geometry-and-trigonometry

[autumn pelican]

opposite ~ -424.58

Seems oddly close to hypotenuse, did I still do somethign wrong

[trail tendon]

uhhhh

let me see

i mean

it shouldnt be a negative length XD

[autumn pelican]

Ohhh

I'm using desmos on the phone and it looks like I was on DEG but it was on RAD cause they make it weird

approx 392.65

[trail tendon]

ohh

[autumn pelican]

Chalk full of mistakes today my b

[trail tendon]

yeah thats right

thats the uh

opposite

using a similar method, do you know how to find the adjacent?

[autumn pelican]

Would we use tangent?

Oh no, we would use cosine correct?

[trail tendon]

yeah cosine

i mean

[autumn pelican]

SO cos(67.5) * 425 = adjacent

[trail tendon]

we COULD use the tangent because now we have the opposite length

but yk

[autumn pelican]

yea

[trail tendon]

yep

make sure its deg and yea

[autumn pelican]

approx 162.64

[trail tendon]

yea

[autumn pelican]

btw, in standard practice do you usually use two or three decimals if the teacher doesnt list for these kind of things?

[trail tendon]

i would think it depends, it doesnt really matter tho

[autumn pelican]

I see

[trail tendon]

like for me specifically i usually write down the amount of digits my calculator shows 💀

but yeah 1, 2, or 3 is fine

[autumn pelican]

haha I guess I could do that too

[trail tendon]

XD

[autumn pelican]

I just realized we completed the question though, thank you so much

[trail tendon]

yeah np!

[autumn pelican]

Appreciate all the help, it's so much nicer than asking GPT only to get 5 different answers from 5 different attempts

[trail tendon]

LOL

[autumn pelican]

True story LOL

[trail tendon]

🤣

gpt kinda sucks with math ngl

[autumn pelican]

I've noticed he has fallen off hard recently, I don't trust anythign from gpt anymore

[trail tendon]

🤣

[upper karma]

!nogpt

[lime crown]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[dark sparrow]

correct decision

[autumn pelican]

gpt would tell me otherwise

[dark sparrow]

last i heard, if you ask gpt whether it's trustworthy wrt math, it will admit that it's not good bc it wasn't designed for such things

[autumn pelican]

Yeah that's true, just a bad joke, it is a language model after all

[autumn pelican]

So... I have another problem that I am wildly confused on

ground, the balloonists simultaneously measure the angle of depression to two
consecutive mileposts on the road below. They find that the angles of depression
are 20ᵒ and 22ᵒ. What is the height of the balloon```

I have come upon this as my answer, to which I got from this video, but I don't even know what I am reading so I expect this to be wrong.

video in question: https://www.youtube.com/watch?v=B5pHWEh4MXc

[ionic timber]

not sure what they mean by "second can be derived from the first using the fact that sin is an odd function"

[dark sparrow]

sin(-x) = -sin(x)

that's what they're talking about when they say sin is an odd function

do you understand Y/N

[ionic timber]

yea

working on trying to understand how to derive it using that fact

[dark sparrow]

sin(x - y) = sin(x + (-y))

[ionic timber]

and if you swap the order its just written as the second one? i think i get that

[ionic timber]

what about the last part when they say we can derive the last two sum identities by using the first two and the fact that cos(theta-90)=sin theta

cant figure that out either

[dark sparrow]

sin(x + y) = cos(x + (y - pi/2)) presumably.

but that derivation is a bit obtuse imo.

[ionic timber]

ok thank you

[dark sparrow]

as far as derivations go i would recommend memorizing the ones for cos(x+y) and sin(x+y) both

[fringe halo]

dilations are similar but not congruent right?

[dense ravine]

Some one explain me trigonometry plz

[dark sparrow]

what like from nothing?

[empty yew]

The every points that located above the line x+y=1 and exterior to the circle of center at (2,1) and radius of 3 units.

Is my description correct?

<@&286206848099549185> anyone help me!

Anyone ensure my description(correct or not)

[dark sparrow]

1. the "does" in your sentence should be "is". affirmative "My description is correct" => question "Is my description correct?"

2. you can graph both inequalities on desmos to verify for yourself. you can simply type them as-is on separate lines and they will get graphed.

and since i could not restrain myself from correcting your grammar:

[empty yew]

what is desmos?

[dark sparrow]

The every points set of all points that located above the line x+y=1 and exterior to the circle of with center (2,1) and radius of 3 units.

[dark sparrow]

http://desmos.com/

online graphing tool

very powerful and useful

[empty yew]

oh thank you so much

[empty yew]

I graphed it in my graph sheet however I want to ensure

[dark sparrow]

again, graph it on desmos and compare it against yours.

[empty yew]

okok

[dark sparrow]

do you know the "give a man a fish" proverb?

[empty yew]

idk

🤔

what is the meaning of that?

[dark sparrow]

Give a man a fish, and you feed him for a day. Teach a man to fish, and you feed him for a lifetime.

[empty yew]

ohhhhhhhhhh

Great proverb

Is this a proverb for me?

[dark sparrow]

delete the second "is"

but yes

instead of giving you a one time check for your graph, i am telling you how you can check all your graphs yourself in the future

[empty yew]

thank you so much!

🙂

how to type a square in desmos?

[glass karma]

hello

can someone help me

[dark sparrow]

^2

[empty yew]

🆗

[empty yew]

how to check is our parabola is correct with focus and directrix?

if a single point on the parabola has equal distance from the focus and the directrix then is our parabola correct?

or do we want to check for multiple points satisfying the parabolic law?

