geometry-and-trigonometry

[upper karma]

no it's not

...

[dark sparrow]

can i see the problem again

i have a feeling there isn't enough data

[little osprey]

Ann

polynomial

[upper karma]

there isn't

[little osprey]

will get angry

if you interfere

[upper karma]

but you pretty much have to assume

[azure reef]

[upper karma]

@dark sparrow here was the problem

#geometry-and-trigonometry message

[silent plank]

the only answer that would make sense here would be the arc measure

[upper karma]

i know

[little osprey]

Which I already mentioned

[dark sparrow]

we don't know how long the hour hand is

[upper karma]

so my answer should be right

[dark sparrow]

there is no right answer to this

[upper karma]

ann yes

[little osprey]

Bruh your answer?

[dark sparrow]

there is no right answer

[little osprey]

You told him to ignore me

[silent plank]

which will be independent on the diameter and radius

[dark sparrow]

simple as that, there is no right answer

[upper karma]

20.9375

was my answer

[azure reef]

the only answer that would make sense here would be the arc measure
why cannot we express hour hand as some function of diameter tho

[arctic vortex]

20.9375
@upper karma that wasnt right unless i had to round to whole numbers

[dark sparrow]

we don't know how long the hour hand is!!!

[upper karma]

What if they want radians

An angle

@arctic vortex how am i supposed to know what you're supposed to do

[little osprey]

Yes we have established that

[arctic vortex]

we don't know how long the hour hand is!!!
@dark sparrow 12 to 6 is 180 degrees

[silent plank]

because we have nfi how long the hour hand is relative to everything else

[dark sparrow]

who's to say whether the hour hand is 5 inches long or 4 inches long or 3 inches long?

[arctic vortex]

180/6

is 30 degrees

[dark sparrow]

do you even read what i'm saying

[arctic vortex]

yes

[dark sparrow]

WE DON'T KNOW THE LENGTH OF THE HOUR HAND

[arctic vortex]

wait

but

[dark sparrow]

WE DON'T KNOW THE LENGTH OF THE HOUR HAND

[azure reef]

because we have nfi how long the hour hand is relative to everything else
well yes, but i mean general form
it will involve arbitrary constant

[dark sparrow]

we don't know how big it is

[little osprey]

lmfao

[upper karma]

...

there is no way it will involve a constant lol

my answer should be right

the question is if they want to round it to 21

@arctic vortex try 21?

[dark sparrow]

aight yall clowns wont listen to me i get it

[upper karma]

ANN

they clearly want a real number

[dark sparrow]

look

[upper karma]

there's no entry box to write "impossible to solve"

I AGREE

WITH

YOU

[dark sparrow]

for all i know, the hour hand could be any length between 0 and 8 inches

[upper karma]

YES BUT THERE'S NOWHERE TO WRITE THAT FACT

[dark sparrow]

it's pointless to try to guess

[little osprey]

aight yall clowns wont listen to me i get it
@dark sparrow Read the chat history.

[dark sparrow]

spiral should email their teacher

[little osprey]

Agree

[dark sparrow]

and be done with this problem

[upper karma]

i guess

[little osprey]

@arctic vortex try 21?
@upper karma Did you keep units

[azure reef]

this problem certainly does not deserve such hot discussion

[upper karma]

@little osprey not sure what you mean by units

[little osprey]

inches

[arctic vortex]

no its wrong

[little osprey]

Yeah

[upper karma]

well no idea then

[little osprey]

Email your teacher

[upper karma]

send an email

[arctic vortex]

ok but

how do i solve this

[upper karma]

LMAO

[dark sparrow]

wow

[upper karma]

you know that dj khaled song

[dark sparrow]

smoothest segue in the west

[upper karma]

another one

im crying

[arctic vortex]

i dont know how to find the tangent

why are u crying

[little osprey]

Suffering from success

[arctic vortex]

what happened

[upper karma]

spiral

does it show you the answer to the previous problem

[arctic vortex]

no

[upper karma]

yeah, i want an update when you email your teacher

[arctic vortex]

ok

i think the hour hand is 30 degrees

[little osprey]

Actually, same

[arctic vortex]

though

[little osprey]

time doesnt go backwards

[upper karma]

For this one, recall a circle theorem

wait

[arctic vortex]

no becasuse 12 and 6 are equal right

like they are opposite

[upper karma]

do they want degrees?

but why would they mention the diameter then?

[arctic vortex]

meaning that 12 and 6 are semi circles

[upper karma]

doesn't make any sense

[little osprey]

huh

[arctic vortex]

and since there are 6 hours in 180 degrees

[little osprey]

d x pi = circumference

[arctic vortex]

180 / 6

so its 30 degrees

each hour hand

[upper karma]

yes

that's correct

[little osprey]

doesn't make any sense
@upper karma What are you on

[arctic vortex]

yeah

[little osprey]

I mean

[arctic vortex]

and you guys ssaid that it was wrong :(((

[little osprey]

like what are you talking about

[upper karma]

denton

it could be the case that they just want an answer of 150 degrees

but that doesn't make sense, since they mention the diameter

to summarize

it's one of the worst questions i've seen in a while

[arctic vortex]

is this asking for arc length or sector

[upper karma]

What do you think?

