thistle]

omg

it’s 5 i’m a pop tart

damn

well i overthought the shit out of that for no reason

[dark sparrow]

@deep stump have you made a diagram yet

[deep stump]

I am doing game development and I had to solve this problem

@dark sparrow How would you solve that?

[dark sparrow]

well my first step would be to make a diagram

[deep stump]

can you help more?

[dark sparrow]

actually heres a more extensively labeled diagram

notice that this diagram is symmetric about both axes of the ellipse so we have that F1 and F2 are the midpoints of AD and BC

taking $AF_1 = 1$ for convenience, we have that $$2a = F_1P + F_2P = F_1B + F_2B = \sqrt{5} + 1$$ and $c = F_1O = 1$

[somber coyote]

Ann:

[elder shuttle]

does anyone know what is arcsin(1 - sin^2x)?

[dark sparrow]

i don't think this can be simplified any further than arcsin(cos^2(x)) @elder shuttle

is there a problem you're doing where you found this expression?

[misty plume]

Anyone has a intuitive method to part a? I ended considering the area of an individual triangle then multiplying it by n sides for A, but would be cool to know if there are faster methods...

[dark sparrow]

nah that's p much the fastest

[royal edge]

I need help with 5

nvm

[arctic vortex]

Can anyone help me with this please

Your explanation is not correct, you state that the given pair of angles that add up to 180 are supplementary In your correction, please state the name of this supplementary angle pair, as formed by parallel lines cut by a transversal?

this is what the teacher said

oh

never mind

i just need to name the angle pair

[royal edge]

Makes sense

[arctic vortex]

yes

@royal edge what kind of angle pair is this??

i think its congruent angles or consecutive interior angles

[royal edge]

I got it, it was SAS

They were corresponding angles Bcs they were parallel lines cut by a transversal

[elder shuttle]

@dark sparrow It is this problem:https://mainapp-frontend-preprod.toppr.com/ask/question/the-diagram-shown-the-graphs-of-y-tan-x-and-y-cos-x/

[dapper solstice]

hi i need some help with trig pls, i dont get when to use or how to see which one to use between cos sin and tan

why do they use tan here and not anything else, also how do they get 23degrees from 0.429??

[dark sparrow]

tan(α) = 0.429

α = arctan(0.429) = atan(0.429) = tan^-1(0.429)

[dapper solstice]

emm

how do i write this in my calc? tan(0.429)^-1?

tan^-1(0.429) this gives me error

also is it ^-1 for all sin cos tan to find its degree

[silent plank]

depending on the calculator, probably a variation of
shift + tan
or
function key + tan

or some trig submenu

[dapper solstice]

hm

i do have mode button but that only changes from radian to degree and vice vers

assume thats the wrong 1

ooooooooooooo i found it n v m

it was just the "2nd" button lul

ty brother

[upper karma]

books answer

[silk crown]

so, what was your approach

you said you used the cosine rule?

[upper karma]

cosine rule

yes

[silk crown]

ok, send me your working/reasoning

[upper karma]

I know my handwriting isnt the best

[silk crown]

let me see

what is the cosine rule @upper karma

you did not write an equation lol, if you did it would have been clearer

[upper karma]

c^2=a^2+b^2-2abcosC

[silk crown]

yes

[upper karma]

yea i was kinda rushing

[silk crown]

then write the equation this time

c^2 = ....

and then at the end take sqrt on both sides

your expression is correct, but the square root of that is the answer

i checked with a calc and it is the right answer

¯_(ツ)_/¯

[upper karma]

I'm gonna check the book's answer

[silk crown]

5.6cm lol

you get 5.59... something

this is why you dont rush

write your equations properly

[upper karma]

yea the books answer is the same as written

I got 5.598 for my solution

Book's answer is 5.64861 which i just checked

[silent plank]

the books crap

[silk crown]

oh lmao i only saw their final answer

[upper karma]

so there answer is wrong?

