help-10

[gilded needle]

yes, what do you get when you add them together?

[wanton dagger]

wait add what together sorry?

also do I just sub in -1 for all the i’s?

ahh actually figured it out

thankyou.

.close

[obtuse pebble]

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[violet hull]

Hey guys can someone please explain how I got this wrong. How do I solve it?

[timber island]

Open the brackets

[violet hull]

2x - 3y - 5x - 4y?

[timber island]

no

[violet hull]

can you please explain further?

[timber island]

what is -1*(-4)?

[violet hull]

4

[violet hull]

how do I apply this in this question?

[timber island]

so why write -4

when u apply distributive

its becomes -(5x)-(-4y)

-1*-4 = 4

so its -5x + 4y

[violet hull]

so uh

2x - (-3y) - (5x) - (-4y)?

[violet hull]

so what is the next step?

[timber island]

no

why 2x - (-3y)?

you are multiplying by 1

that doesnt change anything

[violet hull]

can you expand on that?

[violet hull]

so is it 2x - (3y)?

[sage dagger]

Do you know how distribution works?

@violet hull

[violet hull]

probs not

Can't remember (much)

[sage dagger]

So if I gave you

$ - (a+b)$ do you know how to expand

Tex bad

$-(a+b)$

[warm shale]

waterrbeam

[sage dagger]

There we go

Do you know how to expand that

[violet hull]

-a + -b?

[sage dagger]

Well yes but you don’t have to write the + there

So it’s just -a-b

[violet hull]

I see

[sage dagger]

Do the same logic for $ -(5x-4y)$

$-(5x-4y)$

[warm shale]

waterrbeam

[violet hull]

-5x - (-4y)

[sage dagger]

What does that expand to?

Unnecessary

What’s a negative times a negative

[violet hull]

so it's just -5x - 4y?

[sage dagger]

Answer this

[violet hull]

positive

ah

[sage dagger]

So then

[violet hull]

-5x + 4y?

[sage dagger]

Yeah

[violet hull]

so the 4y was a negative at the start?

if so

[sage dagger]

What do you mean?

[violet hull]

here

4y is a negative?

[sage dagger]

Yes it is -4y because your question here (2x-3y) - (5x-4y) said so

[violet hull]

how does 6 - 6 have a difference answer to 6 - (-6)?

[sage dagger]

6 - 6 = 0

6 - (-6) = 12

[violet hull]

yes

[sage dagger]

So what are you confused about?

[violet hull]

but saying that -5x - (-4y) the - in the middle is unnecessary why?

[sage dagger]

Well it’s not incorrect but you’ll likely confuse yourself more leaving that there

It’s easier to mentally think negative times negative is a positive

So I can just write it as +4y

[violet hull]

how does this compare?

[sage dagger]

Instead of -(-4y)

[sage dagger]

What do you mean? You wanted to know the difference between those two

I’m not sure why you brought up the 6-6 and 6-(-6) because that doesn’t anything to do with this question unless you don’t know how expanding brackets work

[violet hull]

Ah I see now

sorry for misunderstanding

[sage dagger]

so you know how to expand now?

[violet hull]

$(2x - 3y) - (5x - 4y)$

[warm shale]

techroz

[violet hull]

My eyes can see better now

so will that then turn into

2x - 3y - 5x + 4y?

[sage dagger]

Correct

Now simplify that

[violet hull]

so now do we get the like terms?

[sage dagger]

Yes

[violet hull]

-3x - y?

[sage dagger]

No

[violet hull]

hold on

negative and positive make negative right?

-3y + 4y

oh

its just y

Hmm

I just really got confused with the positive and negative equal negative rule

[sage dagger]

Yes

A negative times a negative equals a positive

a negative plus a negative equals a negative

[violet hull]

wait

[violet hull]

-3y + 4y

[sage dagger]

What?

Yeah

Now do that

[violet hull]

not positive 3y

[sage dagger]

Simplify

[violet hull]

its just y

I see

[sage dagger]

Yeah

[limber jungle]

you should take y common now

[violet hull]

but it was negative 3y and a positive 4y

[sage dagger]

What’s wrong with that?

[violet hull]

so when do we apply the "Negative and positive make negative rule"?

[limber jungle]

the y itself isnt negative, it is just multiplied to -3

making it -3y

[sage dagger]

So if I gave you -3 + 7 do you know what that equals to

[violet hull]

4

I see

[sage dagger]

I think he’s confused about why negative plus a positive equals a positive

[sage dagger]

Yeah

Now do that for here

-3y + 4y

[violet hull]

y

Oh I see

[sage dagger]

Correct

[violet hull]

I was getting it mixed up with multiplication and division

[sage dagger]

So do you know what the final answer is

[violet hull]

-3y + y

[violet hull]

I was told the other day the rule still applied to addition and subtraction

[sage dagger]

Correct

The -3x+y is

[violet hull]

anyway

am I trippin or it doesn't show the answer here?

or like the right choice

it's none of them?

