help-23

[lean otter]

you are having

[
\lim_{x \to \infty} 12x^{3}(\frac{1}{x} -\sin(\frac{1}{x})
]

[flat frigate]

♡LexQa♡

[lean otter]

?

[iron glacier]

yea

we use l'h rule for this right?

[split ether]

As soon as you turn that into a fraction, yes

(of an indeterminate form)

[iron glacier]

do I just take the derivative in both the top and bottom?

[split ether]

Yeah

[iron glacier]

I ended up with (-1/x^2 - cos(1/x)/x^2)/(-3/x^4)

[split ether]

Notice that it approaches infinity

Oh wait

Hmm multiply top and bottom by x^2 I guess

(1 + cos(1/x))/(3/x^2)

Which approaches infinity

[iron glacier]

isn't it -1?

[split ether]

No

Oh you had -sin(1/x), right

So it's +cos(1/x)/x^2 there then

So it's (cos(1/x) - 1)/(3/x^2)

Another L'hopital I guess

[iron glacier]

oh wait

there was a hint that said let t=1/x and find the limit in terms of t

[split ether]

So it becomes 12(t - sin(t))/t^3 oh I see now

As t approaches 0

Applying L'hopital's rule once yields 12(1 - cos(t))/(3t^2)
Applying it twice yields -2sint/t

Which approaches -2

[iron glacier]

wait how did we get 3t^2 on the bottom?

[split ether]

Derivative of t^3 is 3t^2

[iron glacier]

I thought the bottom was originally 1/x^3?

[split ether]

This

[split ether]

Turns into that

[iron glacier]

im still confused

[split ether]

Forget about dividing top and bottom by anything

Just take the original limit and write it in terms of t

[iron glacier]

so 12x^3(t-sin(t))

[split ether]

Write x^3 interms of t too

If t = 1/x, what's x equal to?

[iron glacier]

1/x^3?

[split ether]

1/t^3

[iron glacier]

ohhh ok

[iron glacier]

how do we know it approaches -2?

[split ether]

sint/t approaches 1 as t approaches 0, that's a famous limit

[iron glacier]

ok that makes sense

[iron glacier]

and how come its -2 and not 2? when you take the derivative of -cos doesnt the negative cancel out?

[split ether]

Oh wait right mb it's +2

[iron glacier]

ok tysm for the help

.close

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[viral current]

can someone please explain the concept of the gram schmidt process

[viral current]

(linear algebra)

[worthy hemlock]

I think 3 brown 1 blue explains that concept well

[river wharf]

true

but with a TLDR, i can say that process which makes the orthonormalization of 3 vectors (ortho meaning right angles and normalization meaning length 1 vector) divides first vector by it's scalar, after for the second vector you find the 90degree projection from the first vector and subscract the remaining part of that vector, and for the third vector you need to find 2 projections 90 degree from the 2 vectors and subscract what is left from that vector @worthy hemlock

i hope that made sense

oops wrong ping

@viral current

[worthy hemlock]

I don't think it has to be 3, I think the minimum is 2 vectors

[river wharf]

yeah it's for 2+ but usually they ask to do it for

3

[worthy hemlock]

Usually but on the practicing side, I think it starts with two and then you do more. Don't wanna confuse the OP when you state 3+ when technically you can do 2

[river wharf]

without the third, he should read only the 2 first sentences then xD

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[lean otter]

I need help with this question and how to solve it as a matrix. I need to get the answer for my solution in row reduced echelon form. I also attached my work, but I know it’s incorrect. Please help.

[orchid patio]

these are basically all the steps, but you have to rewrite this with your augmented matrix which would be the same steps

but they just also have to be applied to your last column

its kinda messy with the fractions so you could do it in other ways without the fractions, but this method will guaranteed you always get the right answer (granted you dont make arithmetic mistakes)

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[glossy flax]

how to start this one?

[orchid patio]

f(g(y)) just means that the output from g(y) is used as the input for f(u)

so the question can be rephrased as f(1/(y^2 +1))

which you can solve by substituting into the equation

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[willow anchor]

Ive been stuck on this for ages

i dont know how to work it out

@worthy hemlock can you help by any chance?

<@&286206848099549185>

[stiff inlet]

,rotate

[stiff inlet]

which question do you need help with?

[willow anchor]

Could you explain question 1

please

[stiff inlet]

do you know the law of cosines?

[willow anchor]

do you mean sohcahtoa?

like h=opp/tan?

[stiff inlet]

sorry, i mean the law of sines

[willow anchor]

i dont think i do

[willow anchor]

okay

[stiff inlet]

so for that left triangle in problem 1,you can figure out the length of the unlabeled side using pythagorean theorem

[willow anchor]

oh would it be 12?

12cm

[stiff inlet]

yeah

[willow anchor]

now what do i do

[stiff inlet]

now you can use SOHCAHTOA

since you have that one angle (25 degrees) and the opposite length (12 cm), you can calculate the hypotenuse

[willow anchor]

so would i do 12/ sin(25)

[stiff inlet]

yep

[willow anchor]

so x is 28.3

[stiff inlet]

don't forget your units

[willow anchor]

Thank you so much!

