help-13

[median wolf]

nope

[steep steeple]

csc and sec

[median wolf]

csc is 1/sin(x) and sec is 1/cos(x)

they are the inverses of the sin and cos ratios

[steep steeple]

yeah

[median wolf]

there is also cot, or 1/tan(x)

[steep steeple]

yep

[median wolf]

it makes sense that sin is negative in q4, however, because y is negative here

so cos is positive, which means both sin and tan are

[steep steeple]

yeah

[median wolf]

Do you see what tan will be in q4?

tan(x) = y divided by x

[steep steeple]

wait

tan is in q3

[median wolf]

tan is positive in q3

[steep steeple]

yeah

[median wolf]

the CAST system tells you what ratios will be positive in which quadrants

[steep steeple]

ah

[median wolf]

all ratios exist in all quadrants

[steep steeple]

so it'd be -y/x?

[median wolf]

but the CAST system tells you where their output will be positive

BINGO

does that make sense?

[steep steeple]

i believe it does

[median wolf]

yey

[steep steeple]

gimme a sec

[median wolf]

okay

[steep steeple]

imma draw something out just to make sure i fully got it

[median wolf]

Sounds good

[steep steeple]

wait

oh wait nvm thats what A in q1 is

[median wolf]

yea

[steep steeple]

this is in correlation to tan

[median wolf]

So tan is positive in q1

[steep steeple]

and q3

[median wolf]

yes

oh ok i think i see what you've written now

and in q2,4 it is negative

[steep steeple]

so for something like cosine

why would it be positive in q4?

[median wolf]

Because x is positive

and cosine is x/h

[steep steeple]

is h based off of x in if its postitvie or negative?

[median wolf]

h is the length of the hypotenuse

[steep steeple]

oh so h is always going to be positive then

[median wolf]

yea

[steep steeple]

(you said that a while back and now only its making sense 💀)

[median wolf]

It is based off of calculations used to find the trigonometric ratios of all angles within what's called the unit circle

that's fine it is a bit complex

[steep steeple]

yeah

takin pre calc

[median wolf]

Here is a good tool to visually play around with it by the way

[steep steeple]

shits tough

[median wolf]

https://www.desmos.com/calculator/qmzx2skkzy

If you want you can play around with it and see the values

[steep steeple]

yeah

[median wolf]

Anyways, now that we know the fundamentals behind where certain ratios will be positive, we can analyze the original question you had

[steep steeple]

so in relation to the original question...

yeah

[median wolf]

I am gonna try to use the bot one sec

bear with me xD

[steep steeple]

you good

[median wolf]

$[\frac{-\pi}{2},\frac{\pi}{2}]$

rats

[wraith dagger]

PoisedPerson

[median wolf]

Perfect

[steep steeple]

noice

[median wolf]

so the original question tells you to find the value of tan(x) = 1 between those two points

I am assuming you already know how to read angles in radian form?

[steep steeple]

give or take

[median wolf]

Do you know what these angles are in degrees?

starting with pi/2?

[steep steeple]

45, -45

wai

90,-90

[median wolf]

yea

[steep steeple]

45 is pi/4

[median wolf]

so pi/2 is 90, -pi/2 is -90

-90 is the same as saying 270

[steep steeple]

yep

[median wolf]

because 360-90 is 270

So now we know between where the question is asking us to look

And if we want to find a value in those two quadrants, which quadrants are we looking at?

[steep steeple]

q4 and q1?

[median wolf]

yey

Which one will have a positive value of tan?

[steep steeple]

wait

wouldnt it be q3 and q1?

because q4 is cosine

[median wolf]

That is where cosine is positive

tangent ratios still exist there

[steep steeple]

oh

[median wolf]

but what will their output be, since x is positive but y is negative?

[steep steeple]

negtaie

[median wolf]

So if we want a positive value, which quadrant will we look at?

for tan

[steep steeple]

q1

[median wolf]

yey

now, have you been taught the "special angles" by your teacher yet?

[steep steeple]

in regards to trig function?

[median wolf]

yes

[steep steeple]

yes

[median wolf]

Nice

So for instance, what is the value of sin at 60 degrees?

[steep steeple]

however, 90 degrees for tan is undefined

[median wolf]

yea

we're looking for a value that makes tan(x) = 1 between -90 and +90

[steep steeple]

sqrt3/2

[median wolf]

noice

what about sin 30?

[steep steeple]

sqrt2/3

[median wolf]

close

[steep steeple]

wait

1/2 (im stupid_

[median wolf]

that's okay

you're not stupid

it's just a lot of memorization

[steep steeple]

yeah i got it mixed with cosine for some reason lol

[median wolf]

I would even help my teacher correct mistakes

she was a mechanical engineer, we're only human not computers

[steep steeple]

oh damn

[median wolf]

I somehow scored 100 on the identity test, but that's because I had a large coffee that day

370mg of caffeine

anyways

[steep steeple]

nice

[median wolf]

so, what about sin 45?

[steep steeple]

sqrt 2/2

[median wolf]

yey

now, where out of those three special angles would tan(x) = 1?

hint: draw out the special angle triangles for q1

[steep steeple]

45 degrees

so sqrt 2/2?

[median wolf]

ding ding ding

[steep steeple]

woop

[median wolf]

do you know what it is?