@dark sparrow could you please help me?

on the above doubt

[dark sparrow]

no energy sorry

[empty yew]

ok no problem

<@&286206848099549185> anyone help me

[upper karma]

This is against the rules iirc <@&268886789983436800>

[rugged wind]

Thanks

[hardy yoke]

hey, so... if i have a question related to trigonometry, i can send it here?

[upper karma]

Sure go ahead

[hardy yoke]

i am stuck at this since yesterday:
1+cos(alpha)+cos²(alpha)+cos³(alpha)+cos^4(alpha)+... = 5, with 0 < alpha < pi, so, sin2(alpha)m equals to:

like, what the fuck do i do with this information

like, i know that sin2(alpha) is 2 * sin(alpha) * cos(alpha)

but the "sin(alpha)" part doesnt disappear

[upper karma]

… here is infinite summation right

[hardy yoke]

right

[upper karma]

Substitute cos(alpha) with u and evaluate the infinite series in term of u

Note that u would always be between zero and one following from the bound of alpha

[hardy yoke]

between 0 and 90°

[upper karma]

Show me the result of your substitution

It’s straightforward that the series will converge

[hardy yoke]

ok, i dont know if i skipped some classes, but how do i evaluate an infinite series in term of something?

and i didnt skipped any classes (so far)

[upper karma]

https://en.m.wikipedia.org/wiki/Geometric_series#Proof_of_convergence

[upper karma]

That’s a topic that goes in #calculus

The question you sent seems to be from competition math too so #competition-math might be better

[hardy yoke]

i dont even have that channel

[upper karma]

Anyways, the left hand side would converge to 1/(1-cos(alpha))

You should be able to figure the rest out ig

[hardy yoke]

ok, after reading that i have decided tô leave this question on the "complain tô teacher" list and focus on the rest

no high school assignment is worth this much effort

[golden trout]

So I have this problem, that says (It's the literal problem translated to english):

The complex set of numbers C, given the usual complex sum and product λ(a+bi)= λa+ λbi is a vectorial space. Demonstrate that the vector u = 1+i and the vector v = 1-i make a linear span of this vectorial space 

Any idea on how to approach this problem?

[dark sparrow]

from what language are you translating?

there seems to be some hiccups at or around the words "linear span"

@golden trout

[flint ingot]

what is a unit circle how do i use it to find derivatives of trig functions

[golden trout]

from spanish

[dark sparrow]

can i see the original

[golden trout]

yeah, I knew there were going to be some mistakes. In spanish it's called "sistema generador", so I googled that up and translated into english

[golden trout]

yeah sure

[dark sparrow]

"generating system" to be literal

or "spanning set" to be closer to existing english terminology.

in other words, you want to show that every complex number can be expressed as a linear combination of 1+i and 1-i.

[golden trout]

,rotate

[golden trout]

yeah, exactly that

[dark sparrow]

ok

well then do it

it went unsaid (even though it should have been said) that we are looking at real linear combinations (i.e. the coefficients in front of u and v must be real and not complex)

[golden trout]

any idea on how to start?

[dark sparrow]

definition of a linear combination

write out formally what you want to show, like with quantifiers and shit

thats how you start

[golden trout]

alright, I'll try that

thanks

[abstract yew]

Anyone know a way to understand trig identities better? They're my biggest struggle atm

I cant really remember them, even if i can intuit the way they work

[fallen elbow]

which identities?

[abstract yew]

Stuff like the sin addition rule

I cant remember it

And also things like tan/sec=sin, but to a lesser extent

[fallen elbow]

do you know how to remember the sin,cos,tan thing?

[abstract yew]

You mean the unit circle?

[fallen elbow]

i feel if you know that than everything else becomes easier

[abstract yew]

Or that cos/sin=tan

That one is easy

[fallen elbow]

cot*

[abstract yew]

Oops yeah

Maybe i just need to practice ugh

Thx

[fallen elbow]

just write them down on a paper

and then stick it somewhere you can see it on a daily basis

maybe that'd help

[solid needle]

so ive done this exercise with a lot of people on this math discord already

im gonna link some of the sessions i had with some other people

if they are too annoying to follow, feel free to ping me and i can just walk through it with you

#geometry-and-trigonometry message

[abstract yew]

Alright, thanks! Ill check them out next school hour

[brave mortar]

is 24 the opp or the adj

[dark sparrow]

well

is it across from the 47° angle or is it next to the 47° angle?

[brave mortar]

across

so opp

ok

[neon jay]

It’s opposite from the angle

Its opp

[jolly vapor]

is anybody online

@hidden tartan Hi dude, wanted to mention I got an 84 on the test. (not amazing but is pretty good) im not too satisifed because 6 people scored higher than me in a class of 20 and the rest lower.

[hidden tartan]

That's pretty great considering how lost you felt yesterday!

[jolly vapor]

this girl was on a 2 week hiatus and got an 100%

maybe shes just a quick learner

[hidden tartan]

You said you were one of 3 people that didn't understand the content right? going from the bottom 3 to the top 7 is no small feat 🙂

[jolly vapor]

well i was one of the 3 who didnt understand AT ALL

there are def more than 3 that understand 50/50

[hidden tartan]

Well maybe she missed some less important weeks, or she had some stuff to review

Well I'm proud of you 🙂

[jolly vapor]

thank you thank you

I have 5 problems to do

you helped me master finding circum and area of a certain sector for a circle

however sohcahtoa is prolly what i dont understand best in a triangle

[hidden tartan]

Oof, I'd offer to help you out with that but I think that's one of the subjects where I know too much to be helpful

Turns out that sin, cos, and tan are actually very closely related to how stuff works in circles, and by god do mathematicians love circles

Actually... I made a graph a few days ago when I was trying to show my sister some stuff about how right-angle triangles are related to circles

https://www.desmos.com/calculator/8uhevqohcn

idk if you'll find it helpful, but I think it's neat

[jolly vapor]

What if its one or two questions

[hidden tartan]

Oh yeah I can probably do that

[jolly vapor]

I wont be askinng you to do 8 or 9 like yesterday

Infact i got an 100 on that assignment!