[arctic vortex]

i dont know

the difference

ohh

if it says area its sector

[upper karma]

https://www.google.com/imgres?imgurl=https%3A%2F%2Fwww.onlinemathlearning.com%2Fimage-files%2Farc-length-area-sector.png&imgrefurl=https%3A%2F%2Fwww.onlinemathlearning.com%2Farea-circles-sectors-triangles.html&tbnid=5zeJ7XadkyCKZM&vet=1&docid=WQjMAL0Q1p3_hM&w=765&h=489&q=arc sector and arc length&hl=es-ES&source=sh%2Fx%2Fim

Wow long link but yeah click

Lol

[arctic vortex]

my geometry finals are coming up and i think i will fail any tips for me

to pass

[upper karma]

Bruh

[arctic vortex]

or anything i should study more

[upper karma]

yes get questions that make sense

that's step 1

[arctic vortex]

..

[upper karma]

You posted 2 problems and literally skipped them?

[arctic vortex]

yeah this one

[upper karma]

For this one, recall a circle theorem

[arctic vortex]

whats the circle theorem

If inscribed angles of a circle intercept the same arc then they are congruent. This follows from the Inscribed Angle Theorem. 7. If two chords intersect within a circle, the product of the measures of the segments of one will be equal to the product of the measures of the segments of the other.

this one

no wait this is tangent

wait

help

nvm

i think i got it

how do i do this

nvm i got that

[upper karma]

lol

something tells me you're not even trying the questions before asking

[arctic vortex]

i am

but i dont know the formula to solve it

for the rachel one i just solved the area and divided by 6

and did 12pi = 18.85

and 6pi

3pi

pi

[royal edge]

I think my notes are wrong. It says z-scores are the standardized scores for x values. Is that wrong or correct

[small perch]

I don't understand what it wants me to do.
I converted 325 degrees to 5.672320069 radians. However, I'm not sure what it wants me to do next.
Do I divide by pi? - (Express in terms of pi)
Do I insert 5.67 into the s, and 12ft into r, for O=S/R?
Any help would be appreciated.

[silent plank]

don't round until the end, until then keep your values in terms of pi. (just use the expression that you entered in your calc before pressing =)

s is what you're trying to find

the central angle will be theta

and s = r * theta

[dusk stratus]

could someone please explain how they got arctan(1/-1) to be 3pi/4? the answer i got was -pi/4, and it seems like they got the supplementary angle as an answer of the angle that i got?

and the range of arctan lies in the first and fourth quadrant, so that also confuses me

[silent plank]

they're using improper notation / function

[azure reef]

they take main interval as (0, pi)

which is bad since tan is undef on pi/2

[silent plank]

they should be using the atan2 function

theta = atan2(1,-1) = arctan(-1) + pi

[dusk stratus]

but why do they use the atan2 function here? is it because im dealing with complex numbers?

[silent plank]

the arc functions have limited range

it apples to the terminal positions of real points too

you need something to account for things that are in quadrants 2 and 3

[dusk stratus]

oh! so Q2, i add pi, and Q3, i subtract pi

am i right?

or

the complex number im dealing with is -1+i, which would lie in Q2

so it would make sense to add pi?

[silent plank]

yes

technically you don't need to know about the existence of the actual atan2 function.
just know about properties of the arctan function, quadrants and how to perform the appropriate shifts

(so there's no need to memorise it verbatim)

[dusk stratus]

yeah, i havent used the atan2 function for anything in a long time

but i get how it works :) tyy

[small perch]

Sorry, I didn't respond sooner. I am also working and we have been really busy today.

So, i take 5.672320069*12=68.06784083

How would I write this in terms of pi? Do I divide by pi?

[silent plank]

(just use the expression that you entered in your calc before pressing =)

i.e. to convert 325° to radians I'm assuming you did:

$325 \times \frac{\pi}{180}$

[somber coyote]

ramonov:

[small perch]

I'm not sure what you are referring to sorry.
5.672320069 is 325 degrees converted to radians. then I multiplied by 12 (ft) to get 68.06784083.
I'm just not sure where I'm going wrong.

No my calculator did the conversion for me

[silent plank]

and that although unsimplified will be the exact value of your angle in radians

oh...

in that case learn the conversion that pi = 180°

[small perch]

How do I simplify 68.06784083 more than it already is? Well I supposed it can be rounded. So 68.07.

[silent plank]

the instructions are confusing as fk

[small perch]

Yes, they are lol

[silent plank]

i mean you did round correctly to 2dp

[small perch]

I might need to get my teacher to clarify what it actually wants.

[silent plank]

so i'm assuming they actually want you to keep stuff exact

so try not to rely on the calculator for certain calculations

keep everything exact

specifically the angle would be what I wrote above

and multiply that by 12 (ft) for arc length

$s = \frac{325 \times 12 }{180} \times \pi \text{(feet)}$

[somber coyote]

ramonov:

[small perch]

oh okay. I'll try that thank you so much

[silent plank]

and use appropriate tools to simplify that fraction

(express it as an improper fraction because mixed numbers are crap)

[small perch]

great that worked thank you

[dark shuttle]

I really need help

with finding the "domain of validity" of an expression

[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.

[idle bobcat]

Hello quick question: When people calculate area, they say their final answer is in units squared. But isn't the squared portion of the unit not relevant once you reach a final answer? Like say you have a 5x5 square, why does it make sense to say the final answer is 25cm^2?

I would understand if they said it the area was 5 squared, but I feel like I'm missing something

[upper karma]

???

but isn't the square portion of the unit not relevant once you reach a final answer
??

[idle bobcat]

?