[steep steeple]

wdym

[median wolf]

do you understand why tan(45) = 1?

[steep steeple]

yes

[median wolf]

oki

then you're golden

so what answer will be the right one then?

[steep steeple]

i kinda dont, but B?

oh wait

[median wolf]

That's okay

[steep steeple]

e

[median wolf]

do you know how to convert 45 to radians?

[steep steeple]

45 * pi/180

[median wolf]

i should have specified

[steep steeple]

1/4

[median wolf]

yes that is in radians

yea

[steep steeple]

pi/4

[median wolf]

perfect

so which answer contains pi/4?

[steep steeple]

a & b

[median wolf]

yes

but what does b have that's different from a?

[steep steeple]

it also includes an option that would be in q1,

x- -x/4

i think

[median wolf]

are you sure that that's in q1?

[steep steeple]

oh wait

damnit

[median wolf]

that's okay

take your time

i am not in a rush

neither should you

slow and steady wins the race

[steep steeple]

ight

b includes both pi/a and -pi/a

[median wolf]

ye

which quadrant is -pi/a in?

[steep steeple]

4

[median wolf]

is tangent positive or negative in quadrant 4?

[steep steeple]

negative

so that menas it' be a

[median wolf]

🙂

[steep steeple]

woo

[median wolf]

yayyyyyy

CAST method is imperative for questions related to trigonometry

it will be ur fren

[steep steeple]

alr

[median wolf]

any other things you were struggling with?

[steep steeple]

nothin as of yet

[median wolf]

oki doki

Imma see if anyone else needs help, if the channel asks you if your answer is complete, just click the checkmark so it opens up for other users

[steep steeple]

alr

think i can do this also

.clsoe

dang

[median wolf]

yea

lol

[steep steeple]

.close

[cedar kiln]

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[steep steeple]

woop

[cedar kiln]

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[violet delta]

how do I find the smallest x with a given Constant to prove a function is Big O of another function

[cedar kiln]

Please don't occupy multiple help channels.

[dire geode]

Close one of your channels

[violet delta]

.close

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[glacial fable]

Question 10:
If 0 ≤ 𝜃 ≤ 180° and cos 𝜃 = −12/13 then tan 𝜃 = ?

My Answer: c) −5/12
Explanation: Since cos 𝜃 = −12/13 and 𝜃 is in the second quadrant (where cosine is negative), we can use the Pythagorean identity sin² 𝜃 + cos² 𝜃 = 1 to find sin 𝜃 = √(1 - (−12/13)²) = 5/13. Then, tan 𝜃 = sin 𝜃 / cos 𝜃 = (5/13) / (−12/13) = −5/12.

[glacial fable]

Did I do this correctly can someone verify for me

[gusty forum]

seems good

[raw gulch]

I'd even write it in a such way: If 90° < 𝜃 < 180° and cos 𝜃 = −12/13 then tan 𝜃 = ?

[cedar kiln]

@glacial fable Has your question been resolved?

[cedar kiln]

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[jagged stag]

Need some help

[jagged stag]

I have a label of (6,2) for figure 2, need to confirm that's right. I also have answers for the angles which I think are correct. However, I'm confused on the paragraph where it's asking when b=1, a = even, and when b = a/2 as well as the rest of what its asking

[frigid osprey]

hey lemme take a look at this

to break this down, from my understanding, a and b are directly correlated to a fact.
a - points in the shap
b - where the points are joined

Thus if b=1 then all points join at every first point
so b=a/2 must mean that the points join at every nth point according to the about of vertices there are divided by 2.

[jagged stag]

a = the vertices in the shape

like figure 1, a normal star has 5 extruding pieces/vertices so a=5, and in figure 2 a = 6 since it has 6 of these

however, I'm struggling to view a shape that has b equivalent to 1

since b = the amount of lines each vertice has

so it would just be a line with two endpoints when b = 1?

so I'm confused on that

[cedar kiln]

@jagged stag Has your question been resolved?

[jagged stag]

<@&286206848099549185> ^

[frigid osprey]

im here sorry

b=1 just means that the vertices are connected at every point

b= the connection points

[jagged stag]

I think it's a = where points are joined

no?

a = vertices

figure 1 = (5,2)

of (a,b)

[frigid osprey]

a= how many vertices there are

[jagged stag]

mhm

[frigid osprey]

b= where the vertices are connected

[jagged stag]

and b = the amount of lines coming off that point

same thing?

[frigid osprey]

yep

[jagged stag]

k

[frigid osprey]

an array

[jagged stag]

so basically a shape where b=1, it's basically one line

[frigid osprey]

an array

[jagged stag]

how?

cause if there's only one line coming off of one vertice

[frigid osprey]

shoot the term is slipping my mind

a ray

[jagged stag]

well it has to be a shape

and an infinite line isn't really one

so is it just a line?

but I guess that's not a shape either

[frigid osprey]

idk dawg my brain too tired for this-

[jagged stag]

np man

[jagged stag]

thanks for the insight 👍

for anyone else, here's the problem

paragraph 1 explains it, confused on para 2

also if fig 1 is (5,2) fig 2 is (6,2)

correct?

[cedar kiln]

@jagged stag Has your question been resolved?

[cedar kiln]

@jagged stag Has your question been resolved?

[cedar kiln]

@jagged stag Has your question been resolved?