[silent plank]

actually 1 sec

i think the diagram is actually inconsistent

[silk crown]

D and A

[silent plank]

theoretically, both the sine and cosine laws can be applied, but due to rounding issues, the values obtained are off

cosine law only focuses on triangle ADC. calculations using that don't rely on external info which is what you applied and what i'd recommend

although you should make your square root clearer

and don't split equations on two lines

[upper karma]

my bad again i was rushing through this

[silent plank]

(there's rushing and there's being oblivious to math rules/ettiquete)

[upper karma]

true

[silent plank]

based on the values within triangle ADC, angle ACD actually works out to be around 99°

[silk crown]

im pretty sure the book's working is wrong though

[silent plank]

nah, its just that the values are inconsistent

if they were actually consistent, applying the sine law like that would get the same result

[upper karma]

http://mathforum.org/library/drmath/view/61118.html

uhh this is already on the interenet

apparenlty its a common problem

[silent plank]

yeh. over specification is an issue

leads to multiple different answers from "correct" methods

[upper karma]

so which is the correct answer?

I assume no correct answer or both are

[silent plank]

both methods are acceptable

but technically there are no correct answers due to inconsistencies

but i mean if you're rounding to 1dp

you'd get 5.6 for both which is just a coincidence

[arctic vortex]

what is this

[teal stag]

https://en.wikipedia.org/wiki/Aerial_work_platform#Scissor_lift

[wooden current]

bruh

@arctic vortex What's giving you trouble

Unless you have solved it already

[arctic vortex]

o i solveed it

my fault

i forgot that f was outside of the parallelogram

[wooden current]

It's ok. Sorry for the ping 😅

[arctic vortex]

its ok

[upper karma]

just wondering is there a name for when imput to trig funcsions are the same?

[nova prawn]

can you proove that this triangle = formula of circle

https://www.youtube.com/watch?v=GQXUpB2NHvQ

have found it

[upper karma]

like are you looking for a circle that touches everypoint?

[nova prawn]

like are you looking for a circle that touches everypoint?
@upper karma no, i have found what i was looking for, thanks again 🙂

how to prove this formula any ideas?

[upper karma]

Yes

[nova prawn]

how?

[upper karma]

@nova prawn yeah hold up

$m\cdot m_{\perp} =-1$

[somber coyote]

HoboSas:

[upper karma]

Id just write it but its kinda long

Lol

That's a long proof that's fine

I mean if you wanna write it and i remove the video go ahead

Btw there's no proof in the video

Shoot

[nova prawn]

yeah only formula

[upper karma]

https://youtu.be/tngPrM2d-OM

Now lol

https://youtu.be/9ilcXZn9CfE

I mean im not fan of formulas

Here's the geometric proof

[nova prawn]

thanks a lot

[upper karma]

There are better ways of approaching it

But whatever

[nova prawn]

like what?

@upper karma

[upper karma]

Finding perpendicular to the line that goes through the point and the the intersection between line and perpendicular

Then distance between two points

Like for example to draw an auxiliary plane

The connection was bad there and it got sent half an hour later but yeah

Oh i was referring to the auxiliary plane

@nova prawn

May i post another video of it lol

[nova prawn]

ofc

[upper karma]

F there are no english good videos

In my lative there are some

(Spanish)

Shieet

Post anyway

Ig you can put subtitles

What if we speak spanish

https://youtu.be/JRchvOLNCFU

[nova prawn]

https://www.youtube.com/watch?v=iATjsfAX8yc

[upper karma]

Lol

@nova prawn does it cover auxiliary plane method

^ that's what I was trying to say

Sniped headshot

[nova prawn]

what is that

[upper karma]

The method i was all the time referring to is a better method than the formula one

https://youtu.be/JRchvOLNCFU
@upper karma .

[nova prawn]

I will check that as well

thanks but i am actualy working with circle

[upper karma]

Put eng subtitles on as well

Kk

[nova prawn]

like I have the center and the equation of tangent line

how to find radius, hope those will work

[upper karma]

Damn

Yeah they will definitely

What lol @upper karma

Nothing

Great

Nice

[rich mist]

Can someone help me with a trig question ?