[sage dagger]

y - 3x is the same thing as -3x + y

[violet hull]

How would I arrange that?

can you explain?

[sage dagger]

Think about it

you’re simply changing the order of the terms without actually changing the signs or outcome of the equation.

-3x is still -3x even if it’s on the right or left hand side

And +y is the same thing

[violet hull]

I see

and I did not know that

[sage dagger]

if I gave you

2 - 4 and -4 + 2 do you think they would have the same answer or different

[violet hull]

same

[sage dagger]

Exactly

[violet hull]

and also another question

[sage dagger]

Sure

[violet hull]

if y = -2, is 3y = -8?

-2 x -2 x -2

so the first 2 make a positive

then the third then makes it negative again

is that right?

[sage dagger]

I’m not sure what you mean

[violet hull]

like my answer is -8

is that right?

[sage dagger]

No

[violet hull]

so what is it?

can you teach me why?

[sage dagger]

What do you think it is?

You’re thinking of (-2)^3 which is not the same as 3(-2)

[violet hull]

so 3y is 3(-2)?

[sage dagger]

Yes because you replace y with -2

[violet hull]

Ohh

I was thinking of powers

alright

[violet hull]

would that give me -6?

[sage dagger]

It would

[violet hull]

alright

anyway

do you mind if I ask another math problem?

[sage dagger]

Sure go ahead

[violet hull]

alright here it is

@sage dagger

[sage dagger]

Hello

Sorry I was busy

I’m back

There are multiple ways to solve this

Do you know the formula Speed = Distance/Time

@violet hull

[violet hull]

that's okay

[violet hull]

yes

oh

let's see here

the speed of the child climbing the escalator that is not working

is 60/90?

when the escalator is working

the speed of the passenger that is not moving, from bottom to top is 1m/s

[sage dagger]

So notice that the moving escalator a passenger moves 60 metres in 60 seconds

[violet hull]

is the speed of the child 1.5m/s?

wait no

[sage dagger]

That means the person moves 1 m/s on the moving escalator

[violet hull]

yes

that is what I said

Needed that comma lol

grammar

[sage dagger]

Using d=s/t on the child they move 60/90 m/s or 2/3 m/s

[violet hull]

yes

[sage dagger]

What do you think we should do next

[violet hull]

60 x 2/3

[sage dagger]

Wrong

[violet hull]

hmm

60 x 3/2?

[sage dagger]

Still wrong

[violet hull]

hm

oh

60 - 2/3?

[sage dagger]

No

[violet hull]

can you give me a hint?

[sage dagger]

before doing any of that you need to consider the combined speed of the walking child and the moving escalator so 2/3 + 1

[violet hull]

so the speed of the child moving on a already moving escalator is 5/3m/s?

[sage dagger]

Correct

[violet hull]

so what do I do now?

[sage dagger]

Now consider how long it takes for them to cover 60 metres if they move 5/3 m/s

Use the speed = distance/time formula and rearrange to find time

[violet hull]

time = distance/speed?

[sage dagger]

Correct

[violet hull]

so now

[sage dagger]

Plug in your values now

[violet hull]

time = 60m ÷ 5/3m/s

[sage dagger]

Yes 60/(5/3)

[violet hull]

now its 60/1 x 3/5

[sage dagger]

Noj

[violet hull]

mb

[sage dagger]

Yes

[violet hull]

typo

so is that right?

[sage dagger]

Idk perform 60 x 3/5 and see for yourself

[violet hull]

180/5?

[sage dagger]

Simplify

[violet hull]

no like did I arrange it right?

yes

36/1

= 36

[sage dagger]

yes

[violet hull]

wow

thanks

I appreciate it

[sage dagger]

Now do you see how you get the answer?

[violet hull]

yes

[sage dagger]

Nice

[violet hull]

just need to remember the fomula

I guess that's it

if there is any more questions I'll come back

thx

have a good day

[sage dagger]

You too

[violet hull]

thanks

[sage dagger]

Type .close when you’re done

[violet hull]

yup

.close

[obtuse pebble]

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[obtuse pebble]

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[glad jetty]

does this rule exists? If so can it be used under all circumstances?

[trim portal]

sounds like power rule with chain rule, so yeah it should be valid as long as (f(x))^n is differentiable

[glad jetty]

thx a lot; just realized that it wasnt complete

[trim portal]

yeah, this is correct & complete now

[glad jetty]

thx;
.close

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[timid silo]

what part do you find confusing?