[cedar kiln]

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[crystal mason]

how do i calculate $D(g \circ f)$ with and without the chain rule?

[wraith dagger]

Levens

Levens

Levens

[crystal mason]

so what’s the derivative of $e^{\frac{y}{x} \cdot xy}$?

[wraith dagger]

Levens

[crystal mason]

i can only come up with the matrix form

(partial derivatives)

<@&286206848099549185>

man

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[fallen moat]

do you mean jacobian?

if it is, you can just use this method
https://math.stackexchange.com/questions/3560463/how-to-find-total-derivative-of-function-f-from-mathbb-r2-to-mathbb-r

something like that

[cedar kiln]

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[fallen moat]

.close

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[gilded wyvern]

How to demonstrate that a+1>b if a^2 + b^2 > 1

[gilded wyvern]

I meant a^2 + b^2 < 1

[mental trail]

Probably untrue

[gilded wyvern]

Really?

[mental trail]

Yes

Pretend for a sec that we have large inequalities

Then pick a = -1/sqrt(2) and b = 1/sqrt(2)

[gilded wyvern]

Thats equal to 1 tho

[crimson sedge]

That's not allowed

[mental trail]

In fact any a < -1/2 and b > 1/2 is satisfied

[mental trail]

Then pick those

[gilded wyvern]

a^2 +b^2 = 1 in that casz

No?

[mental trail]

Still no

a²+b² < 1/2 maybe

[gilded wyvern]

Hmm

[crimson sedge]

But what if you choose a=-1/2 and b a little less that sqrt(3)/2

Then the sum of squares is less than 1 but a+1<b

[mental trail]

sqrt(3)/2 >1 so no

[mental trail]

Yes, I already said that any a and b satisfying this doesn't work

[mental trail]

So I think this is the condition that might work

[crimson sedge]

Yes so there should be some restriction about a and b being positive or negative or something like that

[gilded wyvern]

Thank you guys

[crimson sedge]

Or this indeed

[gilded wyvern]

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[stray comet]

Do the following Statements apply for every n dimensional vector space V and all subsetts M1, M2 ⊆ V?
1.⟨M1⟩ ∪ ⟨M2⟩ ⊆ ⟨M1 ∪ M2⟩.
2.⟨M1 ∪ M2⟩ ⊆ ⟨M1⟩ ∪ ⟨M2⟩

I realy dont get the diference between the two statements. For me both mean the same. Also how would i proove this?

[sleek condor]

1 is true 2 is false

take R^2

M1={(0,1)}

M2={(1,0)}

[cedar kiln]

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[harsh garden]

Hi

[harsh garden]

Can someone guide me solving this 5. Items?

It's all about simplifying radical expressions

[unkempt wedge]

for the first 2 expressions make them in the same racine

[harsh garden]

Wdym Racine?

[unkempt wedge]

square

[harsh garden]

Wait like those?

The one that I doodle?

@unkempt wedge ?

[unkempt wedge]

yes

[harsh garden]

I'll do the 31 first

May I know why do we make them same square?

[unkempt wedge]

to simply by a common factor betwwen denominator and nomiator

like in the scond one you wille simplify by (x-3)

and (x+y) in the third

[harsh garden]

Next is to remove nth power of radicand?

[harsh garden]

Okay okay

@unkempt wedge

@unkempt wedge

Pls don't gooo😟

[unkempt wedge]

sry i try to solve it

[harsh garden]

Next is to remove nth power of radicand?

[unkempt wedge]

yes

[harsh garden]

Is this good?

And next is to remove the radical from denominator?

[harsh garden]

Is this good?

@unkempt wedge

[unkempt wedge]

no 7*3 is 21

not 23

[harsh garden]

It's 22 and I put the remaining 1 in the radical

@unkempt wedge

😟

@unkempt wedge

. close

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[crimson sedge]

How did they conclude that $\mu_s = \m\tan{\alpha_0}$?

[novel topaz]

because if tan alpha is less then the friction coefficient then the block wont slip

[crimson sedge]

how am i meant to know that

[novel topaz]

its given that block starts sliding when angle is alpha

uhh u could make the FBD

[crimson sedge]

so at $\alpha_0$ we have $f_s = N\cd \mu_s$

[novel topaz]

the force along the line of motion are friction and mgsinalpha

[novel topaz]

yep

[crimson sedge]

right

[novel topaz]

is the physics server ded

[crimson sedge]

and we have[
N = mg\m\cos{\alpha_0} \
mg\m\sin{\alpha_0} = f_s
]

[novel topaz]

yeah

[crimson sedge]

yeah i kinda gave up on it i wont lie

the staff included mechanics in the tags in #1021175428326633542

so like, it is okay to post about it here now so why not

[novel topaz]

yeah its ok

but the people who would help with physics here would be rare ig

[crimson sedge]

oh yeah i see it. so like [
mg\m\sin{\alpha_0} = f_s \q \t{(before motion has begun)} \
mg\m\sin{\alpha_0} - f_k = ma_x \q \t{(after motion has begun)}
]

[crimson sedge]

and then we can divide both equations

[crimson sedge]

i have received quite a bit haha. ig its because this is highschool level

[novel topaz]

ok

[crimson sedge]

an yways thanks

.clos

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[winged pumice]

why is the answers so specific as in decimals

shouldn't we not even have decimal points here

[stiff totem]

the point is that the answers all lie in different bins in the histogram

it's asking which ones are possible to be the median, not what it actually is (as you say, it's impossible to determine that accurately from the histogram)

[winged pumice]

thats true

sheesh

gotta get glasses lmao

aight thanks

😄

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[half smelt]

quick question: can a group be defined by a binary operation that involves both, say, addition and multiplication? so a \circle b = a+2b?

