help-26

[languid mulch]

forgot the final answer

[keen matrix]

well yea Mellow just kept it as such cause I think he was showing conversion

getting a common denominator

[topaz sinew]

@elfin stone Has your question been resolved?

[topaz sinew]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[muted crow]

Is this correct?

[rigid ivy]

did you get your last one correct?

[muted crow]

it dosnt tell me until I submit all of em

[rigid ivy]

What you had selected for the last one did not look correct, btw

[muted crow]

do you wanna do that one first or this

[rigid ivy]

first one

[keen matrix]

select option E "just measure it" but there I'd suggest you use the trig area formulas

[muted crow]

can you help me through it

I think the answer is b now

and 11.4

so b and c

is that correct now

[keen matrix]

how did you get those?

[muted crow]

using sas area formula

[keen matrix]

$A=\frac{1}{2]ab\sin(C)$

[muted crow]

bruh

I think its right

[keen matrix]

oh

i made a sqaure backet

instead of a curly one

mb mb

[rigid ivy]

None of the answers are possible

[muted crow]

how

[keen matrix]

$A=\frac{1}{2}ab\sin(C)$?

[thorny flame]

XxMrFancyu2xX

[muted crow]

yes

and for a I put 7.2 and 5.8

7.2 gave me 9.2

and 5.8 11.4

[keen matrix]

well then ig those are your two areas

[muted crow]

ok

[muted crow]

@keen matrix

how about this one

did I do it right I just guessed idk how to do it rlly

[keen matrix]

We have $\cos(C)=\frac{a^2+b^2-c^2}{2ab}$ and $\sin(C)=\sqrt{1-\cos(C)}$, yes?

[thorny flame]

XxMrFancyu2xX

[keen matrix]

you understand how to get those?

[muted crow]

yes

[keen matrix]

how did you get the second equality?

[muted crow]

using pythag theorm

[keen matrix]

ok good I was jut testing you lol

I just wanted to make sure you weren't just saying yes

[muted crow]

ok

how about 3-5

[keen matrix]

Hence $\sin(C)=\sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}$

[muted crow]

I got that but I dont know how to do what 3 is asking

[rigid ivy]

None of the options make a triangle that correctly match the problem

[keen matrix]

I forgot to square it

[muted crow]

I already submitted it too late

[muted crow]

ok so what would 3 be

[thorny flame]

XxMrFancyu2xX

[rigid ivy]

Note how none of these make an area of 30, or an angle of 65 anywhere

[muted crow]

so what do I do for 3

or did I do it right

[rigid ivy]

@muted crow what is the name of the website where you are getting these problems? Is the company making these problems, or your teacher?

[muted crow]

albert.io

im not sure

he just assigns it

I dont think he would spend time making these

its just this dumb site

can you guys tell me what I did wrong

is 3-5 good?

[rigid ivy]

You did nothing wrong

albert.io is a scam site

[muted crow]

exactly

[rigid ivy]

You can't get these problems right

[muted crow]

@rigid ivy but for this

did I do something wrong

.close

[topaz sinew]

Channel closed

Closed by @muted crow

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[sand mirage]

how are they getting the red domain

this is graphing inverses and other stuff

[grim jacinth]

range of f = domain of f inverse

[sand mirage]

wait what

is going on

i am so confused she ella

shenella

ohhh nvm

how did they get the range of the original

[grim jacinth]

Idk, I can't see that

[sand mirage]

oh bet

Ty

.coose

[radiant tapir]

.close

[topaz sinew]

Channel closed

Closed by @radiant tapir

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[night flax]

I need help

[loud dawn]

Have you sketched it? @night flax

[night flax]

nope

[loud dawn]

Do you know how to sketch it?

[night flax]

u mean sketching the boundary

integral boundary

[loud dawn]

Yeah

x goes from y^2 to 1

[night flax]

after I sketch the graph

what do I do next

https://www.youtube.com/watch?v=NETmfwOAKpQ

I found a video

Patrick Jmt clutched it

😎

[topaz sinew]

@night flax Has your question been resolved?

[topaz sinew]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[drifting wigeon]

Need help with this problem I have work done of how i set up and what equations

and I don’t know what I did wrong

[silk pewter]

section formula

[drifting wigeon]

here’s the formula i used

[silk pewter]

wtf is that

[drifting wigeon]

and here’s my work

my teacher didn’t go over this part in the lesson and says to use to that formula

so i got (1.5,3.5)

[silk pewter]

$(x,y) = \left( \frac{mx_2 + nx_1}{m+n} , \frac{my_2 + ny_1}{m+n} \right)$

[thorny flame]

bettim

[drifting wigeon]

i’ve never seen that equation

[silk pewter]

wtf

[drifting wigeon]

what does n stand for

[silk pewter]

then idk

[silk pewter]

m and n are ratios

1 : 2 means m=1 and n=2

[drifting wigeon]

yeah i haven’t seen that the ss i sent is the equation it says we’re supposed to use but i don’t know

cause my teacher didn’t go over it in class

[silk pewter]

i dont know then

i know only section formula

[drifting wigeon]

thank you for trying then

.close

[topaz sinew]

Channel closed

Closed by @drifting wigeon

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[carmine elbow]

why would it be (1/3)(sin(3x)) instead of 3sin(3x)

the chain rule is f'(g) * g' right ?