[hidden tartan]

That's pretty great! So what do you have?

[jolly vapor]

I would appreciate if you could teach me the basics one or two times then I could do it from there

and 2, is a paper necessary for these types of problems

[hidden tartan]

Doing it on paper isn't necessary, but I do think working through problems by hand will help the concepts sink in a lot better

As for that problem though... what does m stand for?

[jolly vapor]

measure

so its askign what is the measure of abc aka what does abc = in a way

[hidden tartan]

So like the total side length?

[jolly vapor]

yes

total side length of abc only

[hidden tartan]

Is that all the information you have? and is the 80 degrees the angle between lines AB and CB, or showing the relative positions of A and C on the circle?

[jolly vapor]

u mean dont know or know

[hidden tartan]

know, like I could give you some crazy stuff about rotation matrices, taylor series, complex exponentials, and whatnot but you definitely wouldn't find those helpful

[jolly vapor]

i see so your ways of trying to educate me would exceed my brain capacity at my current age [or just not be helpful whatsoever]

is that what you mean

[hidden tartan]

Essentially yeah

Like, the distance from my understanding of sin/cos/tan is so different from the understanding that you actually need that it's hard for me to be helpful

I guess... do you at least remember what sohcahtoa stands for? I could maybe build on that a bit

[jolly vapor]

i do

sohcahtoa means

sin is um

sin is

opposite over adjacent

cah means

adjacent over hypotnuse

and toa means tangent is opposite over adjacent

so its knowing what to divide and overall helps you solve simple geometry

but these are the units i missed (the ones explaining sohcahtoa)

[hidden tartan]

Yeah, though I think you meant opposite over hypotenuse with sin

[jolly vapor]

yeahs thats what i meant sorry

Sine is opposite over hypotnus

Cosine is adjacent over hypotnuse

and tangent is opposite over adjacent

excuse me

[hidden tartan]

Essentially, all three of those functions take one of the two non-90 degree angles of the triangle and show you the ratios between the different sides

Now imagine you had a right angle triangle where the length of the hypotenuse was 1

[jolly vapor]

Ok

[hidden tartan]

if one of the angles is x, can you get the other two sides?

[jolly vapor]

idk

[hidden tartan]

Well sin(x) is the length of the opposite side over the length of the hypotenuse right?

[jolly vapor]

ye

[hidden tartan]

But since we said the length of the hypotenuse was 1, that means sin(x) equals the length of the opposite side

OH! Something you might find useful that's related to all that

Imagine taking any triangle, then making a new one by multiplying all the side lengths by some value

Both the original triangle and the new (scaled) triangle will have the same internal angles

This means if you scale (multiply) all the side lengths of a triangle by the same value, the triangle will still look the same aside from being a different size

Does that kinda make sense?

[jolly vapor]

kinda

Idk

[hidden tartan]

Hmm... well another way of saying it is that the size of the triangle doesn't actually matter, only the ratio between the side lengths matters

[jolly vapor]

yeah

[hidden tartan]

Like.... a/b = 2a/2b

[jolly vapor]

i know

whay would be the steps to solving that tho

[hidden tartan]

Okay cool

The problem you posted?

Honestly I don't understand what it's asking for, I'm sorry

[jolly vapor]

its askign whats the length of abc

[hidden tartan]

Yeah, but do we have the size of the circle? The relative positions of A and B? I feel like there's some additional context I don't know about

[jolly vapor]

apparently thr answer to it is 80 degrees

https://cdn.discordapp.com/attachments/326138757474680852/1230275615870161038/IMG_0576.jpg?ex=6632ba74&is=66204574&hm=dd3f88e8afc1f6a6e2c568a3e329f1114407ffe3d84f581bfd58ef6877099932&

could you infer what the question is asking if the answer is 80 degrees

[hidden tartan]

measured angle of ABC maybe?

[jolly vapor]

yeah

that

[hidden tartan]

I guess could be saying, given that A and C 80 degrees apart on the circle, find the angle between lines AB and CB

Is that enough info? Hold on...

[jolly vapor]

would that match up to 80 degrees

its not a test its a practice to help you so if i dont understand im gonna ask ai

[jolly vapor]

ai is usually inaccurate does 160 make more sense than 80

it is asking for degrees btw

[hidden tartan]

So I started by putting a point D in the middle

The angle between AD and AC is 80 degrees as per definition, which I set to x

[stoic quartz]

Can someone please help me

[hidden tartan]

I can help in a bit, I'm helping w atm though

[stoic quartz]

Thank you so much

[jolly vapor]

are u in 8th too

[stoic quartz]

No

9th

And its due in 9 hours

[jolly vapor]

oh

[stoic quartz]

1

Hour

[jolly vapor]

ur cooked buddy

but lock in

[stoic quartz]

I know blud💀

Im literally in class

[hidden tartan]

So, since the lengths of AD, CD, and BD are all going from the edge to the center they all have the same lengths

[jolly vapor]

is that not the answer

[jolly vapor]

so 80

[hidden tartan]

Working on it

[keen depot]

help

[jolly vapor]

it would be = to my schlong

my shlong is about 7 inches

so multiply that by 500 kilometers

you should get i believe

what is it

[hidden tartan]

Don't be crass

[jolly vapor]

3500 or so

[keen depot]

i need help not this

[jolly vapor]

Aorry

Sorry

[stoic quartz]

@keen depot

[keen depot]

that isnt close to any of my answers

i got 2 answers

1. 650
2. 350

[stoic quartz]

That what it said

[hidden tartan]

Okay the answer I got was actually 40

[stoic quartz]

It was 350

Mb

[hidden tartan]

Do you think you can try deriving that from the diagram I posted?