[subtle raven]

well, I believe you can interpret a binary operation as like a function $f: A \times A \to A$

[wraith dagger]

_Kookie

[subtle raven]

in this case I suppose your function works like $f(a,b) = a + 2b$

[wraith dagger]

_Kookie

[subtle raven]

does that answer your question?

[static owl]

Yup. How you construct that function doesn't matter - all that matters is that it satisfies the group axioms.

And this is in general a very important thing to remember as you go into more abstract mathematics - constructions start to matter less and less, and what properties they satisfy start to matter more and more.

The binary operation inside the group can be anything as long as it satisfies the group axioms - so if you find a way to define it using both addition and multiplication, and it follows the axioms, then it is entirely valid.

[half smelt]

ok, fantastic. thank you both

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[molten schooner]

hello i need help with a math problem

[sudden magnet]

[molten schooner]

its a high school level problem. ellie runs a race. at 13:10 she has run 1/4 of the race. at 13:25 she has run 3/5 out of the race. how long does it take to run the whole race? i got 42,142 minutes. now is this correct?'

im not really sure thats whyi ask. i also have another problem im stuck with. i just want to validate this one or discuss it first

hello is there anybody here?

[fair geyser]

I get 42.86

,calc 15 / ((3/5) - (1/4))

[wraith dagger]

Result:

42.857142857143

[molten schooner]

alright thank you. i have a second problem: a bottle with a capacity of 2/3 liters is filled 1/3. whatever fills it is then poured into a bottle with a capacity of 5/6 liters. how much of that second bottle is now full? how do you solve it

[fair geyser]

i'm not like trying to help, i just hoped you would explain how you got your answer

[molten schooner]

i havent gotten an answer since im stuck

[fair geyser]

i mean the first problem

[molten schooner]

well the only reason for you wanting to know how i got my answer that i can think of is you wanting to help in some way

[fair geyser]

no, i just can't tell how you did it, and i prefer to know instead of not knowing

[molten schooner]

alright i understand

this is how i thought

i started by multiplying 1/4 with 5 and 3/5 with 4. which gives me 5/20 and 12/20. theres 15 minutes between 13:10 and 13:25. this means that it took her 15 minutes to run 7/20 of the race. then i calculated how long it would take to run 1/20 of the race. so 15/7. which i got to 2,142. 13 x 2,142 + 15. which i got to 42,846. (not 42,142 which i wrote earlier. that was wrong. i still dont know how i didnt get the same as you tho

[fair geyser]

thanks

[molten schooner]

what do you think of my thought process

[cedar kiln]

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[hoary silo]

hello i need to solve these two equations by elimination method but it seems that i cant think of the right solution for it

[hoary silo]

x+2y > 8

7x+y>10

the lesson is linear programming

[crimson sedge]

Start by multiplying the second equation by 2 to make the coefficients of y the same in both equations.

[cedar kiln]

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[hoary silo]

how about multiplying the first one by 7?

[digital mauve]

?

[hoary silo]

the first constraint

[digital mauve]

That chat is closed now

[hoary silo]

.reopen

.reopen

?

[digital mauve]

Theres already a new chat now

Too late to reopen

<@&286206848099549185>

?

There was a message here for a second

Helloo

<@&286206848099549185>

.close

[cedar kiln]

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[dull plover]

wanted to ask if f(x) is continuous at -3 and 10

[dull plover]

u cant say that ull take the limit from the both sides such as in x=1 and call it a day because -3 doesnt really have a left side

and x = 10 doesnt have a right side

[crimson sedge]

for minus 3

it starts from -3

so just check at -3 and -3+

for 10

check at 10 and 10-

[dull plover]

but thats the exact same thing

[gusty forum]

which is why you check for one sided continuity instead

[crimson sedge]

same function

[dull plover]

since ur substituting in the same function

[crimson sedge]

which is great

you already know its continuous

[dull plover]

so its continuous?

[crimson sedge]

and by logic it should be

[dull plover]

oh

[crimson sedge]

yessir

wait let me check

still

[dull plover]

huh

[crimson sedge]

one sided

you ignore the side which you dont have

because our doman in restricted

[dull plover]

u can do that and call it continuous?

[crimson sedge]

ofc

if your function begins at 0

[dull plover]

but surely u dont include the point

right?