[woeful drift]

for derivatives, yes

[carmine elbow]

OH

nvm i forgot that im integrating

.close

[topaz sinew]

Channel closed

Closed by @carmine elbow

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[shy oar]

Why is the -4 in ) but 7 in[ in this case?

[drifting swift]

does -4 belong to the domain?

[shy oar]

I know that X=7 for the numerator and x=-4 in the denominator, but how does that relate to the domain again?

[drifting swift]

sounds like some kind of misconception or overfocus on computation on your part

can you describe in your own words what the domain of a function is, in general?

[shy oar]

The set of numbers that the function reaches in the x line in a graph.

[drifting swift]

the set of numbers that are valid inputs to the function.

[junior pumice]

what ann is trying to say is that you're forgetting this is a composite function, the outer function being square root. which has a certain domain that would involve the numerator in this case.

[shy oar]

So since it’s in a square root the numerator is also apart of the domain?

[junior pumice]

what is the domain of square root

[unique seal]

sqrt doesnt have anything to do with this question specifically
-4 isnt part of the domain because at exactly -4 the function is undefined

division by zero

[junior pumice]

i suggest re-reading what they are asking and looking at the domain of this function

square root's domain. that is the answer to the reason why the domain is the way it is.

yes we can't divide by zero, that's the first step.

what else can't we do with square root

look at a graph of sqrt(x) if you have to.

[shy oar]

[o,inf.)

[junior pumice]

right, so in this case, how do we make sure that the value under the square root stays within that interval?

same thing we did to make sure we don't divide by zero.

[shy oar]

Include 7 so [7,inf) since 0/n is still valid?

[junior pumice]

so to make sure we didn't divide by zero, you took the denominator and set it = 0 right?

[shy oar]

Yes

[junior pumice]

so now we take the other piece of the puzzle, the numerator... make sure it's not what?

think of the domain of square root

what is outside the domain

0 separates the number line between what kind of numbers

[shy oar]

Negative and positive

[junior pumice]

right, so what condition must the numerator satisfy here? what does it have to be? think of an equality.

[shy oar]

Does it have to be positive since there can’t be a negative in a square root

[junior pumice]

can't be a negative, right

[shy oar]

Yeah

[junior pumice]

so what equality should we set up

to make sure the numerator stays positive

[shy oar]

Or 0 right

[junior pumice]

we're thinking in equalities here. less than, greater than or equal to

[shy oar]

Greater or equal to 0

[junior pumice]

perfect. so what does the equality look like

? >= 0

[shy oar]

Inf

[junior pumice]

no. remember how we used algebra to make sure we didn't divide by zero?

we can use algebra on inequalities just like equations

[shy oar]

7

[junior pumice]

no. we are concerned with the numerator here aren't we?

this is very similar to making sure we don't divide by zero, but instead we use an equality

whats the numerator

[shy oar]

X-7

Sqr x-7

[junior pumice]

we are dealing with what we input into the square root, not the square root itself.

? >= 0

[shy oar]

X

X>=0

[junior pumice]

use the whole numerator, because that's what determines whether what we give to the square root will be negative or not.

x - 7 >= 0 you can solve this for x. and this gives you the other piece of the domain. then you combine the domains of the denominator and numerator and that is the answer they are showing you.

whatever x may be must satisfy both conditions: no division by zero and no negative numbers.

these are the restrictions put on the function by division and square roots.

i suggest studying more on what exactly a function is. its something that takes an input and produces an output. sometimes a function can only take certain inputs, restricting its domain.

and it can even have a limit to what it outputs, which is its range.

study these terms some more and you'll get it.

[shy oar]

Ok thank you

.close

[topaz sinew]

Channel closed

Closed by @shy oar

Use .reopen if this was a mistake.

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[west isle]

Hi I need help with this question

[west isle]

I have worked out the 1st and 2nd derivative

and found the critical point

but I am a bit stuck as what to do from there

[rain radish]

Could you send your working

[west isle]

yes

[rain radish]

Are you aware of the condition of critical point in the second derivative to identify the minima and maxima

[west isle]

yes

so can I just sub in my critical points into the second derivative?

and that will give me the sign as to if it is a max or min

[elfin stone]

tysm for your help, my wifi turned of so i coudnt reply but i got the question right!

[west isle]

ok so i figured out which is max and which is min

[rain radish]

Now, put them into the first equation and find the values of a

[west isle]

will do 👍

[topaz sinew]

@west isle Has your question been resolved?

[west isle]

Thank you I think I got it

[topaz sinew]

Channel closed

Closed by @west isle

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[glad lion]

so i need help

[glad lion]

how would i convert this to general form

to input inside

<@&286206848099549185>

[sage anchor]

we should fins x value??

[glad lion]

no no xD

convert this equation to general form

ax^2+by+c=0

you know?

