help-28

[minor needle]

uhh

when the matrix is like 0 0 0 1

is that the only case?

[spiral vigil]

yeah, when you can get a line like that after row reduction

[hushed briar]

I'd simply use Rouche-Frobenius instead reduction

[minor needle]

bro idk whats a rouche-frobenius 😭 im at my 2nd week of linear

[hushed briar]

In that case, you might not be supposed to using it. It's based on the ranks of the coefficient matrix and the expanded matrix

[minor needle]

ok...

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[rare sage]

What have you attempted for this problem so far, or what formulas might you be asked to use?

[rare sage]

Oh sorry, hadn't seen

Let's see if that works out.

Henry Neal would begin and Bolt would come in $0.57$ s in, when he now has $9.58$ s left to reach the finish line.
This is exactly the time Bolt needs to go from finish to end. So yes!

[glossy valve]

Drenitor

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[misty yacht]

can someone please help with b on ex 2

when it asks the maximum value on the interval [1,5], idk what it's referring to

and it's also hard to tell with the limited set of values given the table

ping me if you can help

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@misty yacht Has your question been resolved?

[shrewd hamlet]

@misty yacht

U still need help

[misty yacht]

yes

[shrewd hamlet]

Yea so it’s asking for the largest y-value that’s paired with any x from 1-5

So Ye the max value would be 4

[misty yacht]

why woulkd it be 4

@shrewd hamlet

[shrewd hamlet]

What x values are given in the table

[misty yacht]

1, 2, 3, 6, 7

also when it says "interval" it means in terms of x right?

[shrewd hamlet]

Yea

And since our interval is 1 to 5 inclusive, we can only look at 1,2,3

(1,4) (2,2) (3,0)

I think u get the picture

[misty yacht]

ohhhh

so because we cant go past 5

the only value is x = 3 max

and based off of those coordinates

y = 4 is the maximum

[shrewd hamlet]

Yea

[misty yacht]

alright tahnks

[shrewd hamlet]

Np

[misty yacht]

why do translations a) and b) result in the same function?

i dont understand why

cant i just move out the 3 from y = |3x| and make it y = 3|x|?

[deft flower]

Translations a) and b) are technically different. If we set $f(x) = |x|$, the left graph represents $y = 3 f(x)$, while the right represents $y = f(3x)$.

The definition of $|x|$ is:
$$|x| = \begin{cases}
x & x \geq 0 \
-x & \text{otherwise}
\end{cases}$$

Notably, both $x$ and $-x$ are lines with y-intercepts of zero. As is true for any such line, $(3x) = 3(x)$ and $3(-x) = -(3x)$.

[glossy valve]

kiako

[deft flower]

If you'd like, we can look at an example where $k f(x) \neq f(kx)$.

[glossy valve]

kiako

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@misty yacht Has your question been resolved?

[misty yacht]

What

bro im in algwbra 2 not data and statistics

jk but i don't get what you mean

wait let me read carefully

i do not understand the |x| = {} part it's confusing me

could you please explain

[deft flower]

ah

that's a way to describe cases

so the absolute value of x is x when it's positive, otherwise it's -x (since the negative of a negative is positive)

we might say that the function |x| is a piecewise function, since it's best described as a combination of two or more pieces

generally, when we need to look at piecewise functions, we can instead consider the behavior of each part

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[tulip citrus]

distributing

[reef dust]

what do you need help with?

[tulip citrus]

well

it got different with this one

is the -5d still distributing?

[reef dust]

nope

only the 3

then simplify after

[tulip citrus]

so then -5d + 3 first

[reef dust]

no

[tulip citrus]

wait were simplifying?

[reef dust]

you distribute the 3 first

you do the 3(2d - 7) first

[tulip citrus]

alright

[reef dust]

because your multiplying 3 by each value inside the bracket, which comes before the addition (order of operations)

so you distribute first, then simplify with -5d

[tulip citrus]

simplify it with what

[reef dust]

distribute the 3 first, what do you get

[tulip citrus]

6d - 21

[reef dust]

mhm

but now you have to account for the -5d

so you have -5d + 6d - 21

[tulip citrus]

so then thats...

= -5d + 6d - 21

well you said it

but is that the final answe

answer*

[reef dust]

nope

simplify like terms

[tulip citrus]

but what if the paper tells me not to

[reef dust]

what

show?

[tulip citrus]

well im just saying

so this is a must?

[reef dust]

ye its assumed that you should get the simplest form

[tulip citrus]

alright

so then how

because we dont know what the variables equal

[reef dust]

you have like terms

-5d + 6d

[tulip citrus]

ohhh

yeah yeah

[reef dust]

yeap

[tulip citrus]

wait so

[reef dust]

so add those together

[tulip citrus]

thats...