[crimson sedge]

you cant tell whats behind 0

[crimson sedge]

depends on the function

[ include
( dont include

[dull plover]

well in the function i showed u

why cant we differentiate at -3 and 10

[gusty forum]

for all purposes of continuity you only call (-3, 10) continuous (if it is, i haven’t bothered reading)

[dull plover]

if it is continuous

[crimson sedge]

who says we cant

we have to check IF we can

thats a part of differentiability

[dull plover]

please elaborate

[crimson sedge]

if theres a kink, its Non differentiable

[dull plover]

wdym

[crimson sedge]

if the slope changes aburptly you cant differentiate

[crimson sedge]

it will come up

if you hvaent learned it yet

[dull plover]

uhuhh

[crimson sedge]

focus on limits

[dull plover]

well regarding to the picture i sent

[crimson sedge]

yeah

[dull plover]

the differentiation does not exist at -3 and 10

why is that

[crimson sedge]

hm

that seems wrong

how can a constant function be non differentiable at 10

5 is a line, it has the same slope everywhere regardless of x

[dull plover]

isnt the differentiation of a constant non existent

because its 0

<@&286206848099549185>

[crimson sedge]

what's your question?

[plucky merlin]

So if you plug in 10 to the -2x - 2 you get -22 which is not equal to 5, and so it would be a discontinuity as the points do not continually run together

@dull plover does that make sense

[dull plover]

yeah but what about the -3

[plucky merlin]

There is no left side to it

[dull plover]

so what do we say about it

[plucky merlin]

Could be a jump discontinuity

Or a removable?

Wait, it might be continuous just not differentiable?

[dull plover]

idk thats why im asking 💀

[plucky merlin]

I'm also confused, it's a closed point only reason I can think of is that it has no technical left side

[dull plover]

well lets go a step back

say f(x) = x>=2

is it continuous at 2

the point sure does exist

but the limit from the left doesnt

[plucky merlin]

Yes, so if not continuous, it's not differentiable

[dull plover]

but that doesnt make sense because it is actually differentiable

u can differentiate the term at -3

without any issues

[plucky merlin]

So can you at 10, it's derivative is 0. But the rules (no matter how funky) state, that if a graph is discontinuous it is non-differentiable

[cedar kiln]

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[thick mauve]

Determine the real parameters a, b, c, so that the function satisfies the conditions of Rolle's theorem on the interval [-1;1].

??

[cedar kiln]

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[thick mauve]

.close

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[ebon cipher]

Can anyone help walk me through this

[rare vault]

are you familiar with cross products

[ebon cipher]

Yes

[rare vault]

what does the cross product of two vectors compute?

[ebon cipher]

The orthagonal vector?

[rare vault]

okay, now what?

[ebon cipher]

Divide by the magnitude of that vector ?

to make it a unit vector ?

[rare vault]

yes, but there's a little catch

[ebon cipher]

uh-oh

[rare vault]

"find all unit vectors"

[ebon cipher]

Hmmm

So how do you go about that 🤔

[rare vault]

try computing the cross product between A and B

and then try computing the cross product between B and A

make sure they are orthogonal via dot product, and see if they are distinct

[ebon cipher]

Dot product = 0 when orthagonal right?

[rare vault]

yea

[ebon cipher]

Okay so it actually ends up being the unit vector from cross product of A and B and then the one from B and A ?

So there is 2 ( if they're both orthogonal )

[rare vault]

they both are guaranteed to be orthogonal

[ebon cipher]

Ah XD

[rare vault]

there is a very obvious connection between them though

think about what being orthogonal to two vectors means

for example, if my two vectors were (1,0,0) and (0,1,0)

[ebon cipher]

Then it'd be (0,0,1) ?

[rare vault]

and...

[ebon cipher]

(0,0,0)?

[rare vault]

no

well okay maybe

[ebon cipher]

I'm not sure im following on this bit

[rare vault]

(0,0,-1)

[ebon cipher]

ah of course because the direction wouldn't matter

[rare vault]

I guess (0,0,0) is orthogonal to everything so you should include that one too

[ebon cipher]

That is also true

So in reality

I'll get a 2 unit vectors from A and B (both directions) and 2 from B and A (both directions) + (0,0,0) so I'll get 5 in total ?

[cedar kiln]

@ebon cipher Has your question been resolved?

[cedar kiln]

@ebon cipher Has your question been resolved?

[ebon cipher]

.close

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[zinc mist]

there are three letters and three envelopes, how many ways are there to derange the letters such that no letter is in it's required envelope?
My answer:
Number of such ways= no of ways to arrange the letters into envelopes-(no of ways to have the right configuration)
which should be 3!-1=5
but the answer is 2

[olive crescent]

there's 2 ways to arrange the first letter and only 1 way to arrange the second

[zinc mist]

so am i misunderstanding the definition of derangements or am i taking extra cases

[olive crescent]

oh I see the issue

[zinc mist]

could you elaborate a bit more cuz i didnt get that-

[olive crescent]

you've solved for "such that not all the letters are correct"

not "such that NO letters are correct"

[zinc mist]

yaaaa

ohhhh

wait a minute let me compare both the cases

oh yea i got what you mean

[olive crescent]

I guess it's...you have 2 choices for the first envelope, and then only 1 choice for the second and third because you cannot pair up with #1.

so 2*1=2

[zinc mist]

ah yess i understood

damn they didnt specify which one should I be solving for

yep it makes sense now

thanks a lot <3

.close

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[vagrant bison]

say i have the statements

[vagrant bison]

(1) ${f(x) : x \in A \triangle B}$ is a subset of ${f(x) : x \in A} \triangle {f(x) : x \in B}\newline$
(2) ${f(x) : x \in A} \triangle {f(x) : x \in B}$ is a subset of ${f(x) : x \in A \triangle B}$