[keen matrix]

you can multiply $x$ to both sides of the equation and keep the equality:

$4x-\frac{3}{x}=2\implies x\cdot(4x-\frac{3}{x})=2\cdot x$

[sage anchor]

easy question

[thorny flame]

XxMrFancyu2xX

[sage anchor]

4x^2-2x-3=0

[glad lion]

yes

so it will cancel out right

[topaz sinew]

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

[sage anchor]

sry sry

[glad lion]

so

[keen matrix]

nah ur good just not supposed to lmao

[glad lion]

4x-2x-3x=0

ah ok i see

i do need help w this too

if possible

[keen matrix]

this is wrong

[glad lion]

mb i forgot the ^2

[keen matrix]

still wrong

$4x-\frac{3}{x}=2\implies x\cdot\left(4x-\frac{3}{x}\right)=2\cdot x$

[thorny flame]

XxMrFancyu2xX

[keen matrix]

Here we multiply x to both sides of the equation

this keeps the equality, yes?

[glad lion]

yes because 4x-3=2x

[keen matrix]

not because that but because we can divide $x$ out of both sides of the equation

[thorny flame]

XxMrFancyu2xX

[keen matrix]

albeit that result is still incorrect

we multiply x to both 4x and 3/x

what is 4x timex x?

[glad lion]

4x^2 ofc

[keen matrix]

yes

and what about x times 3/x

[glad lion]

cancels out

so -3

[keen matrix]

yes

so we have $4x^2-3=2x$

[thorny flame]

XxMrFancyu2xX

[keen matrix]

now we can get that into the form ax^2+bx+c=0

[glad lion]

ye so 4x^2-2x-3=0

[keen matrix]

yes

I agree with that

[glad lion]

oh so all this time you wanted me to put the zero lmaaoooo

[keen matrix]

well also the x^2 you missed that

like

everytime lol

[glad lion]

lmao so sorry i forgot

i subconsciously wrote it

[keen matrix]

also as for this, do you now how to complete the square?

or do you know what completing the square means?

[glad lion]

i mean i did ques 1 and i did something

[glad lion]

yes

so i moved 7 to the other side

[keen matrix]

Show me your work or logic for q1a

[glad lion]

-x+5x=3

then

-x+5x+(5/2)^2=3+(5/2)^2

[keen matrix]

you start out with the quadratic $x^2+6x+8=0$ where did $-x+5x=3$ come from?

[thorny flame]

XxMrFancyu2xX

[glad lion]

i mean the first

one

im not sure of my ans

for 1

[keen matrix]

ohh 1b

ok

that is $-x^2+5x=3$

[thorny flame]

XxMrFancyu2xX

[glad lion]

yes

[keen matrix]

ok im sorry for interrupting

proceed if you want

[glad lion]

i already did

-x+5x+(5/2)^2=3+(5/2)^2

[keen matrix]

are you stuck there?

[glad lion]

sorta yea

i mean i did expand it but idk if im right

[keen matrix]

of so your first issue

[thorny flame]

XxMrFancyu2xX

XxMrFancyu2xX

XxMrFancyu2xX

[glad lion]

yes

so we dont need to divide it

[keen matrix]

yes

[thorny flame]

XxMrFancyu2xX

[glad lion]

ah

why so?

[keen matrix]

^

[glad lion]

ah i see

[keen matrix]

ok so how do you do that in your quadratic: $-x^2+5x-3=0$?

[thorny flame]

XxMrFancyu2xX

[glad lion]

x^2-5x+3+0

[keen matrix]

= but yes

i agree with that answer

ok now we're all good to proceed, do you know how to complete the square, what's the first step?

[glad lion]

-5/2)^2?

[keen matrix]

in your case yes

but there's another step that must come first

[glad lion]

which is?you mean the factorisation thing?

[keen matrix]

move the 3 over

[glad lion]

ah yes

x^2-5x=-3

then x^2-5x+(-5/2)^2=-3+(-5/2)^2

[keen matrix]

yes

[glad lion]

x^2-5x-25/4=3-25/4

[keen matrix]

yes

now do you know how to do $3-\frac{25}{4}$

[thorny flame]

XxMrFancyu2xX

[glad lion]

12/4-25/4

so 13/4

-13/4

[keen matrix]

yes I agree

[glad lion]

so how do i procceed?

[keen matrix]

now do you remember the perfect square trinomial?

[glad lion]

(x-2)^2= ax+2ax+b?

[keen matrix]

not quite

If I have a quadratic $x^2-2\cdot\frac{5}{2}\cdot x+\left(\frac{5}{2}\right)^2$

[thorny flame]

XxMrFancyu2xX

[keen matrix]

what do you notice about it?

[glad lion]

x+b/2^2-b/2^2+c=0

[keen matrix]

no, we can write it as $\left(x-\frac{5}{2}\right)^2$

[thorny flame]

XxMrFancyu2xX

[keen matrix]

do you see how?

[glad lion]

ye bc factorise

[keen matrix]

ok now solve $\left(x-\frac{5}{2}\right)^2=-\frac{13}{4}$ and we have the answer!

[thorny flame]

XxMrFancyu2xX

[glad lion]

so uhh we expand

[keen matrix]

no

you just factorized

expanding would just make it hard again

[glad lion]

oh

then how

so sqrt it?

and then x-5/2 = +-sqrt-13/4

[keen matrix]

yes

and now it's just a linear equation

[glad lion]

4.303 and 0.697

[keen matrix]

❌

$x-\frac{5}{2}=\pm\sqrt{-\frac{13}{4}}$

[thorny flame]

XxMrFancyu2xX

[glad lion]

yea

[keen matrix]

you have this right?