Like you wouldn't say the answer is 25^2

you would just say 25

But apparently you DO say 25^2 is the final answer

and thats why I'm confused

[upper karma]

Uhh? For a 5cm side length square area, A=l² so A=5*5=5²=25 cm²

Its not the same 25² than 25cm² which is the correct form

[idle bobcat]

why is the unit of measurement squared?

is it because 5cm*5cm = 25cm^2

i guess that makes sense

but using units in a formula is new to me so doesn't feel intuitive i guess

[upper karma]

Yeah fair

[pure pivot]

Can anyone help me out with this graph

[pure pivot]

Anyone

<@&286206848099549185>

[idle bloom]

Everything is right except the vertical shift

Remember the vertical shift is halfway between the maximum and minimum

Oh wait

Amplitude also needs fixing

[pure pivot]

What’s wrong with the amplitude

4-1=3

Ohh

Can you help with me with the vertical shift

[idle bloom]

4-1=3
@pure pivot
how did you get 1

[pure pivot]

4-2/2= 1

D=1

[idle bloom]

yea

you added the max and min

[pure pivot]

4-1=3

[idle bloom]

and divided by 2

and got 1

[pure pivot]

Yes

[idle bloom]

that's your vertical shift

+1

[pure pivot]

For this graph isn’t there a phase shift

[idle bloom]

it's not a phase shift

it's a vertical shift

so instead of +2

you should have +1

you said D = 1

[pure pivot]

Ok I got that

[idle bloom]

D is the variable commonly used for vertical shift

[pure pivot]

Yes

But does this graph have a phase shift?

[idle bloom]

if you write it as a -cosine

it doesn't have a phase shift

[pure pivot]

Cause rn I’m 3cos(pi/4x)+1

[idle bloom]

-3

not 3

[pure pivot]

Why is it -3

[idle bloom]

or if you want to use 3

then there's a phase shift if you want to use 3

[pure pivot]

Ohh nvm

[idle bloom]

you get why it's -3?

[pure pivot]

The graph goes up and it’s cod

Cos*

[idle bloom]

cause we start at the minimum, reach the midline, then the max, then the midline, then back down to the minimum

that's the pattern for -cosine

[pure pivot]

I got it

Thank you

[idle bloom]

👍

[sinful anvil]

Lol

I’m supposed to find a sine and cosine function for the graph. Sorry that it’s sideways lol

[upper karma]

,rotate

[upper karma]

anyway the sine and cos are going to be the same, just slightly different argument, since one will be a pi/2 - x shift of the other one up to other shifts and transformations

[sinful anvil]

It’s in degrees lol

[upper karma]

ok

note that

[sinful anvil]

I didn’t even start the 11th grade lol

I’ll take note

For the sine function

I had f(x)= 4 sin(x - 270*) - 3

[upper karma]

,w sin(x + pi/2) = cos(x)

[somber coyote]

Results provided by WolframAlpha

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

Requested by polynomial#4757

[upper karma]

not sure why

but anyway

pi/2 = 90 degrees

same thing, just wolframalpha doesn't like it for some reason in here

[crystal quiver]

hello

[sinful anvil]

Lol

[crystal quiver]

@upper karma

can u help me

[upper karma]

why would you immediately tag me

[crystal quiver]

to get ur attention

sorry

can u help me pls

[upper karma]

idk

[sinful anvil]

Is f(x) = 4 sin(x - 270) - 3 right?

For the sine function

[azure reef]

wdym

here?

[sinful anvil]

👍

[azure reef]

but why you even need shift of sine here

it is pure sine

graph depicts literally 4sin(x)-3

[sinful anvil]

Oh, I’m dumb af

I didn’t realize that lol

[azure reef]

well, if you want shift you can shift by 2pi and get the same

[sinful anvil]

So I now have the sine function

Now for the cosine function lol

[azure reef]

you want to find f(x)=some function of cosine

with the graph above?

[sinful anvil]

Yeah

[azure reef]

well, you can use that sine is cosine shifted

for example, cos(x)=sin(x+pi/2)

[sinful anvil]

The question was « Determine 2 equations using the parent functions sinx and cosx »

I had an equation for cosine as well

But I think I’m being retarded again

[azure reef]

question is badly worded ig

what does mean determine two equations

"equation" imply that they should be equal to something

[sinful anvil]

For the curve, it wants me to find a sine equation and a cosine equation

Using parent functions

[azure reef]

well for cosine one

parent function is cosine

is it obvious that it is also 4cos-3?

[sinful anvil]

To me it wasn’t obvious. It was my 1st time working with Trig functions

I only guessed the equations based on the amplitude and the horizontal shift

[azure reef]

wtf was that

[sinful anvil]

Idk

[azure reef]

wel for trig remeber

that sine-cosine have amplitude 2

[sinful anvil]

Can you not @pallid sierra

[azure reef]

that sine-cosine have amplitude 2
since they both lie in [-1, 1]

so that's why you need multiply by 4

[sinful anvil]

Oh

[azure reef]

and -3 you need since sine is 0 at 0

but graph shown -3

[sinful anvil]

👍

[forest niche]

Does anyone know how to graph this

[upper karma]

yes

watch this insane shit

,w graph -3csc(2x + pi/4)

[somber coyote]

Results provided by WolframAlpha

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

Requested by polynomial#4757

[upper karma]

insane right

[forest niche]

Lol I have to find the points

[upper karma]

well then

have fun plugging in marginally smaller and bigger values

[forest niche]

[azure reef]