[hidden tartan]

So what have you tried/where are you stuck?

[stoic quartz]

all

My teacher havent done a lecture about it so i dont really much

[hidden tartan]

Hmm well do you know much about using sin/cos/tan with right angle triangles?

[stoic quartz]

Kind of

[keen depot]

did u use chatgpt

[stoic quartz]

Yeah

I know like

How to get those

[hidden tartan]

Lets start with a), do you think you could draw another line on that triangle so that it is made of two right-angle triangles?

[stoic quartz]

Right side?

Is this math topic hard?

[hidden tartan]

Eh, I don't really have the perspective to say one way or the other

[stoic quartz]

For you?

[hidden tartan]

Yeah, I'm pretty deep in the maths

[stoic quartz]

idk im scared i might be cooked

[upper karma]

Can we determine the formulas of a straight line knowing one point of each of them and the angle they create?

[hidden tartan]

Well idk about being cooked, but I think you could probably work though this problem set at least

[stoic quartz]

Do i try to find the line or the angles ?

I keep seeing my classmates find the angles first

[hidden tartan]

Now, imagine putting another point D on that triangle so that the line AD is at a right angle with the line DC

[stoic quartz]

Yes

[limber sigil]

isn't the formula for side a in law of sines a=b(sinA/SinC)

[hidden tartan]

It depends! In 1a) I would probably find two side lengths, in 1b) I'd get an angle then use that to get two sides, in 2a I'd do length then angle, and 2b I'm not quite sure

[stoic quartz]

Do i try finding A

[hidden tartan]

Oh that's also true

[hidden tartan]

Okay, but with point D on the triangle and a line drawn from D to A, the triangle is now made of two right-angle triangles, right?

[stoic quartz]

Yes

[hidden tartan]

So using what you know, do you think you could get the length of AD?

[stoic quartz]

Yes?

is it the half of ac?

[hidden tartan]

Hmm, no not quite

But that is the right direction

You would want to use sin for this, sin(45) will give you the length of AD divided by the length of AC

[stoic quartz]

OHH YEAH

Theta right

[hidden tartan]

Sure, that's a common symbol for the angle

[jolly vapor]

please give me answer and explain this simple question so i can go on and attempt the harder ones

[hidden tartan]

Do you think you could try using this diagram to solve it?

[upper karma]

Have to calculate the area of the trapezium with given data, does the traingle created with defined lines have a right angle?

[stoic quartz]

AD

OH MY GOD

MY CLASSMATE JUST GAVE ME THE ANDWER

Is this right?

[hidden tartan]

Common mistake, since the output of sin(45) is AD/AC, you'd want to multiply it by AC instead of divide

[stoic quartz]

Ohh

9.761?

[hidden tartan]

I got 16.97, did you put it in the calculator right?

[hidden tartan]

Yeye

Now, notice that ABD is also a right angle triangle

[stoic quartz]

Yes

[hidden tartan]

And sin(50) will be AD/AB

[stoic quartz]

Yes

[hidden tartan]

I'll note at this point, your friend's work seems to be right but I'm not familiar with how they did it

[stoic quartz]

What the hell

[hidden tartan]

I need to get going, do you think you can keep trying from here?

[stoic quartz]

Isk how rhey did it either

Its okay i dont want to disturb you, thank you so much for making it more clearer!

[upper karma]

Have to calculate the area of the trapezium with given data, does the traingle created with defined lines have a right angle?

[upbeat plaza]

40

because the AC is 80 and the m ABC is 80/2 = 40º

[sudden shell]

Can someone help me with question number 3

[jolly vapor]

can u teach me pls

im in 8th grade taking geometry

i was on a 3 month hiatus for family reasons

I have good reasons to not understand what sohcahtoa is😭

but i know how to find circumfrence with radius area with radius and i know how to do vice versa

[sand osprey]

Oh thats fairly simple

lemme see the formula

yea

its 1/2(120-50)

Thats the formula for secant secant angle

theorem

so the answer would be 35 I think

[surreal escarp]

this might be a silly question but does "one-half of a meter" mean 1.5 or literally one 0.5

[cunning lion]

one half means 1/2

[surreal escarp]

see but then like using 0.5 doesnt seem to make sense in this question im doin

hold on rq

i got 2.375 as my answer (when i originally used 1.5)

[cunning lion]

the original cube had a side length of 1 meter, and they cut out a smaller cube of side length 0.5 meters

[surreal escarp]

ohh i see

would the volume just be 0.5 then?

[cunning lion]

no. first calculate the volume of each cube (1 meter and 0.5 meters)

[surreal escarp]

the volume of the og cube is 1 and the cube taken out of 0.5 ? cuz isnt the edges n height all the same in a cube

?

[cunning lion]

what is the volume of a cube?

what formula gives the volume of a cube in terms of the side length?