Youve helped me so much

[stiff inlet]

👍

[willow anchor]

I just had one more question

Im really confused on how to work out angles using SOHCAHTOA

[stiff inlet]

just do the same thing you did in reverse

so if you had 28.3 for the hypotenuse and 12 for the opposite angle, you'd know that sin(x) = 12/28.3

or 28.4 rather

[willow anchor]

Oh

[stiff inlet]

so you know that sin(x) = .42

then whatever calculator you're using should have a inverse-sine button

just make sure it's using degrees and not radians

[willow anchor]

so after i do 12/28.4

in my calc

it give 0.42

then i press iverse sin

[stiff inlet]

yeah inverse-sin(.42)

[willow anchor]

which would give 24.8

Thanks Again!

[stiff inlet]

👍

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[ashen estuary]

It's said that this constraint can't be a posynomial, but I'm not sure why that is

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[ashen estuary]

<@&286206848099549185>

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@ashen estuary Has your question been resolved?

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[dull sequoia]

is i a variable

[cobalt bough]

a+bi form

[dull sequoia]

a+bi?

if i = √-1 then all of these numbers are purely imaginary

there's no real part to any of them

i would 100% evaluate the i^n first

[cobalt bough]

?

[dull sequoia]

can you show what the original question is

[molten herald]

How’s this? 😭

[dull sequoia]

#❓how-to-get-help

[dull sequoia]

what's the question

[cobalt bough]

Simplify

[dull sequoia]

ok

i would deal with the i's first

i^2 = -1

2i^2 = -2

[cobalt bough]

So -3

[dull sequoia]

$(3i^2-i^3)(4i^6+9i^7)\
(-3+i)(-4-9i)$

[flat frigate]

Frosst

[dull sequoia]

that's what i'd do first

[cobalt bough]

Ok

[dull sequoia]

now there aren't many powers and it's much easier to see what's what

[cobalt bough]

Yeah

21+23i

Thanks 🙏

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[mild shoal]

The down force part is confusing me, what am I supposed to do?

[iron plume]

Okay so I’m also taking physics

And I might be completely wrong

But my best guess is this

Don’t mind the tan things at the top

$Fd+mg=Fn$

[flat frigate]

esteban

[mild shoal]

that makes sense so i think thats prob right too

ty

[iron plume]

I just used the downward force as a another force

Like the name doesn’t rlly matter is what I think it’s like another forced applied

[mild shoal]

ye

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[lethal tinsel]

hi

[quasi bison]

this problem statement is incomplete

what's B?

[lethal tinsel]

idk it doesnt say😭

[quasi bison]

are you sure it doesn't say?

again,

this is part c of a bigger problem

if you cannot find anywhere above this that says what B is, then take a pic of the whole worksheet and send it here

[quasi bison]

oh, so now you magically know what B is, and it does say.

are you now able to use this knowledge to find the value of k?

[lethal tinsel]

yes

.close

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[faint cave]

Hello

[dusk iron]

Just Ask

[faint cave]

What is the coefficient in this term:

11x

What is the coefficient in this term:

-13bc

What is the coefficient in this term:

-14m2

[dusk iron]

Coefficient is the number next to the variable ig

[faint cave]

I haven't really tried anything, I fell behind just trying to catch up and I don't have much time

[hallow blade]

coefficient is the number

[dusk iron]

So yea.. Tahts it

[faint cave]

so it would be 11 , 13 , 14?

[dusk iron]

Yup

[dusk iron]

-14

[faint cave]

Also this!

[hallow blade]

is that an online quiz?

[dusk iron]

Just google the definitions

[faint cave]

Practice quiz yeah

[dusk iron]

Basically find the number of terms

[hallow blade]

mono- means one, bi- means two, tri- means three, poly- means four

terms are the stuff seperated by the operations

[dusk iron]

A term is a variable or number or number × variable

[hallow blade]

count the terms and you identify what it is

[dusk iron]

After simplifying the entire expression*

[faint cave]

Sorry , I'm still confused on how to get the answer

[hallow blade]

for example the term "2x", this is one term

but if you had, let's say, "2x^2+5x" then that would be two terms

[dusk iron]

Just searching these topics on khan academy will help

[faint cave]

Okay Thank you

.close

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[dry summit]

hello, how do i get help?

[icy lance]

just ask

people will come along if you post a question

[dry summit]

ill translate in a second

The line 2x + y - 4 = 0 intersects the ordinate axis at a point

[icy lance]

yeah its (0,4)

[dry summit]

Which drawing shows the graph of the line 2x – y + 2= 0?

[icy lance]

lets see, y=2x+2, should be the first one

positive gradient and positive y intercept

[dry summit]

is it the first one?