-1d

[reef dust]

nope

[tulip citrus]

bruh

mb

1

[reef dust]

yep

well 1d

then just d

but im assuming you knew that

[reef dust]

but we still have the -21

so our final answer is?

[tulip citrus]

1d - 21

[reef dust]

yes but you dont need to have the 1

[tulip citrus]

d - 21

[reef dust]

yep

[tulip citrus]

alright

hard to remember but ill try

ty

[reef dust]

no worries

[tulip citrus]

could you make like a list in order?

on how to do this type

if not its alr

[reef dust]

its simply

1. distribute
2. simplify like terms

thats it :)

[tulip citrus]

ty

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[cedar mortar]

Hi

[cedar mortar]

I want to ask if

slope of curve =tanx is base on we set the triangle’s opssosite and adjacent

Parallel to x axis and y axis

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[astral sinew]

you just said it

lol

i think i understand that from line 2 to line 3 there was a row reduction of R2=R2-(1/5)*R1

[steep anvil]

What ? 💀

What is this Ohio question ?

Idk bro 💀

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[slate ivy]

i need help with this

[thick hedge]

try factorisising

[slate ivy]

what the x squared

[grave elm]

if it helps, you can let u=x^2, then it will become a regular quadratic which can be factored easily

[slate ivy]

so 2u^2+2u+1

Ah

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[coarse elbow]

u^2+2u+1, without the 2 at the front

[slate ivy]

alr i see it

[grave elm]

and if you were to factorise this, you would get (u+1)^2 = (x^2+1)^2

[slate ivy]

alr ty

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[coarse elbow]

I will note that you need to be a bit careful here, since sqrt((x^2 + 1)^2) = |x^2 + 1|. But since x^2 + 1 is always positive you can drop the absolute value.

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[gray dagger]

I do not know where I went wrong

[gray dagger]

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[torn jolt]

is it possible to find probability of a and b

[torn jolt]

if you have p(a) and p(b)

and p(a|b)

no other info given

i tried doing p(a) x p(b) but that was wrong

so was p(a) x p(a|b)

[fast peak]

do you know a formula for p(a|b)

[torn jolt]

p(b and a) x p(b)?

im just so lost

like the questions are asking for all this information about it from just this 3 points

like if it's dependent and mutually exclusive as well

@fast peak do u know?

[fast peak]

no

look up the definitions of all these terms again

[torn jolt]

it says p(a and b) = p(a) x (pb|a)

i assume u can divide both side by p(a)

but that gives p(b|a)

not a|b

would it be p(b and a)/p(b)? idk

[fast peak]

yes

[torn jolt]

im still not sure how to get a and b

i tried doing p(b) x p(a|b) but that was wrong

[fast peak]

thats correct tho

[fast peak]

check your arithmetic

[torn jolt]

oh my got

im gonna scream

lmao

thx

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[faint iron]

c)

not sure how to do that using a) and b)

a) is 1/2 + sqrt3/2 i

and b) is cis pi/3

[onyx glen]

Hence find sin(pi/3) in surd form
as if this was not already known to be sqrt(3)/2

[faint iron]

lol

[onyx glen]

what a strange question

[faint iron]

maybe if it was like pi/8 would be better

but im guessing they're tryna test my knowledge of something

<@&286206848099549185>

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[faint iron]

oh wait just equate coeffficenitent

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[cloud vale]

conic sections, transforming general form to standard form

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[amber cipher]

I need to find a (trigonometry) but i don't know where to begin

[onyx inlet]

Ever heard of

SOH CAN TOA?

[amber cipher]

yes

i just don't know exactly how to use them

[onyx inlet]

Do you know what SOH CAH TOA stands for

[amber cipher]

Sin opposite hypotenuse. Cos adjacent hypotenuse. And TOA i dont know

[onyx inlet]

Tan = Opposite/Adjacent

[onyx glen]

^

[onyx inlet]

Do you know how to identify the sides?

[amber cipher]

kinda

[onyx inlet]

Which one is the hypotenuse, opposite or adjacent?

Identify it in the 2 triangles for me please

[onyx glen]

here, this might help.

[amber cipher]

?

[onyx inlet]

Yes

[amber cipher]

huh

[onyx inlet]

Great you’re right

[onyx glen]

i gave names to all relevant points.

[onyx inlet]

So now from this, do you know which one to use?

[onyx glen]

you know that it is standard practice in geometry to name points with uppercase latin letters, and to name line segments and polygons by their vertices. yes?

[amber cipher]

no idea

cos

?

[onyx glen]

recall the SOH-CAH-TOA mnemonic.

[onyx inlet]

You’re looking for side “a”

[onyx glen]

X labeled the sides with O, H and A.

can you name which two are relevant to you?