@forest niche wdym graph

i mean which precision of graph you want

[azure reef]

ok so

csc = 1/sin(x)

where do u think there will be asymptotes

[forest niche]

Every pi/3

Pi/b

[azure reef]

what

[forest niche]

Isn’t that the formula to find the asymptote

;-; I’m sorry I’m lost

[upper karma]

Use common sense, if csc(x)=1/sin(x) we'll look at when the denominator equals 0

@forest niche

[forest niche]

Wdym I’m sorry my trig is horrible

[upper karma]

That was not trig really

Uh

[forest niche]

Idk I’m stupid I guess ;-;

[upper karma]

$\csc(t)=\frac{1}{\sin(t)}$ and we want to see where the denominator equals 0, which will give us the vertical asymptotes

[somber coyote]

Al𝟛dium:

[upper karma]

Or better lets just find the domain

[forest niche]

Isn’t the domain all real numbers

[upper karma]

well....

the domain is approximately...

i better not say it, i will anger the ann

[dark sparrow]

too late lmao

[forest niche]

It’s ok I’ll figure it out

[upper karma]

the domain is $\mathbb{R} \setminus {0}$

[somber coyote]

polynomial:

[upper karma]

Oh f i closed the app and the text i was writing got deleted

Ill write it again ig

[forest niche]

It’s whatever I’m gonna watch some more YouTube videos it’s not a big deal

Thanks for trying to help

[upper karma]

...

[somber coyote]

Commander Vimes:

[azure reef]

lmao

[upper karma]

yes

[azure reef]

,w csc(588727723pi)

really?

[upper karma]

see, it's equal to the conjugate of infinity

[somber coyote]

Results provided by WolframAlpha

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

Requested by Commander Vimes#7532

[azure reef]

but it is complex infinity

and infinity itself is NaN

[silent plank]

@upper karma with no respect at all if you're gonna deliberately spout misinformation please fuk off

[upper karma]

😦

tbh i just forgot what function we were talking about lol

[azure reef]

If you do not know what we are talking about, do not interfere

@forest niche

[somber coyote]

Commander Vimes:

$\csc (x) = \frac{1}{\sin (t)}$

[silent plank]

variable fk up

[forest niche]

Wait what I didn’t even say anything

[azure reef]

so as it was said it wail to exist when sine is zero

variable fk up
@silent plank it is plan!

[somber coyote]

Commander Vimes:

[azure reef]

so look

it fails to exists and results in asymptote when sin is 0

because limit of 1/x as x -> 0 is inf

that means that for asymptotes you need sin(t)=0

when sine is zero

do you know it?

@forest niche

[forest niche]

I’m genuinely lost rn

[azure reef]

well what do you know about sine?

[forest niche]

Idk sine=O/H? 😂

[weary drift]

i expected at least 10 copswings on poly's last latex

[azure reef]

well 4

Idk sine=O/H? 😂

[weary drift]

that's better but not enough

[forest niche]

Yea idk what you wanted

[azure reef]

what is O/H

do you know anything about sine?

[forest niche]

Opposite/Hypotenuse

[azure reef]

oh yes

nice

[forest niche]

Idk what else are you looking for

[azure reef]

well

,w unit circle

[forest niche]

Oh

[upper karma]

good one vimes

,w define unit circle

[somber coyote]

Results provided by WolframAlpha

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

Requested by polynomial#4757

[forest niche]

Sin = y cos= x?

[next jackal]

Somebody said that trigonometry is more about circles than triangles

[upper karma]

pretty shit illustration ngl

[azure reef]

Sin = y cos= x?
@forest niche yes

shitty drawing

[somber coyote]

Commander Vimes:

[azure reef]

is angle

@forest niche is this drawing undertandable for u

[forest niche]

Yea

[azure reef]

so look, when sine will be zero then

[forest niche]

It touches the x axis

[azure reef]

or what is the value of angle?

[forest niche]

Idk

[azure reef]

just look at drawing

what angle makes radius with x-axis when y-coordinate is zero

[forest niche]

Cos?

[azure reef]

cos?

what

how cos can be angle

[forest niche]

Phi then? It’s whatever man I give up

This problem is slowly making me mental

[azure reef]

prolly that is better

when sin is 0?

literally, when blue part is absent?

[calm shoal]

hello im new to the server and was wondering if i can get any help for these kinda problems

[azure reef]

channel is clearly occupied

[forest niche]

When sin is 0 cos is 0 what am I missing

[azure reef]

what

@forest niche ok let do that way

draw position of radius so that sin is 0

[long shale]

@calm shoal just break the vectors into components, You have 2 vectors, one of magnitude 353, and an amgle of 229.2, you need to get the x component and the y component, using the cos(a) = x and sin(a) = y

[calm shoal]

so it would be cos(229.2)=353?

[long shale]

have you reviewed what the component method is?

[calm shoal]

not really lol

[next jackal]

@long shale what happens when one side of a triangle =0?

[long shale]

@next jackal are you asking, when your on the axis?

[next jackal]

Ohh sorry, I mistook you for that@forest niche guy

[long shale]

@calm shoal

[calm shoal]

ooo

[long shale]

you can solve for the angles at the origin, use some trig and youll get the componets, do the same for B then add components

[silver fable]

Hi I have this problem I know how to find X but I also need to find the angle between A and B

Is it possible? anybody knows how can I get that?

[long shale]

@silver fable what have you tried

[silver fable]

Its for programming purposes I tried to get the angle of OA and OB and subtract them but doesn't seem right

[long shale]

is OC is perpendicular on AB

[silver fable]

Yes it's always perpendicular

[long shale]

so how can you use that fact?