[keen aspen]

Uhh what are you having issues with

[rich mist]

i sent my question

[keen aspen]

Yes what part of the problem do you not understand

[rich mist]

I just need the answer

[keen aspen]

Bruh

[rich mist]

can you help me its time

timed

[keen aspen]

I can walk you through it

[rich mist]

okay please

[keen aspen]

So just solve the equation

tantheta+1=0

so tantheta=-1

Now think of a theta that when you take the tan of it, it is -1

Notice here that tan is negative, so theta is in the 2nd and 4th quadrant

What theta comes to mind

[rich mist]

what do you mean what theta

[keen aspen]

What value of theta comes to mind such that tan of it is -1

[rich mist]

idk what is it?

bro its timed

[remote heart]

Theta is $\theta$

[silent plank]

timed...

[remote heart]

is this a test @rich mist?

[keen aspen]

Lol

[rich mist]

are there 3 answers

I got theta = (3/4)pi

and (7/4)pi

[keen aspen]

Good

[rich mist]

are those the only two answers?

[keen aspen]

In that interval

[rich mist]

Hello

my answer is x=3/1+4logbase15(3)

^17 not 15

how would I write that with these resources

[keen aspen]

You need it in log base e or 10

[urban egret]

@rich mist PJS said it best you need to take the natural log of both sides. For some reason you don’t have natural logarithm in your final answer.

Review your natural logarithmic properties

[signal hornet]

how do I solve this 0= x + 2sin(x) domain [0, 2pi]

no idea how to get x on its own

I know 0, and 2 pi from just looking at the unit cricle

[umbral snow]

There's no way haha. That one isn't something doable by algebra

[keen aspen]

Yeah

[umbral snow]

,w graph x + 2sin(x)

[signal hornet]

what about with calc 1 knowledge?

[somber coyote]

Results provided by WolframAlpha

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

Requested by Kaynex#7520

[umbral snow]

In fact it looks like 0 is the only solution

[signal hornet]

my guess is to find where f' = 2

but that is probably just as hard

if not worse

[umbral snow]

Honestly yes, the derivative could help. It's pretty easy to see where the mins/maxes are because the derivative is just
1 + 2cos(x)

[signal hornet]

I want to to know x-intercept

[umbral snow]

Plug the mins/maxes into x + 2sin(x), that's enough to show that the only solution is 0

[nova prawn]

I am looking for circle's equation which has:
center on the line y = 3x and tangent to the line x = 2y at the point (2,1)

[dark sparrow]

what have you tried so far

[nova prawn]

Perpendicular line I was looking for

but from point, if i get perpendicular line, it will be perpendicular to line that is on center. I need perpendicular from center -> tangent not tangent -> center i think

I checked internet, could not find anything.

[dark sparrow]

if you construct the perpendicular to x = 2y at the point (2,1) then you get another line on which the center must lie

[nova prawn]

but it also should be perpendicular in the tangent point?

[dusky surge]

perpendicular to x = 2y at the point (2,1)

[nova prawn]

did not get it

[silent plank]

you aren't interested in any lines perpendicular to the line the center lies on

[nova prawn]

yeah, i need line that is perpendicular to tangent right?

[silent plank]

yes

[nova prawn]

then on that line some point intersects from center line

that must be my center

[silent plank]

yes

[nova prawn]

how to calculate this?

[silent plank]

do you know the property of perpendicular lines?

[nova prawn]

yeah, sloped are reversed.

[silent plank]

that's vague

[nova prawn]

like if one them is x, isn't the other 1/x

[silent plank]

no

(for non-horizontal and vertical lines) the slopes of 2 perpendicular lines are negative reciprocals of each other (or their product is -1)

$m_1m_2 = - 1$ or $m_1 = -\frac{1}{m_2}$

[somber coyote]

ramonov:

[nova prawn]

ok

but i need 2 things

[dark sparrow]

there seems to be a language barrier here

[nova prawn]

my english.

is it that much bad?

[silent plank]

you grammar is bad. can you at least clearly understand what we are saying?

[nova prawn]

i can understand you guys, books, videos don't worry 🙂

[silent plank]

ok

[nova prawn]

I just type fast. That's why i guess

[dark sparrow]

then type slower.

and read every single message you write at least 2 or 3 times

[nova prawn]

You are right, i must have thought you before i had clicked enter. Sorry.

[silent plank]

currently you are looking for equation of the line perpendicular to your tangent x=2y at (2,1) right?

[nova prawn]

yes

[silent plank]

what is the slope of x=2y

[nova prawn]

The slope of the equation x=2y is 1/2

[silent plank]

what would be the slope of a line that is perpendicular to that?