[icy lance]

yeah

[dry summit]

ty

[remote lodge]

Help pls) translate: Find the domain of the function of y

[dry summit]

The function is defined by the formula y = -3x+2. If x = 4, then the value of the function is

[icy lance]

[-infinity to 5) as if its 5 its undefined and if greater than 5 the denominator becomes complex

[icy lance]

what do you think it is

[dry summit]

i honestly have no idea, i have 0 math in my brain

thats why im here :p

[icy lance]

what you have chosen is correct

just replace x with 4 and calculate

[charred latch]

-10

[dry summit]

okay thanks

i got that answer from ai didnt think it would be correct

Which of the given points lies on the line

[charred latch]

(0, 1)

[dry summit]

tysn

m

Point K (t; 10) lies on the line 2.5x + y = 0 if t is

[charred latch]

-4

[dry summit]

wow u are so fast

i wish i was smart as you

If a cyclist rides steadily at a speed of 15 km/h, then in x hours he will travel y (kilometers)

[charred latch]

y=15x

[charred latch]

you already have answers

haha

[dry summit]

its for just some of them

cuz the ai worked on only the ones that dont have images

Which drawing schematically shows the location of the line -x – y +5 = 0 in the coordinate plane?

[charred latch]

choice 3

[dry summit]

ty

Given the equation of the circle line. Its center is at a point

the (x-3)2 + ... is supposed ot be in the midle of the thext

[charred latch]

equation of circle is (x-h)^2 + (y-k)^2 = r^2

center would be (h,k)

(3,-2)

[dry summit]

tyty

On the circle line

there is a point whose coordinates are

[charred latch]

0,4

[dry summit]

the next ones is geometry and i need to send pictures that ive written

will u be able to do that?

[charred latch]

lets try

[dry summit]

Write the equation of the line that passes through the origin of the coordinates and has the same direction as the line 2x -5y + 5 = 0.

there might be some words that arent like correct in math language if u know whati mean cuz this is from google translate

[charred latch]

2x-5y=0

not sure haha

[dry summit]

i need to write that on paper, can u type how u came up with that cuz the teacher probably wont accept just the answer

its okay

even if its incorrect,that will be better than nothing

Find the line -x – y + 1 = 0 and

coordinate of the intersection point and draw these lines.

could anyone help with this math problem?

😦

[dry summit]

thanks

[charred latch]

https://www.desmos.com/calculator

[dry summit]

is this for the first or second one?

[charred latch]

2nd one

[dry summit]

ok thanks man

the answer is ( 1 , 0 ) ,right?

Given a straight line y = -2x + 5. Write an equation for a straight line parallel to the given straight line and passing through the point M( 1; 4). Draw both lines.

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[lean otter]

im looking for a more detailed explanation on how they derived the expression

[buoyant shadow]

the first term is j+i−j, then j+i−j−1 down to 0

so i to 0

it's not like a transformation they just looked at it and rewrote

i mean the middle part

[safe radish]

@lean otter Has your question been resolved?

[cosmic grove]

$\sum_{k=j}^{i+j} (j+i-k) = \sum_{k=0}^{i} (j+i-(k+j)) = \sum_{k=0}^{i} (i-k)$

[flat frigate]

Ｈｅｒｅｌｓ

[cosmic grove]

$\sum_{k=0}^i (i-k) = \sum_{k=0}^i i - \sum_{k=1}^i k = i(i+1) - \frac{i(i+1)}{2}$

[flat frigate]

Ｈｅｒｅｌｓ

[cosmic grove]

$=\frac{i(i+1)}{2}$

[flat frigate]

Ｈｅｒｅｌｓ

[lean otter]

Thanks guys

👍

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[frank glen]

Find the values that make the function undefined

And you make the function undefined if dividing by 0 is involved

[stark lintel]

Ah I see thank you

[frank glen]

[stark lintel]

Would you initially find that by solving for the domain?

Or is that step unnecessary

[frank glen]

Yep

You first spot the places where the function becomes undefined

Then after you scan them all, you then conclude the domain

[stark lintel]

that was really helpful thank you!

[frank glen]

Glad to help :)

[safe radish]

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[swift heron]

I did jacobian but y and x just don't disappear from the numerator / denominator of the equation

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@swift heron Has your question been resolved?

[plucky elk]

show your work

[lean otter]

express x and y in terms of only u and v

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[swift heron]

.reopen

[safe radish]

✅

[swift heron]

when i equate them to solve for u and v it just gets rid of x or y

[lean otter]

y=uv ?

[swift heron]

yeah

x = u^2/y and y = u^2/x

when i plug in x into y it just gets cancelled

[plucky elk]

you're supposed to express x in terms of just u and v. no y

x = x(u, v) and y=y(u,v) in the jacobian

[swift heron]

that's what I got from deriving the 2 initial conditions of u = sqrt(xy) and v = sqrt(y/x)

[plucky elk]

i realize that. it's not enough though

[swift heron]

so yeah that's when I plugged in x into y to try to solve for the variables u and v

but it just cancels out thats why im rlly confused

[plucky elk]

Solve for y in terms of x and v. Then plug that y back into your top left equation for x

[swift heron]

ohhhh ok I see it's x = u/v now

ty ty

if my jacobian is = 0 does that mean there's no change to the function when changing the variable?