@amber cipher

[amber cipher]

A and O?

[onyx glen]

these two?

[amber cipher]

yes

[onyx inlet]

No

[amber cipher]

no

[onyx glen]

why is that bottom side relevant?

it is neither known nor sought.

[amber cipher]

dont know just guessing

[onyx glen]

why guess?

[onyx inlet]

you’re looking for side “a”

[amber cipher]

cause i dont know

[onyx glen]

this question has a clear and straightforward answer.

in any problem you have things that you KNOW and things that you SEEK.

literally almost any math problem.

[onyx glen]

yes, ok good

[onyx inlet]

Great

[onyx glen]

so which part of the SOH-CAH-TOA mnemonic kicks in?

[amber cipher]

cah

[quasi marsh]

a is not A!

[amber cipher]

cause (i think) Sine would be used to find 6 and cos would be used to find a

[onyx inlet]

No

[onyx glen]

no

you highlighted the opposite and the hypotenuse

[onyx inlet]

It forms a fraction

You can find either one if you know the other side

[onyx glen]

you're making it all more complicated for yourself.

[amber cipher]

6.3

[onyx glen]

i didn't ask you to blurt any numbers out.

[amber cipher]

6/sin(70) = 6.3

[onyx inlet]

Huh?

[onyx glen]

wow, ok, way to jump ahead..

[torn jolt]

How would you solve 7?

[onyx glen]

@torn jolt please open your own channel. #❓how-to-get-help

[onyx inlet]

Find a different channel

[onyx glen]

how quickly you went from not knowing your way around a triangle to correctly writing down the formula for side a!

[amber cipher]

i need to find b

[onyx glen]

and now we are rapidfiring problems, i guess.

[onyx glen]

without so much as a "thanks, i understand how to do this now" or even a "ok, i'm done with this, on to the next one".

kind of impolite there tbh.

[amber cipher]

sorry, i forgot the english word to go to the next question

[onyx inlet]

I believe you know what to do now

[amber cipher]

didn't mean to be mean

[onyx glen]

what is your native language

[amber cipher]

Danish, but sometimes i forget words in both languages

[onyx glen]

i think it is kind of suspicious that you would also forget the word for "thanks"

but ok

this problem with the green triangle is actually EXACTLY THE SAME as the previous one

only with different numbers

but the model is 100% the exact same.

[amber cipher]

So the answer would be 9.4?

by using the same equation?

[onyx glen]

yes it looks like

do you understand why these two problems are the exact same tho

[onyx inlet]

Yes

[amber cipher]

i know the word "Thanks" and im sorry for being impolite and all that. i've just had a long day and stuff so brain aint working much

[onyx inlet]

Is that all you need help with

[amber cipher]

because this one is basically the same as the last one just flipped vertically?

[onyx glen]

yes

[amber cipher]

no, i have alot more, but i dont wanna just make it look like i need help with my intire homework sheet. and i also dont wanna waste others time.

[onyx inlet]

Well do you know what to do now?

[amber cipher]

not really, but a lil more than before

[onyx inlet]

Well if you’re in doubt just ask

[amber cipher]

well i also need help for another task wich many of my classmates also used alot of time on

like here i need to find a

[onyx glen]

you care about the big outer triangle here

label its sides and you'll win

[amber cipher]

like this?

or do i need to do something else?

[onyx glen]

no

i said big outer triangle

[amber cipher]

ahhhh ok

so do i need to do 5*sin(53) to find a

?

also how is buttom the hypotenuse?

[amber cipher]

which equation should i use in this ?

[onyx glen]

you care about adjacent (A) and opposite (O)

recall SOH-CAH-TOA

(again)

[amber cipher]

cosine

ahhh tanjent

5 tan(53)?

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[spare viper]

Sorry to bring this here again but can anyone tell me where I messed up?

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[hushed briar]

Why don't you calculate the volume of green minus the volume of blue?

[spare viper]

If I was looking for green I'd do yellow minus blue but in this case it seems like it's just asking for the integral from the origin to 9 on the y axis

[dapper garnet]

by looking at the image shown it looks like you're calculating using the wrong reigon... I'm not entirely sure

try 3pi though

it says rotation of r1 which is the blue section, if so then it should be a lot more simple than the solution you've shown

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[limber pier]

how do i do this bruh

i knew how to do this for normal ones like |x-4| but idk about this one

[torn jolt]

what have you tried?

[stark forge]

What is the function sqrt(25 - x^2) graphically look like?

[limber pier]

is it just 0 then?