!nosols

[lime crown]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[surreal escarp]

its width length n height right or just the number times itself thrice cuz its a cube

[cunning lion]

yes

[surreal escarp]

oh wait hold on i think i know where i went wrong

0.5 times itself is 0.25 n then multipled by 0.5 again is 0.125 right
i think i accidentally put 1 instead of 0.25

then just subtract that from the volume of 1

[empty yew]

what is scaling of graph?

Is my answers correct?

[rich scroll]

How do i prove figure 1? Ik the possible values for y but idk how to two column proof it

[empty yew]

@dark sparrow Do you have energy now?

[dark sparrow]

if you have to ask then no i don't

don't rely on me to save your ass every time

[empty yew]

if yes could you please help me?

sorry for always asking help to you

[empty yew]

no one coming so only i am asking you

could you please help me?

[dark sparrow]

no

[empty yew]

ok thank you for replying 🙂

after you regained your energy help me🙂 (if you wish to do)

[trail tendon]

clearly doesnt 💀

[empty yew]

what?

[trail tendon]

no

[empty yew]

i cant understand

[trail tendon]

its alr xD

[empty yew]

which one wrong

[trail tendon]

i'm pretty sure your process is wrong

i don't know if im qualified to explain it because uhh

i don't know the proof behind it, i just know the answer 💀

at least im pretty sure

like when you are scaling a function

[empty yew]

how do you find the answer?

what is scaling?

[trail tendon]

i can...show you

one sec

[empty yew]

ok

[trail tendon]

this is the graph of y = x^2

this is the graph of y = 2x^2

you scaled y = x^2 by 2 and get y = 2x^2

i'm horrible at explaining this 💀

my mind is not in the right place rn lol

[empty yew]

when scaling don't we need to change both the values of x and y?

[trail tendon]

nno

uhm

basically if you're scaling something you either multiply or divide by the term that contains x

which is a horrible definition

[trail tendon]

yeah

do u understand that or nah

[empty yew]

see the second paragraph

why they're changing both the x and y?

[trail tendon]

oh

i mean if you do the same thing to y and x i guess you're technically scaling...just zooming in or out

thats not the scaling im used to 💀

uhh

[empty yew]

🫨

what to do now?

[trail tendon]

do they want you to scale it while preserving the shape or do they just want you to scale it

[empty yew]

@trail tendon

[trail tendon]

uhh

i feel like they want you to scale it

but i'm honestly not sure

like you see the first part

you can scale x, y, or both more or less

i am not being clear 💀

[empty yew]

scaling doesn't preserve the shape but we can preserve by changing both x and y. am i right?

[trail tendon]

that is correct

i'm not sure if it wants you to preserve the shape or not though

[empty yew]

in this question they don't want to preserve the shape

[trail tendon]

yeah

i'd assume they want to change the x

[empty yew]

oh tq

[trail tendon]

it's weird how they say replacing x with ax

1 sec

ohh yeah im dumb

ok

[empty yew]

it's general i think

[trail tendon]

yeah

you replace x with ax for horizontal scaling

and y with ay for vertical scaling

💀

[trail tendon]

31 wants horizontal scaling

i am so dumb ☠️

[empty yew]

for studying calculus is it good to use
Calculus: A Complete Course?

[trail tendon]

i've never heard of it, but its probably not bad

[empty yew]

yep

[trail tendon]

for me textbooks are usually kinda confusing but yeah

[trail tendon]

so what do you do when you want to horizontally scale a function/equation?

[empty yew]

replacing x

[trail tendon]

yes

and not y

so then what do you get

[empty yew]

mmm

one min

[trail tendon]

by the way you don't have to do anything fancy like solve for x

you can keep it in terms of y

[empty yew]

y^2 = 3x - 1

[trail tendon]

i mean thats correct, but it might be better just to have y = sqrt(3x-1)

but it works yea

[empty yew]

i am just slow typing numbers

[trail tendon]

ah xD

i think you know how to do all of them you just scaled both x and y instead of individually x or y, depending on the question

[empty yew]

yessssss

thank you

[trail tendon]

👍

[rugged shuttle]

For vertical shift/compression multiply the entire input by a certain constant

[trail tendon]

sorry for making things more confusing for 99% of the time ☠️

[rugged shuttle]

For horizontal shift/compression multiply the x term (just the x term by a constant)

[trail tendon]

yeah

[rugged shuttle]

Btw hi @empty yew

[empty yew]

hi @rugged shuttle

[rugged shuttle]

I like ur about me

[trail tendon]

thanks ☠️

i like ur pfp

i'm not just complimenting back like i actually think its cool 💀

[empty yew]

entire input with constant?

[trail tendon]

yeah, which is the same thing as replacing y with ay

[rugged shuttle]

Lol thx

[empty yew]

ohh

[trail tendon]

y = f(x) which is the function

[rugged shuttle]

Think like vertical stretch of 2 y=x+2 would become y=2(x+2)

Wait

[trail tendon]

waht

[empty yew]

shift?

[trail tendon]

you mean a vertical stretch of 2 on y = x + 2

oh

[rugged shuttle]

Wait no yeah

[trail tendon]

i thought you meant 2y = x + 2 and i was like bruh what 💀

[empty yew]

what?

[trail tendon]

he meant stretch, not shift

[rugged shuttle]

Vertical stretch of 2 of y=x+2

$y=2(x+2)$

[empty yew]

what's a strech though?

[somber coyote]

Cyberr

[rugged shuttle]

Basically imagine holding the top of the graph and pulling up

That’s a vertical stretch

Anyways I gtg

Lol

[empty yew]

gtg?