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[arctic flame]

can someone help me simplify this

[deep apex]

get a common denominator then sum the numerators

[arctic flame]

mind doing it on paper

from start to the end?

[plucky elk]

you should do your own work

[arctic flame]

i wanna know how to start

and how to do it

i wanna learn

[twin heron]

go on mathpapa

it has a step by step

😩

[arctic flame]

will i be able to learn it easily?

[twin heron]

prolly

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[twin heron]

$3/4 + 1/2 \implies 3/4 + 2/4$

[flat frigate]

AJ_l0l

[deep apex]

this is not a valid mathematical statement

[arctic flame]

how am i supposed to type it in

[arctic flame]

its paid

[twin heron]

@arctic flame

its free bro

https://www.mathpapa.com/algebra-calculator.html

[twin heron]

it is now

so basically you multiple the left term top and bottom by the deno of the right term and vice versa

simplify the expressions once you combine thee fractions (after doing the above, they will have the same deno)

and then you simplify by cancellling out like terms

[deep apex]

the common denominator is simply 2x+6

I don't want to give away the full solution

[twin heron]

oh yea

$x+3 \in 2x+6$

[deep apex]

that's what the computer would do, it's the mechanical solution

[arctic flame]

not understanding shit

omg

[twin heron]

so basically when you have two fractions

with different denominators

you need to cmobine them by finding a common denominator

[arctic flame]

how do i find a common denominator

[twin heron]

for example $2/5 + 3/7 = 14/35 + 15/35 = 29/35$

[flat frigate]

AJ_l0l

[twin heron]

the mechanical solutions, as assassin put it, is just by multiplying one fraction by the denominator of the other and vice versa

for example, $(2/5) * 7 = 14/35$ and $(3/7) * 5 = 15/35$

[flat frigate]

AJ_l0l

[twin heron]

thats for numbers -- but that still holds for variables and stuff. Like in this case, the denominator of the left is x + 3 and the right is 2x + 6

but notice that $2x + 6 = 2(x+3)$. The right side already has the x+3 term of the left sides denominator Thus, we only need to multiple the left fraction by 2

this is like a small shortcut that pertains to your particular problem

cool?

[flat frigate]

AJ_l0l

[arctic flame]

yupp

[twin heron]

EPIC

[safe radish]

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[vocal spindle]

can somen go step by step because im not sure what to do

[low canyon]

What did you try

[vocal spindle]

i did (x+2)(x-2) to get my denominator of my final fraction

then

[novel inlet]

Write all numerators above the least common denominator

[low canyon]

Can you show me what you tried ?

[vocal spindle]

i did x(x-2)-4(x+2)

[novel inlet]

tale photo of your notes

[vocal spindle]

for my numerator

[vocal spindle]

i cant

im on pc

[low canyon]

So you got $\frac{x(x-2) - 4(x+2)}{(x-2)(x+2)} = 1$, right?

[flat frigate]

Fέliχ

[vocal spindle]

yeah

[low canyon]

What can you do now?

[vocal spindle]

then i canceled the x+2 and x-2

[low canyon]

❌ Wrong!

[vocal spindle]

oh

cuz fo the minus right

of

[low canyon]

You do not have a multiplication at numerator

[vocal spindle]

right

[low canyon]

As you just said

[vocal spindle]

so do i expand then

[low canyon]

Before to do so, notice that you have a fraction that you want equal to 1

[vocal spindle]

yeah

[low canyon]

What does it take in general for a fraction to be equal to 1?

[vocal spindle]

what?

[low canyon]

Ok, I will simplify my question

if a/b = 1, what can you say about a and b?

[vocal spindle]

im not sure

a=b?

[low canyon]

✅

So go back to your problem, what can you say?

[vocal spindle]

i times eveything by (x-2)(x+2)?

[low canyon]

Well you could... but following what we just said

Can you tell me which equation can you deduce

[vocal spindle]

thats what i did for a/b=1

[low canyon]

as you have a fraction equal to 1

Oh ok

[vocal spindle]

i times everything by b

[low canyon]

Then do it!

You'll have
numerator = denominator

[vocal spindle]

so i would get x(x-2)-4(x+2)=(x-2)(x+2)

[low canyon]

Right

[vocal spindle]

then do i expand?

[low canyon]

Yes

[vocal spindle]

so i get x^(2)-6x-8=x^(2)-4

right?

[low canyon]

Yes

[vocal spindle]

.close

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[rocky aspen]

would my reasoning be correct for this q? Proportion of people with agreebleness is 0.21 * 284 of which 0.68 are friendly, so 0.21 * 284 * 0.68 = 40.5552 and similarly openness (cautious) results in 0.26 * 284 *0.27 = 19.9368, so number of employees categorised in these two groups would be the sum of these two? =60.492=61?

[safe radish]

@rocky aspen Has your question been resolved?