[silver fable]

Oh right I can make a right triangle with it!

So I can easily get the angle between OC and OA and then after I got that I multiply it by two and I get the angle that BOA makes, is that it?

[long shale]

yes

and are you good for x?

[silver fable]

Yeah I already solved for X it's basic pythagora but sometimes it gives me wrong answers when the arc is so small which I don't think is a problem with the math which brings me here for the second solution

Thanks

[neon sun]

Hi I’m a little stuck on this problem! I’m not quite sure what I can do next and was wondering if I could get some help😅

[upper karma]

@neon sun ?

[neon sun]

The question is in the brackets. To put the expression in terms of a &b when a=sin(x/2) and b=cos(x/2). If you understand what I’m supposed to do, could you give me a hint as to what I should do next?

[upper karma]

@neon sun if you know the formula for sin(2x) and cos(2x)

use the same formula, but do sin(x) and cos(x) instead

as if it were x/2 + x/2

and expand

[neon sun]

I see. Thank you!

[fickle sleet]

Could 2-cos^2x/sinx not be simplified to 1+1-cos^2x/sinx?

Like it makes sense in my head but apparently its not correct

From here i could turn 1-cos^2x to sin^2x

Which makes the whole expression simplified to 1+sinx

The textbook just says cscx + sinx which makes no sense to me

[versed river]

Which makes the whole expression simplified to 1+sinx
this seems to be the part where you made the error

wait also

could you tell me

which parts are actually in the fraction

is it 2-cos^2x/sinx or(2-cos^2x)/sinx

[fickle sleet]

Sorry yeah the formatting is pretty awful

The latter

[versed river]

ok good, the error is where you went from (1+sin^2(x))/sinx to 1+sinx

can you tell me how you got that result?

[fickle sleet]

sin^2x = sinx + sinx

One instance of sinx cancels out with denominator

[versed river]

sin^2(x)=(sinx)^2

=sinx*sinx not sinx+sinx

[fickle sleet]

wait hold up right my bad

[versed river]

would still be wrong but

[fickle sleet]

but still though, that means (sinx)(sinx) right?

[versed river]

yes

[fickle sleet]

A factor can cancel with denominator right?

[versed river]

can you be more specific

[fickle sleet]

For example (x+3)(x+3)/x+3 = x+3

Since one factor cancels

so why not here?

[versed river]

for x!=-3 but yeah

because you have the one here

can you show me how that applies in this case

[fickle sleet]

Isn't every factor that is the same, has the ability to cancel? (3)(3)/3 = 3 for example

I can't think of a scenario where this doesn't apply

[versed river]

yeah its true but what you have is $\frac{1+\sin^2(x)}{\sin(x)}$

[somber coyote]

Sneaky:

[versed river]

which part are you factoring out of the numerator

[fickle sleet]

sin^2x

Since it composes out of 2 sinx's

OH WAIT I see now that 1 prevents this from being simplified right?

[versed river]

$\frac{1}{\sin(x)}+\frac{\sin^2(x)}{\sin(x)}$?

[somber coyote]

Sneaky:

[versed river]

that gives you $\frac{1}{\sin(x)}+\sin(x)$

[somber coyote]

Sneaky:

[fickle sleet]

Right I see now

[versed river]

OH WAIT I see now that 1 prevents this from being simplified right?

yes

[fickle sleet]

I was about to make an analogy with (1+3^2)/3 for a second there and I caught myself

That clears it up thank you for pointing out the dumb mistake

[versed river]

no worries

[fickle sleet]

I get it now too with (1/sinx) + sinx

that's where the cscx comes from

[versed river]

👍

[upper karma]

csc(x) + sin(x) yEet

[solar shale]

How do I tell

[silent plank]

circles are congruent if they have the radii of equal lengths

[solar shale]

Oh ok then it’s not congruent because they would have to overlap

[upper karma]

no

did you read what ramonov said

[sleek thistle]

Am i correct here?

nvm

[upper karma]

Ababababababba

[sleek thistle]

Abababa

[tawdry pivot]

Abababababa

[arctic vortex]

can anyone help me with some geometry problems ??

[upper karma]

@arctic vortex go with them

[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]

ok

:) u are very nice

is this a?

@upper karma

[upper karma]

First, is it a test?

[arctic vortex]

nope

[upper karma]

Ok good

[arctic vortex]

it is homework assignment

[upper karma]

Let me see

[arctic vortex]

my test is on august 8th

its the final exam

i have 15 problems but i only need help with like 6

[upper karma]

Okay yes it is a

[arctic vortex]

yay

how do i do this one

im confused

[teal stag]

You have angles and a circle. I would recommend going through the facts you have about angles and circles, and seeing what they tell you. You might not know which facts will be helpful, but if you go through them you'll have more info than before

[arctic vortex]

okay

i will do that

in the mean time can u check something else @teal stag ?

im confused on inscribed angle and central angle

[teal stag]

The boxes you have look good. "central" means it's at the center of the circle (which is A). "inscribed" means it's drawn on the circle itself (which D is)

[arctic vortex]

ohhhh

so it is correct

ok i thought that central meant the A

[teal stag]

good thinking!

[arctic vortex]

thank you :)

for this one the answer i got was 2/9 = 169pi

how do i make it so that its one of the answer choices

@teal stag

[teal stag]

2/9=169pi is an equation, and I don't think that's what you meant to write

[arctic vortex]

yes

[teal stag]

Both "2/9" and "169pi" are important values here, but what is the area (in terms of pi)?