[glad lion]

thats the answer

for positive and negative

[keen matrix]

not entirely

[glad lion]

oh

[keen matrix]

$x-\frac{5}{2}=\pm\frac{\sqrt{-13}}{2}$

[thorny flame]

XxMrFancyu2xX

[glad lion]

how do you type that omg

[keen matrix]

practice

and failing like 87 times

but practice mostly

[glad lion]

the number of brackets and symbols

that alone gained my respect lmao

[keen matrix]

ohhh just wait till you see matrices in latex 💀

$\begin{vmatrix}\frac{\partial}{\partial\rho}\left[\rho\cos\theta\right] & \frac{\partial}{\partial\theta}\left[\rho\cos\theta\right]\ \frac{\partial}{\partial\rho}\left[\rho\sin\theta\right] & \frac{\partial}{\partial\theta}\left[\rho\sin\theta\right]\end{vmatrix}=\begin{vmatrix}\cos\theta & -\rho\sin\theta \ \sin\theta & \rho\cos\theta\end{vmatrix}$

[thorny flame]

XxMrFancyu2xX

[glad lion]

how

are you guys capable of this wizardy

[keen matrix]

LaTeX is lots of patterns

try typing $x^2$ in $\LaTeX$

[thorny flame]

XxMrFancyu2xX

[glad lion]

$x^2$

[thorny flame]

Adonis

[keen matrix]

now x cubed?

[glad lion]

$x^3$

[thorny flame]

Adonis

[keen matrix]

yes

no what about x^10?

[glad lion]

$x^10$

[thorny flame]

Adonis

[glad lion]

heh

[keen matrix]

see

doesn't work

do you know how that could be fixed?

[glad lion]

nope?

[keen matrix]

in latex the curly braces are king

$x^{10}\neq x^10$

[thorny flame]

XxMrFancyu2xX

[keen matrix]

if you don't have curly braces the LaTeX compilier will only look at the first character and raise that

however if the curly braces are there it will raise everything in the curly braces

so try writing x^(abcdefgh) in LaTeX @glad lion ? (sorry for ping)

[glad lion]

ahh so

syntax

but why this kinda syntax

x^(10)

[keen matrix]

well thats how we write it out on text

but how do you write that in LaTeX?

[glad lion]

x^{10}

[keen matrix]

don't forget your dollar signs!!

dollar signs indicate when the LaTeX compiler should enter "math mode" and when to stop

[glad lion]

why that choice of syntax

[keen matrix]

because besides for money when do you use the dollar sign?

[glad lion]

fair enough

[keen matrix]

so what's x^(10) in latex form?

[glad lion]

x^${10}$

[thorny flame]

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

[glad lion]

x^${10}$

[thorny flame]

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

[keen matrix]

the entire thing should have dollar signs

[glad lion]

$x^${10}$

[thorny flame]

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

[keen matrix]

you have three there

[glad lion]

x^{10}$

[keen matrix]

just at the beginning and end

[glad lion]

$x^{10}$

[thorny flame]

Adonis

[keen matrix]

yes

now what about x raised to the 37th power? (written in LaTeX)

[glad lion]

$x^2{10}$

[thorny flame]

Adonis

[glad lion]

$x^{37}$

[thorny flame]

Adonis

[keen matrix]

yes

very nice!

in latex fractions can be written as: \frac{a}{b} and yield $\frac{a}{b}$

[thorny flame]

XxMrFancyu2xX
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[glad lion]

this is like learning js for the first time

[topaz sinew]

@glad lion Has your question been resolved?

[topaz sinew]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[hazy saffron]

How do I verify this? I got stuck when I substitute it to cos and sin. I started at the left hand side

[hazy saffron]

So, what I did is to convert csc to 1/sin and cot to cos/sin. And in denominator, tan to sin/cos.

[spare smelt]

,w cot(x)csc(x) - (csc(x)+cot(x))/(tan(x)+sin(x))

[thorny flame]

Results provided by WolframAlpha

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

Requested by Mr. Gamer#6658

[hazy saffron]

And in there is where I don’t know what to do.

[spare smelt]

alright so we know it's true.

[hazy saffron]

Yep

[spare smelt]

ok why don't we try writing in terms of sines and cosines

(1/sin + cos/sin) / (sin/cos + sin)

multiply the top and bottom through by sin

[hazy saffron]

I don’t know if mine is correct. I got in the top is 1+cos and below is sin^2/ cos plus sin^2

[spare smelt]

yep

so you have

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

now multiply the top and bottom through by ?

(what will allow you to clear the fraction?)

[hazy saffron]

So, I foil?

[spare smelt]

no

read again

[hazy saffron]

I multiply both below and top by cos?

[topaz sinew]

@hazy saffron Has your question been resolved?

Channel closed

Closed by @hazy saffron

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[thick adder]

I have calculated the skaters velocity using $\sqrt(2\times9.8\times4)$, this will give me the velocity that the skater will have at point O. I have then calculated the loss of GPE that the skater will experience past O, using 4cos(30) to give the small change in height. I am unsure how to find the anser though...

[topaz sinew]

Please don't occupy multiple help channels.

[thorny flame]

totalminer

[topaz sinew]

Please stick to your channel.