[rocky aspen]

<@&286206848099549185>

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[jagged finch]

Can someone explain why ln gives positive and negative value? And why is this the case? Doesnt that only apply to roots?

[plucky elk]

,calc log(0.5)

[flat frigate]

Result:

-0.69314718055995

[plucky elk]

,calc log(2)

[flat frigate]

Result:

0.69314718055995

[plucky elk]

$\log(\frac{1}{x}) = \log(x^{-1}) = - \log(x)$

[flat frigate]

riemann

[plucky elk]

so if log(x) is positive, then log(1/x) will be negative and vice versa

[jagged finch]

Ahh alright

Thankyou very much

Appreciate it ❤️❤️

.close

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[lean otter]

S

[lean otter]

How do we do the lim for arithmetic sequence and geometric

The limit

[misty coral]

to find the limit of an arithmetic sequence it relies on the sequence being convergent, to find the limit of a geometric sequence you can use s->inf = a/(1-r) where r lies between -1<r<1

[lean otter]

What is the a what is the r

@misty coral

[misty coral]

a is the first term in the sequence and r is the common ratio between terms

[lean otter]

So a1/(1-r)

To find the limit

[misty coral]

if -1<r<1 yes

[lean otter]

And if not ?

[misty coral]

the series won't converge

[safe radish]

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[foggy fjord]

someone help a brotha out

[foggy fjord]

<@&286206848099549185>

[twin solar]

"After 15 minutes, feel free to ping helpers", not after 2 minutes

[foggy fjord]

oh my fault

would u like me to delete the message

[twin solar]

no, no worries, youll know it next time

[foggy fjord]

gotcha

[gentle lintel]

At point (-2,2) count 5 units right and one up, the graph is the line connecting these two points

[foggy fjord]

so connect the -2,2 with the 3 -1?

[gentle lintel]

3,3

[foggy fjord]

thought u said go up one

oh wait

[safe radish]

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[worldly bronze]

.help

[safe radish]

Commands:
clopen: .close, .reopen, .solved, .unsolved
consensus: .poll
factoids: .tag
help: .help

Type .help <command name> for more info on a command.

[verbal grove]

.help

[safe radish]

Commands:
clopen: .close, .reopen, .solved, .unsolved
consensus: .poll
factoids: .tag
help: .help

Type .help <command name> for more info on a command.

[verbal grove]

how do i make my own

[verbal grove]

ok

.help

[safe radish]

Commands:
clopen: .close, .reopen, .solved, .unsolved
consensus: .poll
factoids: .tag
help: .help

Type .help <command name> for more info on a command.

[verbal grove]

@warped roost

help me

idk what im doin

[thin bridge]

don't ping mods for help

[junior smelt]

This is your channel now

[verbal grove]

okay

[junior smelt]

has your name on it

[verbal grove]

how do i ask for help with my assignments

[junior smelt]

[verbal grove]

idk wtf im doing in geometry

[thin bridge]

instructions for how to get help are pretty clear and posted in very accessible locations

[verbal grove]

okay okay

butg

how

do i work it

[thin bridge]

work what

[verbal grove]

the bot

[junior smelt]

What do you mean "work the bot"?

[verbal grove]

wait is this server an ai help?

or human

[thin bridge]

no

human

[junior smelt]

Real people

[verbal grove]

wtff my friends said it was ai LMAOO

[thin bridge]

with a bot assistance to help type out math

[static vale]

💀

[verbal grove]

my friends are trolling me

but i do genuinly need help

[static vale]

Bot = fast answer
Human = not as fast as bot but we can explain you how the sum is done.

[thin bridge]

post an actual math question that you're stuck on

[verbal grove]

so like

i mean 6-8

i have

like

no idea what i’m doing

i listen in class

but i have no clue

how to tell which is greater

[static vale]

Here listen

Remember pytho?

In Pythagorean, which side is the largest?

[verbal grove]

c2

[static vale]

Bear with me for a sec

[static vale]

Jesus wtf?

What's c2?

[verbal grove]

C squared

[static vale]

💀

[verbal grove]

💀

[static vale]

Oh ye ya call it hypotenuse

So its because

[verbal grove]

go away

[static vale]

In a triangle, the side opposite to the largest angle is the greatest, and vise versa @verbal grove

[thin bridge]

within a triangle,
longer sides are opposite large angles

[verbal grove]

okay

i understnad that

now how do i pair those two letters together

like

[static vale]

So first

[verbal grove]

#help-23 this is my channel

[static vale]

$$Get \angle ecs out$$

[flat frigate]

anand

[verbal grove]

what

[static vale]

Why no space bruh

[static vale]

Angle ecs bro

[glacial island]

Try taking out outside of the $s

[static vale]

The red one

[verbal grove]

😧

[static vale]

What's the sum of angles of triangle?

[last pilot]

#1021175428326633542

[verbal grove]

the triangle is always 180

[static vale]

Yes

[verbal grove]

is it 180 - 120 - 32

[static vale]

So apply it for triangle CES

Yeee

[last pilot]

wrong wrong go #❓how-to-get-help @patent gazelle

[static vale]

28 ain't it?