[empty yew]

32. $y^2 = 16x-16$

[somber coyote]

KingDanger

[empty yew]

33. $3y^2 = 2x-3$

[somber coyote]

KingDanger

[empty yew]

34. $16y^2=x-4$

[somber coyote]

KingDanger

KingDanger

[empty yew]

@rugged shuttle and @trail tendon do they're correct?

[queen nest]

HATRED

HATRED

HATRED

[dark sparrow]

maybe bc you put an x instead of θ?

[queen nest]

yes exactly

i had the question right but i got it wrong because of that small nuacne

the original question starts with x you substitute in theta so everything from that point should be in terms of theta

thats what got me like 17 different times now

on a 3 question assignment

i am so very tired and also exhausted

[inner vapor]

no worries, take a break, and you will be fine

[upper karma]

Can anyone help me to solve it

I dont even know what to begin with

I can apply sine of sum

[versed path]

for half angle formula do you decide the quadrant based on x/2's value or x's value in lets say cos(x/2)

[upper karma]

But it will be squared and it's fonna be hard

[upper karma]

Are you asking me?

[versed path]

anyone really

i have a test tmr 😭

[upper karma]

Why do you need to decide a quadrant?

[versed path]

the half angle formula has both a positive and negative root

lets say its cosine

[ocean roost]

what grade level is this bro 😭

[upper karma]

Maybe highschool 11th grade

[ocean roost]

oh dang

im 9th bruh doing indices got me all scared now

[upper karma]

Indices?

What are these

[ocean roost]

index laws

exponent? powers? index law

[upper karma]

Gotcha

Powers

[ocean roost]

loll

[subtle juniper]

why is graphing so painful

[versed path]

what you graphing

[subtle juniper]

trig functions

[versed path]

dw i feel you

[ionic crypt]

anyone wanna help me? no payment pls

messege me

[upper karma]

Post the question here

[ionic crypt]

ok

[dark sparrow]

this transparent background makes it impossible to read on dark theme.

@ionic crypt so, do you have any progress on this so far?

if you don't, then say "no progress, i don't know where to begin"

[ionic crypt]

is it 7?

[dark sparrow]

how did you get 7?

[ionic crypt]

.

Uh-

guessed.

[dark sparrow]

bad!

this is not the place to guess.

[ionic crypt]

.

[dark sparrow]

ok, let's start from the beginning.

[ionic crypt]

i'm so dumb

[dark sparrow]

do you know what parallel lines are?

[ionic crypt]

look the same

[dark sparrow]

all lines "look the same", so no.

https://youtu.be/oE98p-0Vw0U

[ionic crypt]

.

[dark sparrow]

give this a watch, so that you are more up to speed

cause you either forgot or don't know

[ionic crypt]

I know what is it.

not helping.

[dark sparrow]

you know what is what?

[trail tendon]

you can't define it tho

[ionic crypt]

Your calling me dumb

[trail tendon]

i am not

lol

[upper karma]

Geometry beef

[trail tendon]

😂

nah dude if you know what something is but you cant define it you should prob learn how to define it lol

[dark sparrow]

he isn't and neither am i

[rich scroll]

It would be 7 if it stated that a is parallel to b

[flint ingot]

find area of triangle BEC i dont know how to do this pls help

[vernal pilot]

What have you tried?

[inner vapor]

I dont get it

interior angles are 180

both abc and dcb are 30

but

[dark sparrow]

what does the text in red say?

i said red.

[merry parrot]

[flint ingot]

wait no

yeah sorry

[dark sparrow]

is ABCD a rectangle?

[flint ingot]

my friend showed me this question and his working was extending the line BE and drawing a perpendicular line from C to the extended line. the perpendicular line is called CF. 0.5xCFxBF=area of triangle BEC. let AB = L. let CF = J. area of triangle BEC is also 1.5L. CEF(angle) = 60. CFE(angle) = 90. ECF(angle) = 30. in imaginary triangle ECF the hypotenuse is double the length of EF. he then used pythagarus theorem and eventually found something like a quadratic equation but u had to find the square root of a not perfect square with irrational number so that doesnt work.

[flint ingot]

yes it is

[dark sparrow]

so that doesn't work.
why not?

is there an explicit ban on using irrational roots?

[flint ingot]

no but you can try it

[dark sparrow]

i got an answer with those

[flint ingot]

oh nice

[dark sparrow]

but now you say that big brother forbids it

or maybe i misunderstood you

[flint ingot]

who is big brother

[dark sparrow]

have you read 1984?

[flint ingot]

?

no

[vernal pilot]

Mediocre book

[dark sparrow]

1984 by George Orwell

oh well, forget it

[flint ingot]

ok

[dark sparrow]

@flint ingot why does the presence of irrational roots make your friend's method invalid?

[flint ingot]

we cant find the answer using irrational roots too hard for us

[dark sparrow]

too hard? or forbidden by authority?

[vernal pilot]

Where did you get the question? That might give you some insight on how you might go about solving this if you aren’t confident in the current method.

[dark sparrow]

the answer i got contains some square roots.

so if you have no access to those, then the problem is impossible.

[flint ingot]

too hard

[dark sparrow]

then this problem is also too hard for you.

[flint ingot]

ill ask my friend

[flint ingot]

ok

[dark sparrow]

your friend and i both reached a quadratic equation.

maybe different ones, since i don't know what he did exactly, but still.

[flint ingot]

did it include defining a variable like n=l^ or something

[dark sparrow]

i don't know what you mean with l^

[flint ingot]

l squared

[dark sparrow]

l^2?

[flint ingot]

but l might be another variable

yes sorry

[dark sparrow]

i didn't make any kind of definition like that no

[sonic bloom]

Can someone prove by HL?