[wraith dagger]

vin

[vagrant bison]

i tried it with f(X) = {1} u X and g(X) = {1,2,3} n X

for union i got that

if A = {2, 3} and B = {3, 4} then
f(A∆B) = {1} u {2,4} = {1,2,4}
f(A) = {1} u {2, 3} = {1,2,3}
f(B} = {1} u {3, 4} = {1,3,4}
f(A) ∆ f(B) = {2,4}

so f(A) ∆ f(B) is a subset of f(A∆B), but f(A∆B) is not a subset of f(A) ∆ f(B)

while for intersect i got that

A = {2,3,4} and B = {3,4,5}
g(A∆B) = {1,2,3} n {2,5} = {2}
g(A) = {1,2,3} n {2,3,4} = {2,3}
g(B) = {1,2,3} n {3,4,5} = {3}
g(A) ∆ g(B) = {2}

so g(A∆B) is a subset of g(A) ∆ g(B), and g(A) ∆ g(B) is a subset of g(A∆B)

so am i correct in saying statement (1) is not true for all functions? or did i do something wrong

[cedar kiln]

@vagrant bison Has your question been resolved?

[cedar kiln]

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[wispy wigeon]

How do I solve 10 and 11

[cedar kiln]

Please don't occupy multiple help channels.

[nimble mountain]

what are those

[hidden escarp]

U gotta post the questions

[wispy wigeon]

Oh

I thougjt I aaploaded it

Mb

[zenith sail]

,rotate

[zenith sail]

@wispy wigeon have you tried anything?

[wispy wigeon]

Ignore y

I need x

[zenith sail]

hm, actually I think we're going to end up getting y first

[wispy wigeon]

😪

Okay

[zenith sail]

do you know what this means?

[wispy wigeon]

The two triangles are congruent

[zenith sail]

right yeah

don't worry about which answer is going to come out first, let's just see what we can figure out

if the triangles are congruent, are there any equations we can write down?

[wispy wigeon]

Uhm

PELASE have mercy on me cuz I suck st geometry honors

Let me think

[zenith sail]

no worries

and no pressure lol

I just mean, if they're congruent, everything in one triangle matches something in the other triangle

like angle A = angle D

[wispy wigeon]

Yea ik that one

Are yiu giving me a hint rn

[zenith sail]

and that is an equation we can use, since we're given expressions for angle A and angle D

[zenith sail]

sorta I guess

[wispy wigeon]

How are they even congruent

[zenith sail]

what do you mean?

the triangles?

[wispy wigeon]

Think I misread something

It’s fine

[zenith sail]

do you mean the angles?

like how do I know that A matches with D?

[wispy wigeon]

Yea

OH

WAIT

I KNOWWW

[zenith sail]

you can actually tell by the way they wrote this

[wispy wigeon]

well

In my way

I list them down

<A Is congruent to <D

<B is congruent to <E yea?

[zenith sail]

yeah

[wispy wigeon]

Okay

[zenith sail]

when they write "triangle ABC is congruent to triangle DEF"

they're also telling you what corresponds to what

since A and D both come first, they're congruent

[wispy wigeon]

Right

[zenith sail]

then B and E both come next

etc

[wispy wigeon]

Mhm

[zenith sail]

for our problem, I think we want to look at <A and <D first

[wispy wigeon]

2y-5=65?

[zenith sail]

yep 👍

[wispy wigeon]

Right

Okay

Y=35

That’s our answer for y

[zenith sail]

yep nice job

[wispy wigeon]

Thank you

Now x

How do we get that thing

Cuz it’s annoying

[zenith sail]

we can do the same thing

[wispy wigeon]

Abd the 2x+y annoying thing

[zenith sail]

since the triangles are congruent, the sides are congruent

[wispy wigeon]

Hiw do I deal with it

Do I js substitute y?

[zenith sail]

Yeah, you already know the value of y

[wispy wigeon]

So

2x+35 = 64.3 maybes

[zenith sail]

not quite

2x+y is a side length

look at the other triangle

[wispy wigeon]

Oh

[zenith sail]

what side does it correspond to?

[wispy wigeon]

OH

90.6

[zenith sail]

yeah 👍

[wispy wigeon]

Okay so

2x + 35 = 90.6

[zenith sail]

looks good

[wispy wigeon]

Is the final answer x=27.8

[zenith sail]

yep

[wispy wigeon]

Wow

Honors class is easy now 😧

[zenith sail]

[wispy wigeon]

Thank you so muchhh

[zenith sail]

no problem 👍 good work

[wispy wigeon]

Thank you

.close

[cedar kiln]

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[cyan void]

it is given that 5x2^t-1=2x5^2t -> (10k)^t =k, find k

[cyan void]

I just get stuck

From the original equation I do 2^t-1 = 2 x 5^2t-1

then do the same for the 2 on right hand side i.e. 2^t-2 = 5^2t-1

[gentle torrent]

idk hwo to say it, but you can use propieties

like square roots, logs...

[cyan void]

this was given as one of the questions in a test - we haven't covered logs in that test

so i can assume that there is a way to do this using surds and indicies rules

[gentle torrent]

oh logs would have made that question easier

[cyan void]

yeah i assume teacher wanted us to do the question without

[gentle torrent]

sowwy w/o logs dunno how to do it

[cyan void]

all g

[cyan void]

what maths exams do you do in your country?

like is it a levels

[gentle torrent]

i had to do that some time ago, but when we did w logs i forgot normal method and did w logs.