[nova prawn]

-2 as you said m1*m2 = -1

[silent plank]

yes

and can you determine the equation of a line with:
slope: -2 passing through the point (2,1)?

[nova prawn]

1 = -2 * 2 + b, then b = 5

y = -2x + 5

[dark sparrow]

congratulations

[nova prawn]

then I will solve it with center line.

[dark sparrow]

now you have two lines, y = 3x (the original) and y = -2x+5 (the one you just found)

and you know the center lies on both of them

[nova prawn]

Yeah, intersection point which is solution set of these 2 equations.

Thanks a lot, it has been like 3 years I didn't spend time with geometry, just started back. Also, sorry for mistakes while typing.

[arctic vortex]

can anyone help me

with geometry

[upper karma]

@silent plank @silk crown @silk crown I have found the issue... the triangle is impossible

I appreciate your help!

[upper karma]

@arctic vortex

[cosmic pebble]

From

#❓how-to-get-help

Rule 1

The help channels are solely for help with math, so feel free to post your question. Asking whether you can ask a question or if anyone knows about some specific topic is unnecessary, so please try to avoid questions of that nature.

[arctic vortex]

Now i am not

But thanks for trying !!

And asorry

[frail rock]

I need help with this problem. I tried setting up the cosine and sine for angle B and I ended up getting stuck.

I managed to get up to this point. I have a feeling that I forgot something here.

[plucky marlin]

where exactly are u stuck

[frail rock]

I have a feeling I've forgotten how to divide an integer by a radical.

I tried looking it up but I got something really weird when I tried to solve it

hold on let me look for my other work

[plucky marlin]

both ur equations for sin and cos are correct

[frail rock]

Ah, I see.

[plucky marlin]

but the thing is, the options rationalized the denominator

so you should do the same and see which one matches up with the options

[frail rock]

To rationalize the denominator, I just multiply it by the square root of 5 right?

[plucky marlin]

yea

multiply top and bottom by sqrt 5

cuz u dont wanna change the value

[frail rock]

Ah ok

Can you simplify 3sqrt5/15 to sqrt5/5?

[plucky marlin]

yeah

ofc

and cancelling out the 3's should be the first thing you do

just to reduce the calculations

[frail rock]

Ah I see. I thought for a moment you could only reduce it f they had the same radical

And I did the one for cosine too

[plucky marlin]

yup perfect

[frail rock]

I think I see now how they got those answer choices. I didn't understand it at first, I guess I just forgot how to divide radicals with integers

So that would eliminate A and C as an answer choice because that's not even the right equation

Then it's either B or D

Wait

Yeah

wait no that would also eliminate D

Ah so it would have been B

I get it now

Thank you for helping me :D

[plucky marlin]

np

[frail rock]

[brisk holly]

How do I solve this?

[ionic bluff]

@brisk holly what does that notation mean

the R

[weary drift]

R_AB is a reflection wrt AB

[austere dragon]

this is a no brainer but are kites parallelograms but not always?

[dark sparrow]

no, there exist kites that aren't parallelograms

[austere dragon]

but some kites are parallelograms though, is it still classified as not parallelograms?

[silent plank]

kites that are parallelograms are parallelograms

kites that are not parallelograms are not parallelograms

[dark sparrow]

[silent plank]

a kite that is also a parallelogram is a rhombus

[upper karma]

Is it?

[silent plank]

the phrasing of:

are kites parallelograms but not always
has bad implications

[upper karma]

Isn't it only squares that are kits and oarlagrams?

[silent plank]

squares are rhombuses

[upper karma]

just wondering is there one triangle you can make on a sphere with three points or are there infinitely many

[sly belfry]

hello everyone! quick question.

can we interchange the ellipse parametrisation? instead of (cos(t), sin(t)), to have(sin(t), cos(t))?

[dark sparrow]

sure? you'll just have the curve get traced out in the other direction

(sin(t), cos(t)) specifically will trace the circle of radius 1 centered at the origin clockwise from (0,1)

[sly belfry]

hmm

I'm not very good at parametrisation, but this seems a little awkward

but yes -- it does make sense

thank you so much.