[arctic vortex]

169* * 3.14?

[teal stag]

Let me rephrase. "2/9 = 169pi" is obviously false because the left side is less than 1 and the right side is over 400. What did you mean to write?

[teal stag]

That's an equation which is false

[arctic vortex]

yes

[teal stag]

The area of the section of the garden isn't the equation "2/9 = 169pi", it's the number: "???"

[arctic vortex]

yes

but how do i find that number

[teal stag]

Let's break this down. What's the area of the whole circle?

I think this is something you already figured out. I just want to highlight the steps of the process.

[arctic vortex]

530.93

is the area

[teal stag]

That's approximately the area of the whole circle. If you just say "the area" you could confuse yourself as to when you'

re talking about the area of the whole circle and when you're talking about the area of the sector

The area of the whole circle is exactly $169\pi,\mathrm{ft}^2$

[somber coyote]

dirib:

[arctic vortex]

yeah im talking about the sector area

[teal stag]

The sector area is much less than 169pi

[arctic vortex]

yes

[teal stag]

That's why I'm emphasizing the distinction

Now, what fraction of the whole circle is the sector/section of the garden? (this is also something you figured out)

[arctic vortex]

so its 2/9 of 169pi

[sinful anvil]

An area of the sector can be found by using the formula theta/360 ^ Pi * R²

[teal stag]

right. And when you have a fraction "of" a number, what do you do.

[arctic vortex]

multiply

[teal stag]

Right

[arctic vortex]

117

[teal stag]

so it's not "2/9=169pi"

[arctic vortex]

.98

[teal stag]

it's "(2/9)*169pi"

[arctic vortex]

ohhh ok

so its 118

rounded

wow u are smart

[teal stag]

You already had the pieces. I just helped you put them together

[arctic vortex]

yes thank you :))

[teal stag]

I have to go now, but if you have more questions, others here can likely help

I was happy to help, though

[arctic vortex]

okay

same i was happy for u to help me :)

ok i will post this one is it 12 ??

because 2 * 3x-2 = 5x

x=4

and then 10 + 2 = 12

[sinful anvil]

I don’t even know vectors

They’re too confusing

[arctic vortex]

oh

[olive solar]

@arctic vortex are you asking for help with this question?

[arctic vortex]

yes

@olive solar is it 12?

[olive solar]

yeah

although id write it as 2*(3x-2) instead

when you write it as 2*3x-2, it looks like you mean 6x-2 rather than 6x-4

[arctic vortex]

yeah true

wait would tihsb e 50

this be *

because 92 is a vertical angle

@olive solar

[olive solar]

are you certain angle RSQ = angle SQT and not RTQ?

look at the arcs formed by each angle

[arctic vortex]

oh yeah

92+50

180-142

38

38?

@olive solar

[olive solar]

yes

[arctic vortex]

is 70 correct @olive solar

because i think its like a triangle

[olive solar]

how did you decide on 70?

[arctic vortex]

180-110

[olive solar]

so the portion of the segment SR that is also within C is a chord, yes?

[arctic vortex]

yea

[olive solar]

and QS seems to be tangent to the circle

so here you would be using the tangent chord angle thm

[arctic vortex]

ohh

[olive solar]

the question doesnt really confirm whether that's a tangent though so

[arctic vortex]

i solved that one @olive solar can u help with tihs?

[livid sequoia]

Is this right?

[olive solar]

@arctic vortex do you know how to find the angle of an internal angle given the measure of its arc

@livid sequoia do you know the classification of a rhombus

[arctic vortex]

i think square is a rhombi

parallelogram, trapezoid, rectangle

i think

[livid sequoia]

a rhombus has 4 congruent sides

so its not A :L?

[olive solar]

can c be considered to have 4 congruent

[livid sequoia]

No

So its B C and F

[olive solar]

yep

[livid sequoia]

thank u

[olive solar]

you are able to find angle C given everything in the diagram@arctic vortex

once you have angle C and angle A, angle B can be found based off the properties of a triangle

[arctic vortex]

64?

[olive solar]

howd you get 64?

[arctic vortex]

47+69

180-116

[olive solar]

remember, C is an inscribed angle

[arctic vortex]

yeah

idk

cna u tell me >

?

[olive solar]

i could tell you the solution but i won't

look at how to get the measure of an inscribed angle from the arc it forms

[arctic vortex]

o

wait

i think i know

wait no i dont

[olive solar]

do you have a textbook?

id recommend that you go through either that or get some proficiency with google

you could search "inscribed angle theorem" and youll get what you need here

[arctic vortex]

o i got it

ty

i finished it a while back actually lol

[sly belfry]

how can I retrieve the general formula of a transformation in the complex plane by having the matrix form?

I'm used to doing it the other way around

even better - I have the general expression in terms of x and y

I think I'll plug in i in the y value?

[dark sparrow]

wym matrix form

[sly belfry]

well

I have general expression f(x, y) of a plane transform

and i want it to be f(z) = cz + d

and I forgot how to do it 😐

[dark sparrow]

how do you know it's even expressible in that form tho

like

[surreal bolt]

Um, can you figure out what the transformations are that get you from start to finish?

[dark sparrow]

it need not be

bc you're being overly general rn

[surreal bolt]

(agrees with ann)

[sly belfry]

ok, here's the exercise

[dark sparrow]

can you maybe be more specific

[sly belfry]

so I know f is a translation after a transformation

with this matrix form (right?)