[flint ingot]

oh ok

[dark sparrow]

what does "HL" mean?

[sonic bloom]

Hypotenuse and leg method

[dark sparrow]

,rccw

[dark sparrow]

can you state the method in full?

[sonic bloom]

Forget it i solved it, it doesnt HAVE to be hl

[sonic bloom]

Wdym

[vocal bison]

help plz I’m confused

[flint ingot]

do u understand that the length scale factor is 2?

the surface area scale factor will therefore be 2^2

so scale factor is 4

so the ratio is 1:4

[vocal bison]

what does that even mean 😭

[dark sparrow]

when you have two similar shapes and their linear sizes differ by a factor of k, then their surface areas will differ by a factor of k^2

e.g. enlarging a shape by a factor of 10 makes its SA grow not by a factor of 10 but by 100

does that make sense to you

[vocal bison]

oh

that makes sense

[dark sparrow]

which of the six subproblems (a through f) do you want to examine?

[vocal bison]

I think f

wait no

e

[dark sparrow]

ok

have you done b already?

[vocal bison]

I got

[dark sparrow]

it will help us to use b to get e

[vocal bison]

8:1

[dark sparrow]

ok cool

V(big cone) : V(small cone) = 8:1

that is correct

now

the frustum is just the big cone minus the small cone.

and its volume is, accordingly, the difference of those volumes.

[vocal bison]

Wait is it 7:8

[dark sparrow]

which letter?

[vocal bison]

For e

[dark sparrow]

yes, e is 7:8

[vocal bison]

Ohh ok

Cool

I think I got it

Now thanks

[lapis junco]

@north kindle for that one area under the curve problem that I think you had, did the solution involve using inequality signs?

It was this

I know that in a college algebra class at my school, you can use inequality signs to either get the area or maybe ‘talk about’ the area under the curve.

[north kindle]

I just missed squaring something, it was the wrong curve

[lapis junco]

Oh that’s unfortunate

[bitter tundra]

Ignore the show algebra part. Why does sin/cos not work when you use it? Correct answer is 56/5 or 11.2

[trail tendon]

why would it work?

[lime dune]

you don't need trig for this LOL

[rugged shuttle]

Can u show work

[trail tendon]

also wait is this 2d or 3d 💀

[bitter tundra]

I know but I was struggling on this problem last night and trig was the first thing that popped up into my head

[hoary totem]

its 2D

this requires absolutely nothing but 1 multiplication and 1 division

with nothing other than the numbers given

[trail tendon]

💀

[hoary totem]

sine should work

[trail tendon]

why?

sine of what angle?

[hoary totem]

sin(BCD) = 8/10 = 0.8
BCD = EBC
sin(EBC) = 0.8 = h/14
h = 0.8 * 14 = 11.2

works

but this is.. u dont need to solve this question this way

[trail tendon]

oh, fair enough

[hoary totem]

there is a far easier option

[bitter tundra]

So in theory, since the angle unerneath number 8 is 90 degrees, the angle on the other side of the dashed line must be 90 degrees also. This makes the triangle formed at the end of the parallelogram a right triangle. Since 8 is one of the values, you can move the 10 on the right side to the hypotenuse of this newly formed triangle, and automatically deduce that the triangle is a 6-8-10 right triangle. Now on top of the number 8, the angle must be a right angle because alternate interior angles are congruent. (The previous alternate interior angle here is 90) The topmost angle of the 6-8-10 right triangle must be complementary to the angle touching it on the other triangle. (180-90) This would mean that since they are complementary it would mean that the topmost angle of the 6-8-10 triangle is congruent to the farthest angle in the triangle on top. I used sin inverse (6/10) and found the topmost angle measure of the 6-8-10 triangle. Then I knew that the rightmost angle on the dotted triangle is congruent to the degree I just found. I used cos(ans)*14 = h and found for H but somehow it was wrong. @rugged shuttle

[hoary totem]

theres ur problem

u should have used sin(ans)

not cos

[hoary totem]

see this

[bitter tundra]

how would it have changed what I got

[hoary totem]

cuz sine is not cosine?

[bitter tundra]

nvm I don't think i asked the question right but whatever

thanks though I understand

[hoary totem]

np

but

ur not interested in the far more easy way to solve this?

[bitter tundra]

you draw a line to make another rectangle?

[hoary totem]

no

[bitter tundra]

im interested

[hoary totem]

ur given a side (14) and a height (8)
thus the area of the parallelogram is 14 * 8 = 112

now ur given another side, which is 10
the side * height for this should also be 112
so the height must be 112/10 = 11.2

[bitter tundra]

ok that's how my teacher explained it

but she drew a rectangle to help us visualize it

thanks

[hoary totem]

np

[wary swift]

can someone help me

ik this is easy for everybody else

but like im in 7th grade

how tf do i find RUS

angle UST is 15 right?

[trail tendon]

I wouldn't say that

[trail tendon]

how do you get that?

[wary swift]

idk

was there like

something abt a chord

and like the angle corrisponding was like half the angle of the chord?

[trail tendon]

um

i know how to do it but this has always been my weak spot in geometry 💀

if you have an angle created by a chord and the diameter

then the arc length created is 2x that angle

basically

[wary swift]

oh

what is the theorem called again

[trail tendon]

no idea, havent done it in a while 💀

[wary swift]

wait

you mean 2x the angle between the chord and diameter

so

i have to find chord RS

and then half it?

[trail tendon]

nah

1 sec

[wary swift]

kk

[trail tendon]

its like this

um

thats an x

its very messy

x is the angle, 2x is the arc length

ri

[wary swift]

oh

so

is UST 15?