[arctic vortex]

can someone help

[upper karma]

What makes a parallelogram a rhombus?

For example perpendicular diagonals

So you can use a famous theorem....

[plucky marlin]

a homeomorphism

[upper karma]

@arctic vortex

[plucky marlin]

enuf shitposting

i watch movies now

[arctic vortex]

diagonals bisect each other @upper karma

[upper karma]

For example perpendicular diagonals

This is more useful

Any ideas?

Consider triangle MSP

[arctic vortex]

wait ohhhhh

the pyhagoren thoerem

[upper karma]

Yes

[arctic vortex]

because right angles

wow u are genius

[upper karma]

Yes

If pythagorean theorem were verified, then the parallelogram would be rhombus

[arctic vortex]

also this one

i forgot how to do this

i knew how before

i forgot

[upper karma]

What's the value of the angles of a square?

[arctic vortex]

90

45 45 90 triangle

but i dont know how to get x

[upper karma]

You said it

The angle must be 45

So 45=6x-9

[arctic vortex]

ohhhhhhh

ok

this is the last one

i do not know at all how to get this

[acoustic bone]

PM=13

[arctic vortex]

its only lm

[upper karma]

First of all

MG is parallel to AL

so?

What's the value of <MLA

[arctic vortex]

is it 82

[upper karma]

Nope

MG and AL are parallel

[arctic vortex]

but its right

[upper karma]

What is right?

Im talking about <MLA not <MAL

They are alternate angles...

So they are equal

But AM=AL, triangle MAL is isosceles so <AML=<MLA

Your turn now

@arctic vortex

[arctic vortex]

im back

its 82 because 49+49=98

and 180-98 = 82

[upper karma]

Yes

[livid sequoia]

is this SAS?

[upper karma]

SAS

[livid sequoia]

these both are not congruent?

ok I got 30 its AAS

[upper karma]

30 and 31 are yes

[livid sequoia]

both are AAS and congruent

[upper karma]

Yes

[mortal mesa]

Hi,
Having troubles solving the following equation:

$(1+\cos(4x))\sin(2x)=\cos^2 (x)$

[somber coyote]

idan:

[upper karma]

@mortal mesa you have to prove it or solve it

[mortal mesa]

@upper karma solve it

[upper karma]

Ok hold up this is an ass-long problem @mortal mesa

[mortal mesa]

@upper karma tell me about it 😩

[upper karma]

Just expand cos(4x)

Hahahah

@upper karma have you finished solving it?

I have not even tried

Kinda stuck lol

My black ink is gone and i can't see shit

Hahah

What if you use complex numbers

Should be juicy

I won't try THAT hahah

Why noy

Not

2 lines of solution

I don't think he wants that method

[somber coyote]

HoboSas:

[upper karma]

https://youtu.be/EZbzM1nIYSU

[mortal mesa]

It's for my students. I got stuck trying to solve in class lol

Unfortunately, complex solution won't help

[upper karma]

Damn

Wait I have an idea

Express cos^2 in terms of cos(2x)

Then t=2x

$$(1+\cos(4x))\sin(2x)=\cos^2 (x)$$ $$(1+\cos(4x))\sin(2x)=\frac{\cos(2x)+1}{2} $$ $$
(1+\cos(2t))\sin(t)=\frac{\cos(t)+1}{2}$$

[somber coyote]

HoboSas:

[upper karma]

Is that allowed

That's how i started

[somber coyote]

HoboSas:

[upper karma]

@mortal mesa you can guess what to do now

[mortal mesa]

@upper karma @upper karma
Nice, tnx.. Sorry, but I'm not sure how to go from here. I still have that 1/2 on RHS.

[somber coyote]

HoboSas:

[upper karma]

I guess

then cos(t) = helicopter

[stone adder]

Is $1-sin^2(2x)=cos^2(2x)$?