[stark forge]

if you're riding a skateboard on that and you're at 5

are you doing down fast

[limber pier]

you die bc the tracks over

[stark forge]

from the left at x = 5, you're going down infinitly fast

so you're deriviative is -infinity

The derivaitive at x = 0 is zero because you're at rest on the skateboard

The limit on the right at x = 5 is not defined

there is no function on the right at x = 5

[limber pier]

wouldnt it be 0 for coming from the left?

bc ur approaching y=0 as you approach x=5

[stark forge]

No because the derivative measure the slope of the function

[limber pier]

o

[stark forge]

Consider the tangent line at x = 4.5

it's facing down right?

pretty steeply

now imagine it at 4.9999

[limber pier]

okok

how do i do this algebraicly tho?

[stark forge]

Ok, let's prove that the limit as x-> 5 from the left of the function f(x) = sqrt(25 - x^2) equals -infinity

Do you know the formal definitions?

[limber pier]

uh

not off the top of my head

[stark forge]

Ok, when you say algebrically what do you mean then?

[limber pier]

just how do I show my work

this is the work showed for this probelm

am i able to do the same for my problem?

[stark forge]

Let me see what I would do

[limber pier]

well i guess you actually could

[stark forge]

0 / 0 is not DNE

it is indeterminate

You can use l'hopitals

Do you know it?

[limber pier]

kinda

not really

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[vital sinew]

Jordan rolls eight fair dice, computing the sum of the numbers on the top of the dice after the first three rolls. What is the probability that Jordan is able to find at least one way to pick exactly three of the latter five dice in such a way that the sum of these three dice is the same sum as that of the first three dice?

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[nimble current]

interesting problem

casework should work, but would take quite some time and you may have to use PIE for the latter 5 dice

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@vital sinew Has your question been resolved?

[vital sinew]

can someone help

[vital sinew]

<@&286206848099549185>

[vital sinew]

yeah casework would be way too tedious

[gritty rose]

That's better than not doing it

[vital sinew]

there’s no better approach?

[vital sinew]

do u see anything better?

[gritty rose]

Yes

[slate thunder]

whats the source btw?

[vital sinew]

I just thought of it

[vital sinew]

share then pls

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[worn matrix]

kx_1=20 and kx_2=24 and kx_3=40/3
how can we find ratio of x1 x2 and x3

[worn matrix]

and we dont want K in the ans

[tranquil cave]

k×(x1)=20 k×(x2)=24 k×(x3)=40/3?

[worn matrix]

yes

[tranquil cave]

Just divide them

k(x1)/k(x2)=20/24

[worn matrix]

and x3 also

then k would come in our answer

[tranquil cave]

No because it cancels out

[worn matrix]

why?

we want x1:x2:x3

not x1:x2

[tranquil cave]

Ok but first find the ratio between x1 amd x2 then x2 and x3

[worn matrix]

i did

[tranquil cave]

k is both in the numerator and denominator

[worn matrix]

i divied x1 and x2 then x1 and x3 and x2 and x3

[worn matrix]

not would be the case whn we gonna divide it with x3

[tranquil cave]

But x1:x2:x3 if talking about ratios is not x1/(x2)(x3)

[worn matrix]

what?

if you divide

x1 and x2

then we got x1:x2 where we dont have k

but when you will divide this

with x3

then k would come in our ans

can you show that in latex

[tranquil cave]

Ill explain wait a sec

So x1:x2 is a ratio you dont divide that with x3 because that would be the ratio of the (ratio of x1 and x2) and x3

Instead we need to find the ratio between x2 and x3

[worn matrix]

so what according to you should be the ans3ew.

answer?

[tranquil cave]

x1:x2:x3 = 20:24:(40/3)

[full forum]

@worn matrix Has your question been resolved?

[worn matrix]

<@&286206848099549185>

[gritty rose]

Did you try using this

[worn matrix]

yeah

but

how can i simplify it

[full forum]

@worn matrix Has your question been resolved?

[worn matrix]

@wicked seal

tf

sry

<@&286206848099549185>

[celest oasis]

what?

[worn matrix]

.close

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[worn matrix]

.close

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[cloud shore]

Based

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@lilac parcel Has your question been resolved?

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@lilac parcel Has your question been resolved?

[tulip marlin]

when in doubt, I would say

• use a venn diagram?
• expand completely in terms of unions and complements
• show set equality by showing subset relation in both directions

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@lilac parcel Has your question been resolved?

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@lilac parcel Has your question been resolved?

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[tepid root]

I don't know if I need to find the equation of the circle in order to do this but that's what I'm trying

so

x^2-6x+9 = (x-3)^2

and I have the first part of the equation of the circle there

[potent loom]

are my answers right?