[trail tendon]

yes its correct

i was simply asking how you got there because

[wary swift]

and does that only apply when it's with the diameter

[trail tendon]

you can do the same thing for the other angle

[wary swift]

yeah

[trail tendon]

yeah

[wary swift]

so

RSU is 25

[trail tendon]

yeah

[wary swift]

OH

I THINK I GET IT

30+180

105

105+25

=130

180-130=50?

[trail tendon]

...what

[wary swift]

dman

dman

damn

im lost asf

[trail tendon]

XD

nah bro

if you have a diameter and a chord

[wary swift]

yep

[trail tendon]

the angle between them is half the arc length that they create

[wary swift]

yeah

[trail tendon]

yeah

XD

[wary swift]

RSU is 25 right?

then

[trail tendon]

yeah

[wary swift]

do i find angle R

and just find RUS that way?

[trail tendon]

ohh i see you were doing that part

[wary swift]

wait

but i can't divide 210 for R

i have to do smth else

[trail tendon]

why did you add 180 and 30

[wary swift]

im stupid

bc

for some reason i thought that there was a diameter

and i could just divide by 2

i dont even know

[trail tendon]

lol

uh

honestly bro i need to review geometry

i don't know how to solve for RUS

wait

maybe i do actually

[wary swift]

what

do i do

[trail tendon]

you can use the same thing to solve for angle U yeah

[wary swift]

rlly?

[trail tendon]

you know that the line US is the diameter

so the arclength is 180

[wary swift]

ooh

i have to find chord RS

OMGH

[trail tendon]

and 50 degrees of that is yeah

[wary swift]

that makes so much more sense

[trail tendon]

👍

[wary swift]

50+180=230

115

[trail tendon]

you'd actually subtract ri

[wary swift]

oh yea

im stupid

260-230

[trail tendon]

bro wut

[wary swift]

30/2=15

wait

i give up

[trail tendon]

one sec

the blue is 50 right?

[wary swift]

yeah

i have to find half of chord RS right

180-50

[daring pollen]

Is it possible to learn all this in one night?

[wary swift]

130/2

75

wait

65

why tf am i so dumb

[trail tendon]

i would say its more of a familiarize thing, it might be possible to LEARN in one night theoretically but i would suggest reviewing it for a longer period of time

[trail tendon]

nah u got it tho XD

[daring pollen]

cause i may or may not have a test on that tmrw

[wary swift]

i got it?

yay

[trail tendon]

um

[wary swift]

25% done with my hw!!

[daring pollen]

and im lost

[wary swift]

if u dont know any of this idk if you can learn it all

cuz like

[daring pollen]

well i know a bit of it all

[wary swift]

took my 2 years to learn that

maybe im just slow

[trail tendon]

i would say work on what you don't know and just familiarize yourself with it

[daring pollen]

mmm

[trail tendon]

i would say it heavily depends

[wary swift]

nice username btw

[daring pollen]

thanks

[trail tendon]

to short term pass a test? maybe you can learn enough in one night

[daring pollen]

my account got hacked

[wary swift]

oh lol

[trail tendon]

to actually be able to use it in applications

[daring pollen]

and i got banned from mathematics

[trail tendon]

u might need more uhh

familiarity

[daring pollen]

hmm

[trail tendon]

☠️

[daring pollen]

because i got hacked by a "nsfw bot"

[wary swift]

banned?

😭

[trail tendon]

i would say if you gotta cram then cram 🤣

[wary swift]

i have an 89 in math rn 😭

i better ace this next test

[daring pollen]

i had a 86 last snapshot and got 2 55s in a row on circles

.......

now i have a 78

[trail tendon]

why are you crying thats pretty good 🤣

[wary swift]

its horrible

[trail tendon]

i mean its not the best obviously but its not bad lol

[daring pollen]

i might be bad at circles

[trail tendon]

bros a perfectionist

[wary swift]

and i have a hard test next

im prob gonna get 80 on it

[trail tendon]

study up and don't overthink it 👍

[wary swift]

then im screwed

[wary swift]

it's on counting

i suck so much at counting

[trail tendon]

what kind of counting

[wary swift]

idk what types there are

[daring pollen]

im kinda screwed. I need to get a 90 this semester and im at a good 78.....

[wary swift]

like combinitorial stuff

[trail tendon]

wait why do you need a 90

[wary swift]

i NEED AN A

[trail tendon]

oh

[wary swift]

or A-

[daring pollen]

to get into honors algebra next year

[wary swift]

im in honors alg 2

[daring pollen]

nice

[wary swift]

its sucks ass

i hate parabolas

[daring pollen]

my teacher thinks i should do it

[wary swift]

what grade

8?

7?

[daring pollen]

I think he's gonna let me take the class either way

[trail tendon]

just make sure you know whats going on

[daring pollen]

yeah

[wary swift]

real

[daring pollen]

im freshman

[wary swift]

it sucks when i am lost in class

[trail tendon]

honestly i find that i usually overthink my material and then sometimes i don't and it becomes much easier lol

[wary swift]

sir

[daring pollen]

i took algebra in 8th grade

[wary swift]

is X 53

[daring pollen]

Wait fuji you're in geometry or algebra?

[wary swift]

both

[daring pollen]

both

[wary swift]

and i take an MCP class

on top of it all

im only in 7th grade

[daring pollen]

what is mcp

[wary swift]

comp class

but i still suck at geo

[daring pollen]

so ur cracked at math

[wary swift]

not really

a kid at my school does calc

IN 7TH GRADE