[empty sail]

Don't open multiple channels

[thick adder]

There 2 different questions

They're

[empty sail]

Even so, you don't open multiple channels

[thick adder]

I will close the other one and come back to it later.

[topaz sinew]

@thick adder Has your question been resolved?

[topaz sinew]

@thick adder Has your question been resolved?

[full ravine]

its just pendulum motion

[thick adder]

but i havent studied that in this chapter... ? it wants me to use energy equation

[neon iron]

Idk man

I also skipped this chapter

Harmonic motion

@thick adder

[thick adder]

I DIDNT SKIP IT I HAVE WORKED THROUGH EVERY CHAPTER CHRONOLIGICALLY

@neon iron i didnt skip it :(((

[neon iron]

Lol dude same well kinda , i regret skipping

I eventually have to DO IT

Wait is there a

discord.gg/physics

Oh damnnnn

I DIDN'T KNEW THAT EXISTED

@thick adder

Go there

I'll join too

[thick adder]

i have to verify my phone number tho....

[neon iron]

Its literally 1:1 to maths server

Exactly same

Just channels are different

Kinda dead too

[thick adder]

I DIDNT SKIP NOTHING >...... ::(((

ITS ON THE LATER CHAPTER

I CAN SHOW U MY TEXT BOOK

u mus be able to solve using mgh ??

[neon iron]

velocity will be calculated by formula root2gr-rcostheta

[neon iron]

All I know is

Somehow

Curve is same as

A straight free fall

If smooth

Frictionless

And all it does is

Change its velocity from vertical to horizontal

@thick adder so yes u should be able to apply mgh till the 90° arc

But that extra 30°

I'm not sure how

To apply it

Probably will have

2 different components

[neon iron]

Lol u are ahead of ur current topic , and here i literally skipped it

[thick adder]

i have only done projectile motion

not only

[neon iron]

Circular motion??

[thick adder]

but only thing i think is similar

no not yet thats on literal next chapter

after momentum chapter

[neon iron]

U should try vertical circle questions to understand this one better

Vertical circular motion i mean

[neon iron]

I gave you the direct formula

root 2g r-rcostheta

here theta is 30

can u

Theoretically explain

oh ohk

How it measures

That extra 30°

Ik till 90°

like the change in height is r-rcostheta

The first 90

[neon iron]

Hmm i see

So it dosent matter what angle it is

Wait it does

From the formula

yh

But generally from mgh

We only need to know the height

Cant we use trignometry somehow

wait let me recheck my formula I actually rote while preparing for jee

let me verify it

To find that end Hight

[neon iron]

Ok

hey it will be just rcostheta

you just need to see the chnge in height

between two points

which is rcostheta by geometry

[thick adder]

I FOUDN IT

before u saided that

I calclauted mgh using 60g4

[neon iron]

jee aspirant?

or ap student

[thick adder]

then calculated mgh using 60 * g * (4-4cos(30))

[neon iron]

ahh yess lol

[thick adder]

took eq 2 from eq 1 an applied ke

[neon iron]

I just rote that formula

lol

for fast calc

[thick adder]

i normaly dont like telling my story cus it is embarrasing

[neon iron]

not embarassing

[thick adder]

i dropped out of college

[full ravine]

now i know your age

[thick adder]

oh nooooo.....

i doxxed myself

cheers for help

.close

[topaz sinew]

Channel closed

Closed by @thick adder

Use .reopen if this was a mistake.

[neon iron]

I am 17 too

hehe

[thick adder]

dont drop out

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[copper delta]

Try pulling out all common factors first

[elder glacier]

it wants me to factorise and shorten it as much as possible

[copper delta]

Yeah so (15a+3b) can be written as 3(5a+b)

Eventually there will be terms that cancel out and give you your answer

[elder glacier]

@copper delta

,rotate

[elder glacier]

Is this correct?

[copper delta]

Yup, is it one of the answers?

[topaz sinew]

@elder glacier Has your question been resolved?

[elder glacier]

yeah

can you help me with this one tho?

@copper delta

[copper delta]

Its a similar question, for example 4xy+8x^2y^2 could be written as 4xy(1+2xy) since 4xy is a common factor

[neon iron]

take a=1 b=1 see which option matches

thug lyf

[neon iron]

take x=1y=2see which option matches

[elder glacier]

6 can not be written with 4 though

do i change the factor?

[topaz sinew]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[lyric spruce]

Hii

[silver edge]

How do I solve this with a coordinate system

for the first one ive got y= -1/6x +1/3

but if I put put 0 on the coordinate system my y is 1/3

[grim jacinth]

Solve for x or y in one equation. Plug that expression into the other equation. That new equation will have just one variable in it you can solve for

[silver edge]

it needs to be graphically

and my coordinate system is a bit messed up

[grim jacinth]

Then plot both graphs

[restive inlet]

messed up how?

use a better scale for more precision

[silver edge]

I didn't knew the term
it was a table of values

like if I put 0 on x my y would be 1/3
but we can't use decimal numbers

[restive inlet]

there are values of x that'll give integer values for y

combining
-x/6 +1/3 into a single fraction may make them easier to identify

[silver edge]

-x/6 +2/6?