[verbal grove]

yeah

[static vale]

So now the triangle is formed

Yk the side opp to the largest angle is largest

[verbal grove]

it would be e rigfht

since e has the 120 degree

[static vale]

So here the side (cs) opposite to angle e is the largest

Yes

But we don't need that 💀

Now which side is 2nd largest?

@verbal grove

[verbal grove]

ce?

[static vale]

Exactly

Smallest?

[verbal grove]

CS

[static vale]

Ya got 1st answer ig

[verbal grove]

okay thanks for that

I have one more

[verbal grove]

EXTREEM question

[static vale]

No 6 done

Ok slide

[verbal grove]

do you know how to do proofs

[static vale]

Ye i probably should I did it when I was in grade 8-9

[static vale]

Now I'm in 12 I should not have forgotten those hellish theorems tho

[verbal grove]

i already know given

[static vale]

Which one?

[verbal grove]

but when i try anything else

its always wrong

skill 13

30

numerb 30

[static vale]

Yee

Oh easy

[verbal grove]

wait rlly

[static vale]

What all all the theorems we can prove it with just remind me

SSS was there ig

And?

[verbal grove]

sss, sas, asa, aas

[static vale]

Ok I see

We can use ASA here

Easily

Side RC common we can see that?

Ya there?

[verbal grove]

yeah

mhm

im listening

[static vale]

Ok so

RC is common

And since RK and 0C are parallel

We can apply alternative interial angleasss

[verbal grove]

ok

[static vale]

Then it says $$\angle KRC = \angle OCR$$

Ya get me?

[verbal grove]

yes

thats the part i gotta prove

which i dont get the steps leading to it

[flat frigate]

anand

[static vale]

Its straight away a property bo

Lemme find it way

Wait*

[verbal grove]

okay

[static vale]

Here

Imagine the two lines parallel here to be kr and oc

And then a line is cutting through em

Compare this property with ur diagram

[verbal grove]

ok

ok i am comparing

i can imagine them being the smae

same

[static vale]

Yee

Then just simply apply alternative internal angles

Angle 4 equal to angle 6

Angle 3 equal to angle 5

[verbal grove]

okayu

but how do i make that into a statment and also put it as a reason

[static vale]

diagonal acts as a transversal and creates alternate interior angles with the parallel sides.

Ya can use this iv

Ig*

[verbal grove]

this upcoming final isgonna be the nail in this stupid fucking coffin 😭

thx for your help

[static vale]

https://caddellprep.com/subjects/common-core-geometry/parallelograms/

@verbal grove

Watch dis

[verbal grove]

okay

brb i need to got o the bathroom

[static vale]

💀

[verbal grove]

k i watched the video, but it still doesnt explain how i gotta give reasons such as "defenition of congruency" or "substitution" or "subtraction"

@static vale

[static vale]

Those are properties

Proven ones

How ya gonna prove proven stuff

[verbal grove]

im supposed to give a statement

then prove that stament

in which i have to use properties

to prove the statement

thats what confuses me

the thing i can hope for on the final ig is that i know everything else and get something higher than an F on it

thx

for the hel

help

you did your best

who am i kidding i’m finna fail it 😭

i’m going to kms

[static vale]

💀

@verbal grove if you're still confused close this

And reopen it

[verbal grove]

why

[static vale]

Then resend ur question

Someone else will help ya

Maybe he can explain it better?

[verbal grove]

you did well

cya

.close

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[lapis cove]

I am trying to double check my answer, I got 612500. Is that correct?

[lean thorn]

hi how did you get that answer? Can you show your work?

[safe radish]

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[lapis cove]

hey im having trouble punching this into my calculator ( ti-83 )

[lapis cove]

I guess im doing something wrong but im not sure what

[stray bluff]

log is referring to the natural log (base e) in this context

So use ln() on your calculator

not log base 10

[lapis cove]

nope

it still wont work

i got it, i just needed to add parantheses

sorry for the trouble

[stray bluff]

👍

[safe radish]

@lapis cove Has your question been resolved?

[safe radish]

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[placid root]

I dunno where to start

I know the erf(x) =(2/sqrtpi)*integral from 0 to x e^-u^2du

In erf(x^2) do i just change the boundary of the integral from 0 to x to 0 to x^2? But the e^u2 also cannot be differentiated

[clear blade]

chain rule and FTC

[static vale]

Erf full form?

[clear blade]

The first part of the FTC tells you that the derivative of $\text{erf}(x)$ is $\frac{2}{\sqrt{\pi}}e^{-x^2}$, then use chain rule

[stray socket]

Fun da mental there um of calc lou luss

Too late bud

[clear blade]

this channel is occupied

[stray socket]

Find a new channel

Just copy and paste and put it onto anotjer

[gritty ember]

Oki doki 🙂

[stray socket]

But delete that

Because it's also just a text wall rn

[clear blade]

rip

[stray socket]

Anyways this

[gritty ember]

oki doki ❤️

[flat frigate]

tushar

[safe radish]

@placid root Has your question been resolved?