[tepid root]

but then I don't know what to do with the y^2 +10y

[tepid root]

create another channel 😛 please

[stark ruin]

do you knoe what the general equation of a circle is

[tepid root]

yea its (x+h)^2 + (y+k)^2 =r^2 where (h,k) is the center of the circle and (x,y) are the points of the circumference

[stark ruin]

yeah so when u expand that

u have

x^2 +y^2 +2gx+2fy+c=0

here

-g,-f is centre

and root over(g^2+f^2-c)

is radius

[tepid root]

let me check

[potent loom]

claimed?

BRUH WHICH CHANNEL IS FREE

[tepid root]

there is like a category

that separates them

so the ones in the top are free

(the red box)

[tepid root]

I got it thanks friend

👍👍👍

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[twin wolf]

So I’m stuck here trying to prove

I reduced it to sin/cos but that’s not useful to me

[nimble crane]

1-cos^2 is sin^2

[twin wolf]

Oh yeah

I forgot

Okay but after that

I’m not sure

[nimble crane]

do you know how tan is defined?

[twin wolf]

sin/cos

[nimble crane]

exactly

[violet anvil]

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

[nimble crane]

^^^

[twin wolf]

ah okay let me try something with that

Okay cool thanks guys

It was just the squared that messed me up

[willow sedge]

you can not cancel sin(theta) and cos(theta)

and even if you mean to write it as tan(theta), the red mark is just unnecessary and could be misleading

[twin wolf]

How come

[violet anvil]

You have the right idea it’s just you’re not cancelling them so using those red marks is misleading notation

you’re rewriting them

[twin wolf]

I turned the sin/cos into tan

[violet anvil]

Yeah and that’s right, it’s just not cancelling

[willow sedge]

you TURNED not cancelled

[twin wolf]

so the denominator becomes 1 while the numerator is tan*sin

ok

So it’s right just the notation is wrong

?

[violet anvil]

only thing that’s wrong is the red marks

[twin wolf]

Yes

$\frac{1-\sin^{2}x\cdot\cos^{2}x}{\sin^{2}x}=\cot^{2}x+\sin^{2}x$

[glossy valve]

water beam

[twin wolf]

how do I approach this?

so I can turn 1 - sin^2(x) into cos^2(x) but will that help me here

i dont think so

what about turning cos^2(x) into 1 - sin^2(x)

@violet anvil

[narrow helm]

the sin^2 is in multiplication with cos^2

so you cant convert it

[twin wolf]

ooh what about what about

converting the 1

into sin^2(x) + cos^2(x)

[narrow helm]

or just distribute the denominator to 1 and sin^2cos^

both works

[twin wolf]

$\frac{\sin^{2}\left(x\right)+\cos^{2}\left(x\right)-\sin^{2}\left(x\right)\cdot\cos^{2}\left(x\right)}{\sin^{2}\left(x\right)}$

[glossy valve]

water beam

[twin wolf]

okay uh what do i do here

@narrow helm

[narrow helm]

divide by sin^2

[twin wolf]

wot

its already divided

im not sure what you mean here

[violet anvil]

It’s like how if you had 8/2 you can make it 4

you can “apply” 1/sin^2 to everything in the numerator

[twin wolf]

theres nothing i can cancel though

because its subtracting

[violet anvil]

(8 + 2)/2 = 4 + 1

Division distributes across addition or subtraction

[twin wolf]

im not sure i understand

[glossy valve]

Dyssrupt

[twin wolf]

so you want me to split the fraction

[violet anvil]

If it helps you understand the concept better

[narrow helm]

yes

[twin wolf]

hows this

[narrow helm]

the first one too

[twin wolf]

which one?

[violet anvil]

(sin^2x + cos^2x)/(sin^2x)

[twin wolf]

you want me to simplify that into 1/sin^2(x)?

[violet anvil]

Split it into (sin^2x)/(sin^2x) + (cos^2x/sin^2x)

[twin wolf]

oh

ok

this is what i got

[narrow helm]

cos/sin is tan?

[twin wolf]

oops

[narrow helm]

also how did cos^2 change to cos?

[twin wolf]

oh i forgor to write it

[narrow helm]

cos/sin is?

[twin wolf]

ooh

cot

omg

i see where youre going now

[narrow helm]

and 1 - cos^2 is?

[twin wolf]

here we go

very noice

thanks

[narrow helm]

@twin wolf do you need help in anything else?