[restive inlet]

denominators are the same
now combine the fractions

[silver edge]

-1x + 2?
don't know what you mean with combine

[restive inlet]

don't overthink

write as a single fraction

instead of a sum/difference of two fractions

[silver edge]

ah ok

[silver edge]

can you help me to do the table of values

I really don't know how to do it (i mean I know but the fraction is just struggling me

[restive inlet]

did you combine the fractions

[silver edge]

I did

[restive inlet]

what's the combined fraction

[silver edge]

-1x +2 /6

im probably wrong

[restive inlet]

missing ()

to indicate the numerator

the 1 next to the x isn't necessary either

y = (-x + 2)/6

[silver edge]

oh so If I combine a fraction with the same numerator

it needs ()?

[restive inlet]

writing on paper assuming you write it out properly like
$$\frac{-x +2}{6}$$
parentheses aren't needed. \
on text however it's different

[thorny flame]

ℝamonov

[restive inlet]

writing \verb|-1x +2/6| following the order of operations will be interpreted as
$$-1x + \frac 26$$

[thorny flame]

ℝamonov

[restive inlet]

anyway continuing

y = (-x + 2)/6
this will be an integer if the numerator is a multiple of 6 (which are divisible by 6)
so try finding such values

[silver edge]

-1x .6 and the 2.6
6x + 12 ?

[restive inlet]

what

no

that's far from what I said

[silver edge]

I thought I have to get rid of the denominator

[restive inlet]

did I say get rid of denominator anywhere?

[silver edge]

no

[restive inlet]

do you agree that
(multiple of 6)/6 will be an integer

[silver edge]

yes

[restive inlet]

so if -x+2 is a multiple of 6
then
(-x+2)/6 will be an integer
giving you an integer value for y

so try finding values of x where
-x+2 is a multiple of 6

you could pick a multiple of 6,
set -x+2 equal to that
then solve for x

[silver edge]

-6

[restive inlet]

no

I didn't say replace x itself with a multiple of 6

[silver edge]

-x/6

[restive inlet]

no

I didn't say wipe the 2 from existence either

[silver edge]

-x/6 + 1/3

[restive inlet]

it's like you're not reading what I'm saying

now you're going backwards

do you agree that
(multiple of 6)/6 will be an integer

you agreed to that right

[silver edge]

yes I do it's just understaning in another language is a bit hard

[restive inlet]

following that, do you have an issue with

so if -x+2 is a multiple of 6
then
(-x+2)/6 will be an integer
giving you an integer value for y

[silver edge]

-x +2

I have a feeling its wrong again 💀

[restive inlet]

that was a yes/no question

[silver edge]

oh

I think its yes

[restive inlet]

yes you agree or yes you have an issue/problem with what i said

[silver edge]

I agree

[restive inlet]

so try finding values of x where
-x+2 is a multiple of 6

and to do that you can

follow these steps one at a time

you could pick a multiple of 6,

[silver edge]

can I just pick 6?

[restive inlet]

yes, 6 is a multiple of 6

now set -x+2 equal to what you just chose

i.e.
write the equation
-x+2 = 6

then solve for x

[silver edge]

-2

[restive inlet]

no

Huw are you getting -2 for x

[silver edge]

-4

[restive inlet]

that's better

and this is one of the values of x where y is an integer

to get others, choose another multiple of 6, then same idea

[silver edge]

so this is the easiest way?

[silver edge]

like I haven't seen to do this

[restive inlet]

this idea would be applied in some way

[topaz sinew]

@lyric spruce Has your question been resolved?

[silver edge]

wait one more question
How do I have to write it down?
Do I have to write -x+6 =6 separately?

[restive inlet]

why are you doing
-x+6 = 6

[silver edge]

you said to write down this equation
im confused where to put y or something

[restive inlet]

solving this equation gives you the value of x,

the y value will be (whatever you chose)/6

as
y = (multiple of 6 you chose)/6 = (-x+2)/6

[silver edge]

how would it look like?

[restive inlet]

how would what look like

continuing from
y = (-x+2)/6

when -x + 2 = 6
x= -4,
y = 6/6 = 1

[topaz sinew]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[restive inlet]

. reopen

.reopen

[topaz sinew]

✅

[restive inlet]

just realised you aren't the official owner of this channel

[silver edge]

I was 1 minute later and he just said hi

[topaz sinew]

@lyric spruce Has your question been resolved?

[topaz sinew]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

Channel closed

Closed due to the original message being deleted

[gilded shoal]

.close

[topaz sinew]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[smoky pasture]

hey simple question, im curious how you can find the normal vector using gradients to the function -3x+4y

i was taught we have to put the function into 1 higher dimension, then take the gradient and that is our normal vector, ie -3x+4y=z then take gradient

what i dont understand is how you actually take the gradient, taking it like i would normal, id get <f_x,f_y,f_z>=<-3,4,1> but that doesnt make sense as the normal vector to a 2d function cant be a 3d vector right?

[neon iron]

if you are embedding into 3d, say its 0z

[keen venture]

In 2d it's easy.

If your gradient vector is (1,2), then the normal is (-2,1)

Something something take the negative reciprocal

[smoky pasture]

So -3x+4y=0z

[bronze bison]

i would do it like with planes
in a plane, like -3x+4y+5z=2
your normal vector is (-3,4,5)
so here in 2d we could do the exact same

[smoky pasture]

But howd u do that?