[placid root]

Ahh so like this, i forgot about ftc lmao

Sorry but im still a little unclear about erf(x^2) how can the chain rule be used?

[clear blade]

$$\frac{d}{dx}\text{erf}(x^2)= \frac{2}{\pi}e^{-(x^2)^2} \cdot \frac{d}{dx} (x^2)$$

[flat frigate]

tushar

[placid root]

i am so sorry but where did the value for erf(x^2) come from

[safe radish]

@placid root Has your question been resolved?

[placid root]

Would it be like this?

[clear blade]

yes

[flat frigate]

tushar

[placid root]

thanks so much!

.close

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[safe radish]

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[open abyss]

Hello

I've worked out the answers to the question

but it's in a format that the website does not accept

for the question that is missing a response, i got: -98 sixth root of x^5

but there is no root symbol with a space for an exponent

so i think that they want it in a different format

[proper moat]

hi

[open abyss]

hey

[proper moat]

isn't the answer -98x multiplied by sixth root of x^5?

[open abyss]

yeah

[proper moat]

and for the format, I have no idea. I guess you can just put 0.8333 for the power

[open abyss]

well this unit we've been using fractions for the powers

but i don't know how to convert it into a fraction

[icy lance]

you could put -98 * x^11/6

[open abyss]

ah

could you tell me how you got that answer so i can use the same method

[icy lance]

sure you have $(7x^{\frac{3}{2}})(-14x^{\frac{1}{3}})=(7*-14)(x^{\frac{3}{2}+\frac{1}{3}})$

[flat frigate]

Azo

[icy lance]

$=-98x^{\frac{11}{6}}$

[flat frigate]

Azo

[open abyss]

ahh

i see

thank you

[icy lance]

np

[open abyss]

:)

have a nice day

.close

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[uncut cloak]

Attempting this question I’ve done the first part but didn’t know how to start second

[halcyon carbon]

The key observation for this question is that f(n)/n counts the number of 1s in the binary representation of n

[uncut cloak]

How are you meant to observe that

[toxic stratus]

i mean

[uncut cloak]

Yeah I did that for the first part

[toxic stratus]

then you should see that its just counting 1s

[safe radish]

@uncut cloak Has your question been resolved?

[safe radish]

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[plush juniper]

Hi there, I'm doing statistics and trying to integrate the joint pdf of a function which is a double integral.

By looking at the** support **how do I know which random variable is dependant and which is independent?

Thanks!

[velvet nebula]

hmm, the question seems suboptimally phrased

let X be the random variable for x and Y be the random variable for y

are you asking if (X, Y) is independent/dependent

(individual random variables aren't independent/dependent))

[plush juniper]

yes, sorry, I thought showing the support was enough

[velvet nebula]

if they are independent, then what happens?

[plush juniper]

then the joint pdf is just the product of the marginal pdfs

[velvet nebula]

what does the support tell you about the value of the marginals?

[plush juniper]

I suppose, it tells me that the marginal pdf of X can take values between (0, ∞) and same for the marginal pdf of Y.

Hmm, actually instead the marginal of X could take values between (0, y) and the marginal of Y could take values between (x, ∞)

I'm unsure which case I should consider. I guess the following domains (0, y) for marginal pdf of X and (x, ∞) for marginal pdf of Y would be better bounds because they are tighter.

Is this what you mean?

[safe radish]

@plush juniper Has your question been resolved?

[velvet nebula]

consider when x > y

Suppose the marginal pdf of X has support including the open interval (a, b) where 0<a<b. Show that the support of y is constrained in (0, b). Hence show that 0<x<y is not the support of the product

[plush juniper]

Hmm not sure, but maybe y can't be the upper bound of x because it needs to be a constant, so that when we integrate we get a number value?

[velvet nebula]

yeah y isn't a constant, so that's why I'm considering an open interval (a, b)

[plush juniper]

well I guess we can say y is constrained in (0, b) and then take the limit of b to ∞

[safe radish]

@plush juniper Has your question been resolved?

[plush juniper]

Sorry, I'll have to think about this some more. Thanks for the assistance though! 🙂

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[edgy rover]

can someone explain how you find a range of values of x of a function?

[edgy rover]

i got the first part, but im not sure how i would go and find the second

[lapis shadow]

What did u get for the firt

First*

[edgy rover]

i just changed f(g(x)) into 1/root(5-x)

[lapis shadow]

What are the values of x, that when u put in here, the expression makes no sense

[edgy rover]

nono, i mean for part a, i dont know what to do for b

[lapis shadow]

Yeah b is asking u for this

1/root(5-x)

For certain values of x here, u just can't find an answer when u put it there

[edgy rover]

how do i know which values will make the expression not work? as all i know is that x is less than 0 for one function, and x is a whole number for another

[lapis shadow]

What r u typing so much lol

[edgy rover]

i kept thinking then erasing

[lapis shadow]

Oh

[edgy rover]

rereading your statements

[lapis shadow]

What happens when x = 10 for example

[edgy rover]

(im terrible at functions)

okay so

wait i should sub into the solved one right?