[twin wolf]

actually yeah i wanted to look at an equation and see if my thinking process was right

hold on

Solve $\cos\left(\frac{3x}{2}\right)=\frac{1}{2}\ \operatorname{for}\ -\frac{\pi}{2}\le x\le\frac{\pi}{2}$

[glossy valve]

water beam

[twin wolf]

okay so

i should take arccos(1/2) to be pi/3

as base angle

so

$\cos\left(\frac{3x}{2}\right)=\frac{\pi}{3}$

[glossy valve]

water beam

[twin wolf]

then $\left{\frac{\pi}{3},\frac{5\pi}{3}\right}$ are the two solutions

[glossy valve]

water beam

[twin wolf]

then we have to solve for the arguement

$x=\frac{2\pi}{9},\frac{10\pi}{9}$

[narrow helm]

wrong

[glossy valve]

water beam

[twin wolf]

aw

[narrow helm]

that should be cos(pi/3)

not just pi/3

[twin wolf]

okay

wait

so its still like

sorta correct kinda maybe

[narrow helm]

no

[twin wolf]

is the stuff i did afterwards still valid

[narrow helm]

also 5pi/3 is not a solution

the interval given is -pi/2 to pi/2

[twin wolf]

i know im planning to cut my solutions out of the domain after

i do that last

[narrow helm]

which doesnt lie in that interval

[twin wolf]

but im asking if my steps are correct

besides the pi/3 thing

[narrow helm]

ok, yeah

[twin wolf]

cool

cool cool so the only solution for this is 2pi/9

leshgo

[narrow helm]

no

[twin wolf]

what

what what what

what did i do wrong

[narrow helm]

https://tenor.com/view/yoda-there-is-another-star-wars-jedi-gif-5140031

[twin wolf]

impassible

wait

no hold on let me cook

ok so i got an idea right

cos is periodic every 2pi

[narrow helm]

doesnt help

[twin wolf]

frick

[narrow helm]

hint : ||even function||

[twin wolf]

-2pi/9

?

[narrow helm]

yup

you should've included -pi/3 at the start to avoid this hassle.

[twin wolf]

so if it was sin instead of cos it wouldnt be -2pi/9?

[narrow helm]

yup

[twin wolf]

okay cool

wait so -pi/3 is a solution

okay

[twin wolf]

woah no no no

i did something wrong here

i did something wrong from the beginning

this is all wrong im pretty sure

[narrow helm]

[twin wolf]

so i get pi/3 as base angle

ok and i find where cos is positive

pi/3, 5pi/3

ok nevermindf

and because even

pi/3, -pi/3

then solve for arguement

ok nevermind

im just genius

[narrow helm]

yes you are!

[twin wolf]

thank you

[narrow helm]

yw

[twin wolf]

OKay i have another question

Let $f\left(x\right)=\sin\left(\frac{2\pi x}{3}\right)$.

Solve $\sin\left(\frac{2\pi x}{3}\right)=-\frac{\sqrt{3}}{2}\operatorname{for}\ 0\le x\le3$

[glossy valve]

water beam

[twin wolf]

is this a similar approach as i just did

im just skeptical with the 2pi*x there

[narrow helm]

dw about it.

[twin wolf]

okay

[narrow helm]

just solve like you normally do.

[twin wolf]

$\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3}$ then $\left{\frac{2\pi}{3},\frac{5\pi}{3}\right}$ are my solutions

[glossy valve]

water beam

[twin wolf]

right?

wait no this is really sad my pi units cancelled

[glossy valve]

water beam

water beam

water beam

[twin wolf]

@narrow helm can you checky check

[narrow helm]

wrong

[twin wolf]

whered i go wrong

[narrow helm]

sin(2pi/3) is?

[twin wolf]

sqrt(3)/2

i thought i just had to do arcsin(-sqrt(3)/2

how come that method no worky for this question but work for other one

[narrow helm]

also arcsin ranges between -pi/2 to pi/2

[twin wolf]

ples tell me why it dont work

why no arcsin (-sqrt(3)/2)

[narrow helm]

are you using a calculator?

[twin wolf]

yes

yeppeee

wait this was the answer and i got 5/2 correct

but not 2

@narrow helm how come their base angle is + pi/3

but when i input into calculator

mine is -pi/3

why is that

[narrow helm]

i told you

arcsin has range of [-pi/2, pi/2]

[narrow helm]

idk what the term base angle means tbh

[twin wolf]

its like using the inverse trig to find the first angle

and then using that angle to find the rest

[twin wolf]

i dont understand

oh yeah i see

[narrow helm]

have you studied inverse trigonometric functions before?

like in depth

[twin wolf]

yeah i remember the inverse sine being trunctated

when i was looking at trig derivatives

[narrow helm]

lol

[twin wolf]

but why does that make -pi/3 positive pi/3

[narrow helm]

the inverse of a function exists only if the function is bijective.

so we restrict the domain of sinx to make it bijective and get its inverse as arcsin

so its range becomes the restricted domain of sinx => [-pi/2, pi/2]

[narrow helm]

wdym

[twin wolf]

calculator gives -pi/3

when the answer says +pi/3

[narrow helm]