[cyan void]

oh alright

[gentle torrent]

im not sure but i think you cant do that

[cyan void]

that's what i thought during the test, i was just kinda stuck

but surely there's a waty

[gentle torrent]

cuz 2^t-1 its not the same 5^2t

if u had 2^t-1 and 2^t+-xyz

then yes

[cyan void]

.

<@&286206848099549185>

[late sky]

is x there multiplication or a variable?

[cyan void]

multiplication

mb probably should've used *

[cyan void]

have you been able to get somewhere?

I haven't been able to get a correct answer as of now even after the test

[late sky]

I was typing something big in a different channel so i havent actually gone through it yet

[cyan void]

oh nw

[late sky]

is the -> just mean that both of those are equations are true, or that you simplified the first equation down to that?

wait, it couldnt have been simplified, theres no k in the first one

but it does just mean theyre both true right?

[cyan void]

yeah both equations are true

the arrow ig means hence/implies that (10k)^t=k

[late sky]

yes, but i felt like an implication in that sense felt like a weird way to describe it since we are already kinda assuming the first part is true, we dont have the false case of the statement to worry about

[cyan void]

idk why they decided to present the question in that way, but it's safe to assume both equations will be true

[late sky]

what level of math are you doing? i need to know how much exactly to expect you to already know, like logarithms or even just algebra and substitution

cuz i dont want to just give you the answer

[cyan void]

im at maths tutor in y12 uk

im doing maths and further maths a level

and we havent covered logs yet

i assume they expect us to solve using surds and indicies rules

[late sky]

uhh, whats year 12 in uk converted to american?

[cyan void]

11th grade?

yeah 11th

[late sky]

i havent heard of the name surds and indicies rule, what is it about?

[cyan void]

It's like the rules for indicies

like a^2 *a^2 = a^4

[late sky]

what level do you tutor?

[cyan void]

supposedly at same level as y12 uk

[late sky]

oh do you mean exponents? I do comp sci, so an index to me is the spot in a list

[cyan void]

yeah basically

in uk we call it surds and indices

[late sky]

wait, so are you in y12, or do you just tutor y12?

[cyan void]

in y12

and i get tutored at y12 standard

[late sky]

wait do you tutor or are you tutored?

[cyan void]

am tutored

this was one of the questions my tutor gave in a test

this was question 2 😭

[late sky]

ohhh, okay, that makes more sense now, i was confused how you could be tutoring your own grade level logistically

[cyan void]

some of my friends were doing that but that's besides the point

[late sky]

so where do you get stuck? do you have an idea of where to start?

[cyan void]

if you read up

@late sky

then i kinda just get stuck midway through after that

i can't get anything to an appropriate form

[late sky]

well it looks like that isnt quite right, you seemed to have divided by five on both sides but forgot to do the same to the constant term

[cyan void]

Wdym constant term?

[late sky]

do you know what a term is? i might just need to describe it better

[cyan void]

yeah

you mean the (10k)^t = k should be divided by 5 as well?

So it looks like 2^t x 5^t-1 x k^t = k?

[late sky]

no

omg wait, is it 2^(t-1) or (2^t)-1 in the original problem?

[cyan void]

the former

2^(t-1)

uh oh

[late sky]

that chages things

i thought the -1 was its own term

[cyan void]

Nope soz

uhhhhh is it possible to dm the workings to said solution to me? its a big ask but im tired and its 00:32 here

if not it's fine, if you point me in the right direction it'd helop

[late sky]

so then the what you had so far was right, i just misinterpreted it

[cyan void]

What do you suggest doing from there then?

[late sky]

well im kinda working it out live, i did it on my own first with logs

[late sky]

do you know how to change the base of an exponent?

[cyan void]

no wasn't taught

the videos i see rn use logs or the examples are bases with a power i.e. 16^x = (2^4)^x = 2^4x

[cedar kiln]

@cyan void Has your question been resolved?

[late sky]

hmm, i dont understand how they would expect you to solve it with out logs or change of bases

[cyan void]

Me neither

What is the answer that you get from logs?

[late sky]

well the examples showing you 16^x = 2^4x, is that something you would be allowed to use or is it something youve found on your own?

becuase if you would be expected to know that, i doubt they only expect 'basic' examples, you should be able to understand the rule in general to use it here, i can explain how to generalize it if you need

[cyan void]

Uhh sure generalise it for me

[late sky]

well for any y^x such that y=a^b can be rewritten a^(bx)

[late sky]

however, 2 and 5 are prime, so theres not really a rational b that lets 2^b=5, so you would have to use logs, but you did say there were videos about logs, were those found on your own or included in the ciriculum so far that you would be expected to know?

[cyan void]

I found them on my own

Also it's way too late for me rn so unfortunately I'll have to call quits on thus

Thanks for the help anyway

Gtg sleep now

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[civic heath]

hi is this where u get math help?

[plucky owl]

Yes just post the question you need help now and someone who can help will stop by to help you

[civic heath]

alr thanks

I don't get how to do this

[astral steppe]

you could draw a line from each letter to D then for each line draw one of the same length at 120deg

from D

[civic heath]

So however long A is from D, I'd draw that length from D at 120 degrees?

alr i got it thanks!