[bronze bison]

we have the line -3x+4y=0
then (-3,4) should be the normal

[keen venture]

Oh yeah, there's that too. The coefficients spell out the normal

[bronze bison]

other than that, we could do:
-3x+4y=0
y=3y/4
y'=3/4
the slope of the normal is the negative inverse of the slope of y
so we get for the normal:
n'=-4/3
integrate
n=-4x/3+b

the b can be whatever, depends on where you want your normal

b=0 if in origin of course

[smoky pasture]

that seems different to what i was taught but gives the same answer, im just wondering about the way i was taught with the raise the dimensions then take gradients

or is the way ur doing it the only logically consistent way to do it?

[bronze bison]

i guess there are more ways than one

i guess you could take your 3rd dimension to be 0 and then take the gradient

[smoky pasture]

do you think you could try to explain the dimension way, otherwise youd have to teach me the whole concept of ur way

[smoky pasture]

so like 0z?

[bronze bison]

yeah i guess

[smoky pasture]

ah ok

[bronze bison]

but that is pretty weird

cause

-3x+4y+0z=0

take the gradient of that

you could just take the gradient of
-3x+4y=0

[smoky pasture]

yeah its the long way, im just learning it so i want to understand whats happening behind the scenes, i can optimize speed once i understand it

but the 0z makes sense

ty for the help!

[bronze bison]

i dont see what taking it to a higher dimension should give us though

as that does not increase our information

the only possible reason would be to use a function that is only defined in 3d

[smoky pasture]

well the way i was taught is that gradients are perpendicular to a level curve of a function

[bronze bison]

at least thats the only thing i could think of

[smoky pasture]

thus every function is a level curve to another

thus find that function that turns into the level curve equal to our function then take the gradient of that

[smoky pasture]

functionally it does nothing

so your totally right

but it just helps build my intuition for why something is occuring

anyways, thanks for the help, have a good one!

.close

[topaz sinew]

Channel closed

Closed by @smoky pasture

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[neon iron]

How do I factor the left side

[neon iron]

I don’t know how to simplify it down

Would it just be

x(x-3)(x+3)

Over 2x

[copper delta]

x^3 +9x^2 can be written as x^2(x+9) and x^2 - 81 can be written as (x-9)(x+9) using difference of squares

Then they just cancel

?

[rapid marten]

guys we have a comedian

[empty sail]

<@&268886789983436800>

[rapid marten]

hes jealous hes failing maths

[copper delta]

The (x-9)'s cancel

[neon iron]

Thx

Does the 2x change at all

[topaz sinew]

@neon iron Has your question been resolved?

[topaz sinew]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[radiant lion]

There are five white balls and three red balls if we put these eight balls in a row with no adjoining red balls the number of arrangements is? And, if we put this eight balls in a row with adjoining red balls the number of possible arrangements is?

[glad cliff]

Are you familiar with the "balls in bins formula"?

[radiant lion]

No

What is this topic called? You helped with a similar question before

[glad cliff]

Combinatorics

[radiant lion]

I'll google it

Thank you

.close

[topaz sinew]

Channel closed

Closed by @radiant lion

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[spring ivy]

someone help me please

[neon iron]

(also post the problem)

[spring ivy]

sorry lol

[frank torrent]

hi yuta

so the general equation of a circle is

[spring ivy]

Idrk I missed school yesterday and I don’t have the notes so idk how to do this really

[safe burrow]

have you done completing the square before?

[frank torrent]

$x^2+y^2+Cx+Dy+E$

[thorny flame]

MochaOhwelp

[spring ivy]

yeah

[safe burrow]

well the good news is that you're basically doing that

but now there's circles

[frank torrent]

the center will be

$\frac{-C}{2}, \frac{-D}{2}$

[thorny flame]

MochaOhwelp

[frank torrent]

I forgot the bracket

anyway

[spring ivy]

i’m so confused😣

i’m gonna get in trouble for not doing this homework

[glass badger]

8 is easier than 7, try that first

[spring ivy]

Ok

[glass badger]

Do you remember Pythagoras theorem? For example, x^2+y^2=2^2 gives all the points that are 2 units of distance away from a center

It makes a circle

[spring ivy]

yeah

a2+b2=c2

that’s what my teacher told me

[glass badger]

In question 8, what would the radius of the circle be?

[spring ivy]

4?

[glass badger]

the radius is c, not c^2. So it would be 2

[spring ivy]

oh yeah

so 2

[glass badger]

do you know how graphs are shifted along an axis?

For example, f(x-3) shifts the graph of f(x) along the positive x axis 3 units. Same applies here for the center of the circle

So (x+3)^2+y^2=2^2 moves the circle 3 units to the left and vice versa

same for y too

Try graphing 8 now

[frank torrent]

actually its just pythagoras theorem

[glass badger]

ye

[frank torrent]

or distance formula

but basically the same thing

[spring ivy]

oh

[glass badger]

did you graph it?

[spring ivy]

no i had to eat sorry

i’m back now

so

(x+3)^2+ y^2=2^2

?