+pi/3 is not a solution

ig they mean for +sqrt(3)/2

they are taking it as a reference

what mostly everyone does

[twin wolf]

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

ok

wait i have another question but its about different thing

$\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}$

[glossy valve]

water beam

[twin wolf]

and im looking for solutions $\left{-\pi\le x\le\pi\right}$

[glossy valve]

water beam

[twin wolf]

the original question is $\sqrt{3}\sin x=\cos x$

[glossy valve]

water beam

[twin wolf]

@narrow helm i was trying my usual method where you find where tan > 0 but it didnt work

because i get 7pi/6 which doesnt exist

so i only have pi/6

how do i find the other one

[narrow helm]

is tan odd or even?

[twin wolf]

odd

[narrow helm]

so our solution doesnt lie between -pi/2 to 0

now what's left?

[twin wolf]

wait

[twin wolf]

idont get

[narrow helm]

is tan positive or negative in 0,pi/2?

[twin wolf]

it will be positive

because its in first quadrant

[narrow helm]

as tanx is odd

so tan will be positive or negative in (-pi/2, 0] ?

[narrow helm]

use this result.

[twin wolf]

whats -pi/2,0

[narrow helm]

interval lol

[twin wolf]

uh

how>

[glossy valve]

Dyssrupt

Dyssrupt

[narrow helm]

example ^

[twin wolf]

so

[twin wolf]

tan(pi/6) = -tan(pi/6)

[narrow helm]

um yeah, but not what I'm trying to make you understand

[twin wolf]

i didnt have to worry about this even odd stuff for a lot of other questions i could just use quadrant

im very confused now why it wont work

is it because of domain restriction?

[narrow helm]

do you know the graph of tanx?

[twin wolf]

yes

[narrow helm]

can you picture it in your mind?

clearly*

[twin wolf]

yes

[narrow helm]

is tan positive or negative in (-pi/2, 0]?

[twin wolf]

negative

[narrow helm]

and your problem wants you to find when tan is 1/sqrt(3)

which is obv positive

[narrow helm]

so does our solution lie in this interval?

[twin wolf]

wait but i dont understand why we look at this interval specifically

[narrow helm]

tanx is changing its sign at -pi/2

you said you could picture the graph clearly

[twin wolf]

yes

[twin wolf]

no

cuz we got + pi/6

[narrow helm]

ignore that for now

so now, is tan positive or negative in [-pi, -pi/2) ?

[twin wolf]

Positive I believe

[narrow helm]

correct

so your solution may lie in that interval

[twin wolf]

Okay

[twin wolf]

So is it theta - pi?

[narrow helm]

corrrrrrect

[twin wolf]

Ho Ho ho ok

[narrow helm]

im more happy than you lol

[twin wolf]

I’m making study notes for an upcoming exam so do you mind if I clarify some things

[twin wolf]

Haha

[narrow helm]

sure (if I know)

[twin wolf]

So for previous questions when I had cos, sin and tan equations for domain restriction usually 0,2pi or -pi to pi

I was able to rearrange using algebra to start. Then with my overpowered calculator I take the inverse angle and use that as a base

For example sin(pi/3)

Then I find the quadrants for where sin is positive and add or subtract whatever is necessary

If I have an arguement in my equation I can solve for x.

This is the general method I use

But when other special factors like a different restricted domain other than -pi to pi or 0 to 2pi I have to switch things up a bit

[narrow helm]

so what do you need my help for?

[twin wolf]

Hold on I’m getting to the part

Just trying to figure out how to word it

For special cases where the domain accepts negative angles and depending if the trig function is odd or even I have to use reverse quadrants or think in reverse @narrow helm not sure if this makes sense to you

Would this be correct

[narrow helm]

yeah it doesnt make sense to me

[twin wolf]

Okay

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@twin wolf Has your question been resolved?

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[jolly terrace]

,rotate

[zenith axle]

.close

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[fluid prawn]

Help

[onyx glen]

sounds like they want you to do ± that, ignoring the very obvious physical origins of this equation

[fluid prawn]

Idk what that means

[onyx glen]

you forgot $-\sqrt{\frac{GmM}{F}}$

[jolly terrace]

r is distance

[glossy valve]

Ann

[onyx glen]

yes as i said, maybe the problem wants you to ignore that.

[jolly terrace]

i'm just replying to what he said

[fluid prawn]

That’s still wrong

[onyx glen]

what did you input

[fluid prawn]

This

[onyx glen]

take a picture of your input please

[onyx glen]

you need both of them...

[fluid prawn]

Wdym

I type both of them in there?