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[hexed vortex]

Why is this wrong

[pastel vault]

you forgot the absolute values

25 ln |x|

try with that now

[hexed vortex]

won't let me use it, says ln uses parenthesis for it's argument

[pastel vault]

yeah, so put the || inside the brackets

[hexed vortex]

oh that worked

thanks

[pastel vault]

no worries

[hexed vortex]

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[frozen mountain]

hello can someone help me with the next 1 following number for these 2 questions pls

[frozen mountain]

its a pattern recognition thing

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[frozen mountain]

.reopen

[cedar kiln]

✅

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[crimson sedge]

.ask

how do i do this

[crimson sedge]

.ask \left(\frac{x^{\frac{1}{2}}y^{-2}}{x^{-\frac{7}{4}}y}\right)^4

jesus

[acoustic grove]

What?

[crimson sedge]

i dont know how to write it down

[acoustic grove]

\left(\frac{x^{\frac{1}{2}}y^{-2}}{x^{-\frac{7}{4}}y}\right)^4

[rain drift]

wrap it in dollar signs

$\left(\frac{x^{\frac{1}{2}}y^{-2}}{x^{-\frac{7}{4}}y}\right)^4$

[wraith dagger]

MellowDramaLlama

[crimson sedge]

what the

[acoustic grove]

Thonks

[crimson sedge]

yeah

so uhm h o w

that needs to be simplified

[acoustic grove]

Lemme recheck my calculations

Yeah

Seems correct to me

[crimson sedge]

is that the simplified vers?

[acoustic grove]

Yeah

Only x and y are remaining, can't simplify further as they are variables

[crimson sedge]

im so

confused

[crimson sedge]

is that an x

[acoustic grove]

Yes

[crimson sedge]

is that a g

why is there a g

[acoustic grove]

Its y

[crimson sedge]

x g

[acoustic grove]

Pls don't bully my writing

[crimson sedge]

im just trying to understadnfusdf

[acoustic grove]

Okay lemme write neatly

This is the best i can write

[crimson sedge]

really?

thank you anyway ill try my best to understand

[acoustic grove]

U can post this on algebra channel

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[twin tinsel]

for estimations using simpsons rule, midpoint rule, and trapazoidal rule. when spliting up the bounds into my n's, do I us the lower bound or the upper bound if it divedes into an odd number

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[dire geode]

Show an example what you mean

[twin tinsel]

this but trap rule and n=5

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[devout mantle]

this is one of my college assignment questions

and i am just..

frustrated

because i feel like this question is not well structured and i dont understand it

anyway, as it carries marks

i did something

and when i go check the rubric it asks to use differentiation to solve the problem.

and i just feel like this question does not make any sense

i want to know whether its normal for me to feel this way and that the question is stupid.. or i am stupid?

Thank you

[ocean mural]

hey idk the solutions but I can say ur definitely not stupid if ur doing maths like that lol

[mighty kraken]

Also, I don't understand why you would use differentiation here.
I mean, there is no restriction to the height, so cylinder will obv be better in this case, while sphere will be restricted to it's diameter. It's logical i believe, plus you found it yourself as well

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[royal loom]

** Is this transformation simply the substitution x=(u-v)/sqrt(2), y=(u+v)/sqrt(2) ? **

[astral bay]

yeah it looks like it is
if it isn't then they're doing some pretty confusing notation with the "x(u,v)" and "y(u,v)"

[royal loom]

Okay

I have questions on what this does to our bounds then

So first we are integrating from y=0 to y=x

if y=0 then this implies that v=-u

and if y=x then this implies v=0

likewise I can do something similar for the x-bounds

is that what I should be doing?

Because once I do that, I get that x=0 implies u=v and x=1 implies u-v=sqrt(2)

and I'm not sure how to express these new bounds in the uv world

because they both involve variables

one isn't a constant to a constant

unless I can say that u going rom x=0 (u=v) to x=1 (u=sqrt(2)+v) is the same as u going from 0 to sqrt(2)

Does that make sense what I am asking or should I rephrase?

[royal finch]

It is easier to follow what is going on if you write inequalities in terms of x and y, then substitute in u and v and draw a picture

You have that 0 < x < 1 and 0 < y < x, now focusing on the first inequality, you have 0< u-v < sqrt(2) or 0 < u < v + sqrt(2) so u is bounded by the lines u = 0 and u = v + sqrt(2)

Plot those on the u,v plane, and then repeat for 0 < y < x and plot those too. Then determine how you want to set up your integral

[royal loom]

like this?

and our region is that little triangular region

[royal finch]

yes, but I find the way you organize it like that to be hard to read

[royal loom]

Oh sorry

I tried to make it organized

Okay so we are going to have to split it since it is a triangular region

the lower bound will be a different function once you cross that middle point

[royal finch]

Hopefully you can see why this is easier to condense the information:

0 < x < 1
=>
0< u-v < sqrt(2)
v < u < sqrt(2) + v

[royal loom]

indeed

[royal finch]

It's the same as what you did in anycase

[royal loom]

Okay thank you

I think I have it set up properly now

what a nasty integral

I have no idea how to integrate that

this is sort of a different question now so I am going to open a different channel

ty for your help!

.close

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