[lapis shadow]

What do u get when u put x=10 into that expression

[edgy rover]

1/root(5-10)

[lapis shadow]

YeH into 1/root(5-x)

[lapis shadow]

And simplify the thing in brackets

[edgy rover]

so just, 1/root(-5) but you cant have negative roots

so that means it doesnt work?

[lapis shadow]

Exactly

Sp for 10, it doesn't work

It's not only 10

It's asking u for all those numbers for which it won't work

[edgy rover]

wouldnt it be any number greater than 5?

[lapis shadow]

Yeah

[edgy rover]

Ohhhh

wait that makes sense

so thats why its written out then, x>_5

[lapis shadow]

"Undefined" means something like '"doesn't work"

[edgy rover]

okok i didnt know

ill note that too

Thanks a ton

[lapis shadow]

Yeah

[lapis shadow]

I meant to reply to this

[edgy rover]

?

[lapis shadow]

No I just replied to the wrong msg, forget it, u understood the problem

[edgy rover]

okay, thanks anyway, see ya!

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[eager prawn]

I am looking for a matrix transformation that can help me solve this problem:

I am creating a picture out of a nxn matrix. There are multiple observations that let me reduce the problem to one of the corner block matrices of size n/4 x n/4.
The picture is symmetrical if you mirror it along the y axis centered in the middle of the picture.
The picture is also symmetrical if you mirror it along the x axis centered in the middle of the picture. (Best example, consider a circle centered at (0,0), you only need one quadrant to create the full "circle matrix".

I am looking for mathematical operations that can mirror a matrix along its y axis and x axis. (diagonal quadrants are simply the transposed matrix of the other matrix). Can someone point me into the right direction?

[safe radish]

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[eager prawn]

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[lean otter]

what order do I do the derivative rules in here?

[lean otter]

do I use the product rule then the chain rule?

[worthy terrace]

see

if you wanna use chain rule

itwill be

8x(4x+3)^4(14x+3)

[lean otter]

okay

but I dont use the product rule?

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[deft cliff]

can someone help me solve this

[deft cliff]

6+3(x+4)=24 my notebook has so many answers i can’t tell which one is the correct one

,rotate

[atomic geode]

@deft cliff first multiply the 3 with the bracket

[deft cliff]

Yeah

let me show you what i did

i stopped because i saw a different answer

[atomic geode]

ur doing it right

[deft cliff]

does x = 2

[atomic geode]

yes

[deft cliff]

i found another answer in the notebook that’s why i wanted to check

also bodmas/ pedmas or whatever u grew up with, they apply in linear equations right?

nvm i googled it

thank you @atomic geode !

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[lean otter]

Real quick, but I forgot why the antiderivative that give arcsinx is $\arcsin(\frac{x}{a}) + C$ and the anti derivative that gives arctan is $\frac{1}{a} \cdot \arctan(\frac{x}{a}) + C$

[flat frigate]

♡LexQa♡

[lean otter]

As in why the extra 1/a product

[junior smelt]

Is that for the antiderivative of (e.g. for $\arctan$, $\frac{1}{x^2 + a^2}$) you mean?

[flat frigate]

chartbit

[lean otter]

Yep

[
\int \frac{1}{x^2 + a^2} \ddx= \frac{1}{a} \cdot \arctan(\frac{x}{a}) + C
]
[
\int \frac{1}{\sqrt{a^{2}-x^{2}} \ddx = \arcsin(\frac{x}{a}) + C
]

[flat frigate]

♡LexQa♡
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lean otter]

Oh well whatever I'm too tired to find out what's wrong with this

[junior smelt]

Literally did this a few mins ago 😂

The bottom you'd factor out an $a^2$, then it turns into the integral $\frac{1}{a} \int \frac{1}{1+u^2} dx$

[flat frigate]

chartbit

[lean otter]

Ohhh

Rightt

[junior smelt]

Mind you, I never remember any of the integrals anyway, when I was doing that q I had to derive everything from scratch 😂

[lean otter]

Haha

Oh yeah

Do you remember the derivation of like

[
(\arcsin(x))' = \frac{1}{\sqrt{a^{2}-x^{2}}}
]

[flat frigate]

♡LexQa♡

[junior smelt]

I guess implicit differentiation? The way I learned was ``let $y = \arcsin(x)$, then $x = \sin(y)$, and $\frac{dx}{dy} = \ldots$"

[flat frigate]

chartbit

[lean otter]

Oh woww I forgot about that

[junior smelt]

Wait why the a here though?

[lean otter]

Although I also never understand the inverse function derivative thing

[lean otter]

Oh whoops

[junior smelt]

Which one, that formula thingy you derive from $f(f^{-1}(x)) = x$?

[flat frigate]

chartbit

[junior smelt]

Or the other way round idk