[onyx glen]

yes

with a comma inbetween

i hope you have at least 4 more attempts

[fluid prawn]

Wtf I ran outta attempts

I didn’t even know there were

[onyx glen]

rip

now we will never know

[full forum]

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[sturdy summit]

A question given for the Conic Section of the Parabola; Latus Rectum Joining (-3,1) and (1,1)

[sturdy summit]

The part that I dont understand is whether the opening would be Left or Right

Even using the Midpoint and Distance formula only helped me determine its Focus and measurement coefficient

[sturdy summit]

But I do not know how to find its opening because of the question given

[gritty rose]

show the original question

[sturdy summit]

,rotate

[sturdy summit]

Its seriously just like that.

[gritty rose]

yea just put both answers

they differ only by a - sign anyway

[sturdy summit]

See??

Both answers final

.close

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[late echo]

do i just write my answer as f(x)=1?

[twilit leaf]

x=0, so f(0)=1

[late echo]

ohh okay thank you

.close

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[barren iris]

I'm not following the remark the author left.

[barren iris]

So, the strips have a height of 2v and not v -w. I understand the height should be v-w but maybe I'm not seeing the 2v?

Is that just saying 2v(x) ?

Oooooo wait

The bludge is the red area

To the left

Why is from 0 to 1 / root 2 better though?

[gritty rose]

where does it say this is better

[barren iris]

The former gives the wrong answer.

[gritty rose]

do you know the height of the bulge?

[barren iris]

So that's what I mean by better

Cause I assume the later gives the right answer

[gritty rose]

where does the author say 0 to 1/sqrt(2) is correct?

[barren iris]

He said the area is pi/8. He doesn't say it specifically. But he doesn't say it's wrong

Oh wait

Is it cause the first question isn't for the whole area between the circle and the line

That must be it

Yeah that's it

.close

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[spark vapor]

I know I is the identity matrix, but what is subscript 4

[spark vapor]

is that just the row?

[eager obsidian]

its generally the number of rows/columns. so you'd have a 4x4 identity matrix here

[eager obsidian]

are these E_13 the elementary matrices btw?

[spark vapor]

yes

[eager obsidian]

because if they are, the dimensions feel weird

unless its zero indexed

[spark vapor]

no, its not 0 indexed

[eager obsidian]

and E_13 is a 3x3 matrix?

just wondering if it should be I_3 on the right side instead

[spark vapor]

all 4x4

[eager obsidian]

ah nvm it's 3 am and I confused myself

all good, idk what I was thinking

[spark vapor]

also in LU factorization, you can get both the upper and lower triangular matrix by the elimination right

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[vapid schooner]

i need help with this problem I have tried working it would but i cant figure it out

[granite finch]

have you tried expanding out the top?

[rough tundra]

first given $f(x)=4x^2+2$ do you know what $f(-1)$ would be?

[glossy valve]

MrFancy

[vapid schooner]

6

[rough tundra]

yep! and what would be $f(h-1)$?

[glossy valve]

MrFancy

[vapid schooner]

-1

[rough tundra]

h is a variable

[vapid schooner]

that is where i get caught up

[rough tundra]

do you know how you got the 6?

[vapid schooner]

plugging -1 into f(x)

[rough tundra]

yep so we do the same thing with h-1, so what does that give? :)

$f(\textcolor{red}{-1})=4(\textcolor{red}{-1})^2+2$

[vapid schooner]

i understand it now

[glossy valve]

MrFancy

[rough tundra]

hope the colors help (just learned how to do them ), so now can you plug in the h-1? :)

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[haughty bear]

Need help on this

I got stuck

[jolly terrace]

multiply with the conjugate

[haughty bear]

I tried

I got stuck

I have square x +4x -64

I think that’s wrong

[jolly terrace]

Can I see your work so far, if you don't mind?

[haughty bear]

I erased it

I multiplied to get rid of the bottom

Then simplified

And I got that

[jolly terrace]

multiplied with what expression

[haughty bear]

The bottom

[jolly terrace]

Can you please be explicit?

[haughty bear]

Square x +4

-4*

[late palm]

yeah and multiply top with that too

[haughty bear]

I did

[late palm]

and what did u get

[haughty bear]

-square x^2 +16 square x + 4x -64

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[haughty bear]

<@&286206848099549185>

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[placid oar]

Very sorry about the image quality. Basically, new to complex analysis, cauchy residue theorem and whatnot. Tried this integral, and the result I get is not the same as wolfram's. So I need help understanding what I did wrong. Don't hesitate to ask me to clarify bits of the image as it's not very good :)

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[sinful canyon]

Can anybody hop in a call with me and help me understand non linear and linear equations

[empty sapphire]

can't hop into a call but a linear equation is literally a line (like a straight line). Non-linear lines are lines with curves. There are some nuances that help more granularly define the difference