[glass badger]

That one's just an example

X+3 shifts a circle center along the x axis in the left direction

Note that the directions are switched

So x^2+(y-1)^2 shifts the circle 1 unit upwards

And vice versa

You can graph every circle possible with this info

So your example, what would the x axis of the center be?

[spring ivy]

it would be at 5

?

[glass badger]

It's x^2, so it would be at 0

If it was (x-5)^2 , then the center would be at 5 on the x axis like you said

[spring ivy]

so i got it correct?

[glass badger]

no

[spring ivy]

oh

[glass badger]

ok, How about (y-1)^2? Where along the y axis would the center of the circle be?

[spring ivy]

1

[glass badger]

Yep

[spring ivy]

bc -1^2 is -1 • -1 and that 1

so up 1 unit

[glass badger]

Yep

[spring ivy]

ok

[glass badger]

7, you have do some algebra to try and fit the equation into this form

In question 7 I mean

[spring ivy]

ok

so

the center would be (6,-4) and radius 3?

[glass badger]

Question 8 or 7?

[spring ivy]

7

[glass badger]

It's false unfortunately

What did you do?

Did you complete the square?

[spring ivy]

Um

I just put the 6 and the 4

[spring ivy]

then did this

like plug in the number

[glass badger]

You can't do that since you haven't generalized the equation yet

[spring ivy]

oh

ok

[glass badger]

Add 9 and 4 to both sides of the equation

You get two factorable quadratic equations

[frank torrent]

For me I'll remember this to calculate faster

Tho it's not good

[glass badger]

Ye it is faster

[spring ivy]

how do you get the 4 and 9 tho

[glass badger]

Do you know how completing a square is done? You take half of the coefficient of x and square it

[spring ivy]

yeah

[glass badger]

What would half of 6 be?

3

[spring ivy]

3

[glass badger]

Then square it

[spring ivy]

then squared is 9

[glass badger]

Then that's what you need to add

[spring ivy]

oh

ok

[spring ivy]

here u would put everything under a like check mark thing and take off the ^2

?

[glass badger]

The center would be at (3,-2)

Radius would be 4

[spring ivy]

ok

this is not all tho

[glass badger]

?

[spring ivy]

This also

is it showing

[glass badger]

Yes hold on

OK I figured it out

It's been a long time since I studied analytic geometry so I don't remember much. Sorry lol. Hold on a moment so I can show you how to do it

[topaz sinew]

@spring ivy Has your question been resolved?

Channel closed

Closed by @spring ivy

Use .reopen if this was a mistake.

[spring ivy]

.reopen

[topaz sinew]

✅

[spring ivy]

ok

[glass badger]

OK sorry for the delay I was just making sure

Do you know how to solve systems of equations?

And you subtract the second equation from the first one to solve for a. Then you subtract the third equation from the second to solve for b

After you solve for a and b, plug them in in any of the 3 equations and solve for c

Try it with the question 27

[topaz sinew]

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

[glass badger]

Oh alright

[spring ivy]

can u remind me again

I know how but I like kinda forgot

[glass badger]

Well, you try to add or subtract to equations you have to eliminate unknown variables. When working with only one unknown variable, you can solve for it

[spring ivy]

Oh

Like the elimination and substitution methods?

[glass badger]

kinda yeah

For example you have 2x+y=1 and 3x+y=5

You can very easily get rid of one of the unknown variables

Try subtracting them from each other and solve for x

[spring ivy]

?

[glass badger]

Yeah like that

[spring ivy]

Alr

[glass badger]

You need to write out three equations to solve for the points of the circle

Recall the general form of a circle

You were give 3 points, so 3 values for both x and y

Though you don't know a and b

Try writing it out like how you would solve a system of equations

[spring ivy]

but it’s just 3 points not like an equation

A: (1,6) B: (5,6) C: (5,0)

[glass badger]

They're represented with the x and y coordinates

[spring ivy]

yeah

[glass badger]

There are x and y in our general form of the circle, no?

So for (1,6) x=1 and y=6

[spring ivy]

yes

[glass badger]

I showed you the general form of a circle before right?

[spring ivy]

Um

[spring ivy]

this?

[spring ivy]

or this

[glass badger]

This one

[spring ivy]

oh

[glass badger]

We we have three pieces of information about x and y, their coordinates

(1-a)^2+(6-b)^2=c^2 and so on

[spring ivy]

ok

[glass badger]

That's for point (1,6)

Try solving for a and b and tell me how it goes

That's all the info you need to solve this question

Reminder that you have three points

[spring ivy]

like this?

[glass badger]

Yep exactly

Now, try using systems of equations. You don't have to add ALL of them. Try going 2 at a time

Excuse me, I said add. But you can also subtract

[spring ivy]

oh

ok

wait

but it’s different

[spring ivy]

this has three terms

[spring ivy]

and this is like in parentheses and squared

[glass badger]

Look carefully. What happens when you subtract them?

Subtract the second and first, see what happens and count how many unknowns you have left

[spring ivy]

well

for the first it would be 1a^2+6b^2=c2

[glass badger]

it shouldn't be that way

[spring ivy]

oh

[glass badger]

Look, don't square them then subtract. You're complicating it. Do the first and second both have (6-b)^2? And do they both have c^2? Then what happens if you subtract them from each other?

[spring ivy]

it would be (1-a)^2 above (5-a)^2