I suppose it is an homotety

because the ratio is not 1

and it's given that f(x, y) is a transformation, if that helps...?

[dark sparrow]

uh

well i guess you could write $f(x,y) = \mat{2\3} + \mat{\sqrt{15}/10 & -\sqrt{10}/10 \ \sqrt{10}/10 & \sqrt{15}/10} \mat{x \ y}$

[somber coyote]

Ann:

[dark sparrow]

for the actual "matrix form" of the function

[sly belfry]

(yes, that's what I did, but it's not terribly useful)

[surreal bolt]

[a, y] * (2 x 2 rotation matrix) + (x_0, y_0) = (xfinal, yfinal) ? 🙂

[dark sparrow]

and that makes it more clear that this actually CAN be expressed as an affine map on C

[surreal bolt]

sorry I got the order wrong

[dark sparrow]

$f(z) = 2+3i + \frac{1}{10}(\sqrt{15} + \sqrt{10}i)z$

[somber coyote]

Ann:

[sly belfry]

hmm

how?

y = i?

[dark sparrow]

multiplication by $a+bi$ considered as a linear map on $\bR^2$ has the matrix $$\mat{a & -b \ b & a}$$

[somber coyote]

Ann:

[sly belfry]

oooh

actually pretty straightforward

if I want to calculate the image of any vectorial eq.

in this case, something like (0,0) + (a, b)

is it okay to plug in z = a + bi in the f(z)?

[dark sparrow]

i mean...

yes

[sly belfry]

sorry ann

i promise I'm way better in some other stuff 😦

[nocturne dove]

could anyone help me with this? Im not sure where to start

[little osprey]

What have you tried

[nocturne dove]

Well its a proof so all I've really done is write the givens lol

tried looking online for some help

didn't really know what to look for

[little osprey]

Is this a question from your teacher or

[nocturne dove]

ye

[little osprey]

ah ok

Well

@nocturne dove Anything?

[nocturne dove]

Nothing unfortunately, Im asking around with my friends but none seem to remember

[little osprey]

Have you used any circle theorems

[nocturne dove]

I'll go check some more out

[little osprey]

Hint: right angle

[nocturne dove]

?

going on a whim here

[little osprey]

Join AP and BQ

[nocturne dove]

ah

[little osprey]

Anything?

[nocturne dove]

sorry nothing's clicking for now, but thanks though!

[little osprey]

Lol

Well you could stay here

and I could try and guide you to the right direction

[nocturne dove]

that'd be great lol

[little osprey]

Well what happens when a radius is drawn to a tangent point

I want to help, but you have to put some effort in :/

[nocturne dove]

Yeah, sorry man, I not too good at this as you can see

I remember something about it being perpendicular or something like that

[little osprey]

yes

right angled

so which angles would be right angled then?

[nocturne dove]

I would assume these four

[little osprey]

Ok, correct, but remember AB is also tangent

[nocturne dove]

Sorry for delayed answers Im just looking at definitions at my other screen to piece things together

[little osprey]

No worries dude take your time, I understand

[nocturne dove]

well I did find this

is it applicable to this or na

[little osprey]

PAM and QBM will form right angles as well?

[nocturne dove]

now that I wouldn't have gotten

[little osprey]

Because they are tangent

[nocturne dove]

yep

ah

[little osprey]

Do you understand the reasoning

[nocturne dove]

a bit but could you explain why it being tangent forms right angles

[little osprey]

The proof?

[nocturne dove]

actually

nah its ok

I think I can handle the rest with my friends seeing as theyre awake now

thanks for all your help bro

I really appreciate it

[little osprey]

No worries dude, sorry I couldnt explain it well enough

[nocturne dove]

nah its alright Im just not good at these things

[upper karma]

nah its alright Im just not good at these things
@PrivateFishy#2029
What's the problem??

Could i try to help

[little osprey]

He left the server

I did try and help him

it was a geo problem

[sleek thistle]

can i get some help?

[long shale]

do you know what a sine or cosine is?

[sleek thistle]

so with part b, would y be tan(50)=x/y or cos(50)=y/12?

[long shale]

ya

[sleek thistle]

or can it be both

[long shale]

so its both, since x = 12sin(50)

they are equal

[sleek thistle]

okay yeah i just wanted to know if one of them was more

right i guess

i know it’s easier with cos so i’ll just put that

[long shale]

👍

[sleek thistle]

is this correct?

[neon sun]

I don’t think you proved triangle AEC=DEB. If I remember correctly from my geom, that should be your last statement if you are trying to prove it.

[sleek thistle]

ooh yeah you’re right

[upper karma]

Pretty sure that's a typo

[sleek thistle]

yeah i meant to put AEC congruent to DEB

oh yeah i also need help with a trig problem

hold on

for c would i be right in saying i can’t use the tangent ratio?

and sorry for the bad hand writing, i could clarify if needed

[glacial haven]

no

you can use tangent after solving for one of the unknowns

[sleek thistle]

oh wait let me clarify the question

the question says would you be able to use the tangent ratio to solve part a-b

[glacial haven]

tan(50)=12sin(50)/12cos(50)

you can

[sleek thistle]

so if you use tangent to solve a you can get x?

wouldn’t that give you tan(50)=x/y?

[glacial haven]

yeah

and x=12sin(50)

[sleek thistle]

but when i put it into my calculator it gives me x=1.19175359y

so wouldn’t you need at least one value before solving with tangent?