[somber coyote]

Wilston Lynx:

[upper karma]

Yeah

Yes

[stone adder]

Okok

[upper karma]

Use \cos and \sin to make it look better xD

[stone adder]

Oh yea lol

I always forgot about that

[upper karma]

$2u^2\sqrt{1-u^2}=u/2+1/2$

[somber coyote]

HoboSas:

[upper karma]

$\sqrt{1-u^2}=\frac{u+1}{2u^2}$

[somber coyote]

HoboSas:

[upper karma]

$1-u^2=\frac{u^2+2u+1}{4u^4}$

[somber coyote]

HoboSas:

[upper karma]

Nice u^6

[stone adder]

Oh god u^6\

[upper karma]

:(

$4u^4-u^6=u^2+2u+1$

[somber coyote]

HoboSas:

[stone adder]

Btw I tried the question and got, $$2 \sin{2x}[1-\sin^2{2x}]=\cos^2x$$

[upper karma]

Damn son

[somber coyote]

Wilston Lynx:

[upper karma]

@mortal mesa

$u^6-4u^2+u^2+2u+1=0$

[somber coyote]

HoboSas:

[upper karma]

,w u^6-4u^2+u^2+2u+1=0

[somber coyote]

Results provided by WolframAlpha

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

Requested by HoboSas#3200

[stone adder]

Oof

[upper karma]

Wait

It's wrong

Yeah lol

,w u^6-4u^4+u^2+2u+1=0

[somber coyote]

Results provided by WolframAlpha

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

Requested by HoboSas#3200

[upper karma]

Still fuck

Try putting the trig

Back?

Like what you said u is

$u^6-4u^4+u^2+2u+1=0$

[somber coyote]

HoboSas:

[upper karma]

Like u=cos(t)?

This is shit

Like what did you sub for u

Ikr whatever

Like u=cos(t)?
Doesn't look good

Doesn't look good
@upper karma i mean, what did you u-sub on

Yes but

No bueno

Rip

[solar shale]

question says find length bz

I got 51 is that right?

[dusky surge]

I'm quite sure it is not 51

[solar shale]

How would I go with solving it

[silent plank]

what were you doing to get 51?

[solar shale]

I got two different types of problems mixed with this question

[drifting drift]

12/9=x/8.5

[solar shale]

So cross multiply 12 and 8.5 then divide by 9?

[silent plank]

not a fan of how the term cross multiplication is being thrown in here

firstly do you understand how to get that initial equation?

[solar shale]

Sort of

[silent plank]

properties of similar figures

as for isolating x, simply multiply both sides of the equation by 8.5

[sleek thistle]

am i supposed to make it equal to 9 since it’s equilateral?

[dark sparrow]

make what equal to 9?

[sleek thistle]

3/2x+30

[dark sparrow]

does an equilateral triangle have angles equal to 9 degrees?

[sleek thistle]

equal to 180 right

[dark sparrow]

what's equal to 180

[sleek thistle]

the entire triangle

[dark sparrow]

the SUM of the ANGLES in the triangle is 180°

but what's the measure of ONE angle in an EQUILATERAL triangle?

[sleek thistle]

60

i think

[dark sparrow]

60 degrees yes

[sleek thistle]

so it’s 20

okay thanks

[mortal mesa]

@upper karma @upper karma
Thank you for your help!
It looks unsolvable using regular methods.. Sorry it was shit

[upper karma]

don't worry we had fun trying to solve it 👍

[austere dragon]

is a right triangle always scalene?

[weary drift]

https://www.voovers.com/wp-content/uploads/2020/05/45-45-90-triangle.jpg

[austere dragon]

wow awesome

o yea i forgot about the phyton theory sry bout that

[dark sparrow]

the what

[upper karma]

o yea i forgot about the phyton theory sry bout that
huh?

[glacial haven]

python coding theory

[upper karma]

My answer here is 198.585 sq.units but the books says the answer is 103.2 sq.units how is the area of this irregular figure id 103.2 if the area of rec only 9 x 15 = 135 is already larger than 103 ?

[dark sparrow]

the circle isn't part of the shaded region

you're supposed to subtract away half its area from that of the rectangle

[upper karma]

Ohhhh so thats why thanks.

My answer here is 32.0632sq .units and in the book is 38.24 sq.units i use 3.3 as the base of the rectangle and use 6.5 height for the height of it. The diameter i use is 5.2 divide 2 for the radius and i use 1/2 pi x r squared

Epic

[dark sparrow]

the lower part is not a rectangle.

[upper karma]

Is it a trapezoid?

[dark sparrow]

yes