[glacial haven]

if you use tangent you're gonna have to use cosine and sine

but its still possible

[sleek thistle]

oh okay

i just think the question is asking whether we can use tangent for both before using anything

[glacial haven]

do you have the question?

[sleek thistle]

yeah

they should’ve clarified more i think

[glacial haven]

uh that question isnt clear

but you most definitely can use the tangent ratio

[sleek thistle]

yeah i’m just gonna ask my teacher

“Are you able to solve a or b with the given info using tangent as opposed to how you actually did it? That’s what the other student is claiming.”

i think i’d be correct right?

cause you can only use tangent when you solve one of them with sine or cosine

[solar shale]

Would that be considered an inscribed angle or no? Bc if so then woudlnt 1=21.5?

Or would I do like 92-43=(1)

[silent plank]

neither

[solar shale]

I honestly don’t know what I do I thought I was an inscribed

[silent plank]

1 isn't at the circumference

look up something like
outside angle theorem for circles

which can be derived from inscribed angle theorems

[solar shale]

Got it 24.5

92-43 then divide by 2

?

[silent plank]

yes

[solar shale]

Thanks also I need help with this question could you like tell me with theorem or what rule I use so I can look it up

[silent plank]

cyclic quadrilaterals and alternate segment theorem

[solar shale]

Ok thank you

[silent plank]

whoops i mean inscribed angle theorem / central angles instead of cyclic quad

[solar shale]

Ok

Oh so the answer would just be 85

[silent plank]

yes

[shrewd wasp]

Any idea to solve acos(x)+bsin(x)=c

[upper karma]

Looking for x?

[shrewd wasp]

yes

I found an identity

[upper karma]

Ok

Epic

[empty hornet]

I got this question as homework a couple days ago and I've been confused since. I got no clue where to start.

[upper karma]

What's the general formula for the volume of a prism?

[little osprey]

Annulus

or how you spell it

[empty hornet]

Its just A x H

[upper karma]

Ye

[little osprey]

How would you go on calculating the volume

[upper karma]

You have both A and H in this case

[little osprey]

oh

he does

[empty hornet]

yeah that would equal 910cm

cubed

[upper karma]

I thought the radii were given too

[empty hornet]

sadly not

thats why im confused

[upper karma]

That's the correct answer

[empty hornet]

that should make more sense then

i though it meant to calculate the actual opening

[upper karma]

It has to be the right answer

Because that's the only volume you are able to calculate

[empty hornet]

yeah

[arctic vortex]

ok so im taking my geometry finals soon and i want to ask if there are any tips u guys wanna give me

[upper karma]

Draw labeled and precise diagrams

Use various colors

[little osprey]

Also

Have fun

[silent plank]

don't forget about the ambiguous case with sine

[little osprey]

Good tip

[charred zephyr]

hey can someone give me a little hint.
ABCD is a parallelogram and I is midpoint of line BC
Line HD is perpendicular to AI with H a point of AI
Show that CD = CH
here's the sketch i have drawn:
idk what should i use, but i think it'll be about angles. i tried to find similar triangles but couldn't.

maybe proving HCD is isoceles

[little osprey]

shall we chase angles or shall we not?

[charred zephyr]

idk

maybe$

so we can get HDC = DHC

[little osprey]

yeah maybe, I'll have a go

I cant think of anything rn...

[charred zephyr]

me 2

😦

[livid sequoia]

help

[little osprey]

@livid sequoia What have you tried

[livid sequoia]

I subtracted the 2m and 1m but I dont know what the

I confused about the 2 and 8

[little osprey]

Do you know what the pythagorean theorem is

[livid sequoia]

yea

[little osprey]

and why did you subtract?

[charred zephyr]

apply pythagoras two times

[livid sequoia]

to find the other leg fo AD

[little osprey]

why would you add

You have the hypotenuse and leg

[livid sequoia]

you told me why did I subtract

o

[little osprey]

I said why did you subtract because you didnt square them

[charred zephyr]

you must square sqrt(2) and 1 and then substract

[little osprey]

yes but remember

[charred zephyr]

yes

[little osprey]

its c^2

so right now you have

c^2 = 9

Do you get what I mean

[charred zephyr]

add a last step

[livid sequoia]

i dont know what to do next

OH

I

[little osprey]

yes

youre done

[livid sequoia]

thank u

[little osprey]

no worries

[upper karma]

@charred zephyr I'll take a look at your problem layer, Imma @ you if I find something interesting

[little osprey]

Hell yeah HoboSas

[upper karma]

Ok that probably took too much time

||Let G be the midpoint of side AD||

||For Thales theorem we have that DM=MH (you can also check this noticing that triangles DMG and DHA are similar and their "ratio of similarity" is 1/2)||

||Consider now triangles DMC and CMH, they are congruent for SAS (DM=MH, right angle and a shared side). In particular CD=CH.||

@charred zephyr

[lusty quest]

how would i find the area of this triangle?

[little osprey]

@lusty quest Heron's formula

[lusty quest]

ahh ok thanks

[little osprey]

Have you heard of it?

[upper karma]

definite integrals

[lusty quest]

yes, i just havent done this in a while

[little osprey]

Oh wait

Distance formula

then herons

[rigid thunder]

You can measure area of planar object by weighing it on scales: cut it out of some tin (to scale, for example in cm), weigh 1 square cm piece of tin, then weigh the triangle, and you get numerical value for the area from weight