[royal basin]

do you have its statement on hand?

write it out

in full

so we are on the same page

are you sure you did not make any typos?

so you are not sure.

i think there still is one.

would have expected to see

a = bq + r

where:
a is the dividend
b is the divisor
q is the quotient
r is the remainder
and r lies between 0 and b-1

do you know how integer division works?

do you maybe remember how it worked in school before you learned about fractions

where instead of, say, 37/5 = 7.4 you would write "7, remainder 2" or something to this effect

okay yeah

this a=bq+r is that but written in a more systematic manner

must have been school

don't remember

wdym by "fully"

idk what "divide it once" is supposed to mean

can you do both "divide once" and "keep going" on paper so i can see what you mean and tell you which one it's supposed to be

194/15

do that

1. please don't call me bro

okay you edited that just in time

that's one long-division step

you should keep going with that until you get a remainder less than 15.

and THEN stop

"I got 14"

you should have gotten two numbers, and be able to clearly say which is the quotient and which is the remainder

i'll be back in a few

[timid silo]

you should clearly say which numbers correspond to which thing in the original statement ^

[high lily]

what's 5 supposed to be for / represent

for the purposes of what you're doing, you went too far in your division

you should keep going with that until you get a remainder less than 15
and THEN stop

and if you plan to divide further, actually write in those 0s in the dividend, and also keep stuff aligned

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[floral dagger]

Can I get help? "let it be >insert the vectors< calculate the score"

[shy vigil]

!status

[obtuse pebble]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[floral dagger]

3

[shy vigil]

Can you show your work?

[floral dagger]

i did it mentally kinda

b at least

i got -15i + -15j which is wrong

the book says the answer should be just 12

but im confused now

[shy vigil]

do u know abt dot product of two vectors

[floral dagger]

i forgot X__X

it's been months since it's been taught

i'm doing missed work

[warm shale]

zplus11

[shy vigil]

i1 i2 ... i_n are the dimentions
which in this case are i and j only, so only 2 dimentions instead of n

[floral dagger]

aaaah

what do the numbers mean

below a for example

[shy vigil]

nothing, just the number of dimentions
we have 2 dimentions in question, so a_1 and a_2 will be the two coefficients

just a way of representing

in the question we have dot product of two vectors

• first one is "a"
• second one is "a + b"

Do you remember how to add two vectors?

[shy vigil]

in simpler words, just take corresponding coefficients of both vectors, multiply all pairs and put them with their respective dimentions

[floral dagger]

do I multiply i with i

or do they cross

omg im so lost i dont remember anything lmao

do i multiply everything with eachother

[shy vigil]

yes

all corresponding numbers of both vectors

just multiply them

did u do first part?

[floral dagger]

i got -10i + (-4j)

answer is 14

[shy vigil]

there won't be any i or j in there haha

[floral dagger]

but i didnt get just 14

aaa

😭

[shy vigil]

see product is a scalar

lets start with part a.

[floral dagger]

aaaaa

so only numbers

[shy vigil]

$a = 5i + j \ b = -2i - 4j$

[warm shale]

zplus11

[shy vigil]

take 5 from 5i, take -2 from -2i
multiply them
-10

take 1 from j, take -4 from -4j
multiply them
-4

the numbers are -10 and -4
now add them
-10 + (-4) = -10-4 = -14

that's our answer

[floral dagger]

aa

u simplified it so well

lmaoooo

aaaa okeeei

[shy vigil]

as for the second part, we have a.(a+b)
first vector is clearly a, second is a + b

so we just need to find a+b, then multiply it with a just like we did in part a

now to find sum of two vectors
take corresponding coefficients, again, and now instead of forgetting about i and j, put the added coefficients with i and j

[floral dagger]

will it be the same?

[shy vigil]

do u understand?

[floral dagger]

a*-14

[shy vigil]

not a

the coefficients were -10 and -4

[floral dagger]

ohhh

i get it

yeah

-10 + 4 * -14?

[shy vigil]

noo

[floral dagger]

😭

LMFAOOOOO

[shy vigil]

u have to put the coefficients with dimensions (i and j)

[floral dagger]

aaaa

[shy vigil]

so it'll be -10i-4j

[floral dagger]

when do we not need them

in the answer?

[shy vigil]

in dot product

[floral dagger]

aa okei

[shy vigil]

because dot product means a number without any dimentions

just a number

[shy vigil]

anyway, so a+b turns out to be that.

now u can dot multiply it with a

preferably try all parts out again on a paper

let me know if you run into any issue.

[floral dagger]

5i+1j * -14?

ohh

5i+1j*-10i+-4j?

IM SO LOST

answer is 12

[proper kelp]

whats the issue?

[floral dagger]

how am i supposed to multiple

[proper kelp]

r u trying to find the dotproduct of these two vectors

[floral dagger]

yep

[proper kelp]

right

[floral dagger]

he told me how but i got lost again

[proper kelp]

basically you see the numbers in front of the i and j

[shy vigil]

Oops, just a small correction. As said here, we need to add the coefficients in case of vector addition, not multiply. So please correct the sum before moving onto multiplication

[floral dagger]

-5+(-3)

[proper kelp]

no

[shy vigil]

-3 and -3

[floral dagger]

bye😭

[proper kelp]

its not that complicated i think your just overcomplicating things

[shy vigil]

https://media.discordapp.net/attachments/903430332324405288/1123244470050230372/912969666295627787.png

[proper kelp]

its literally multiplication

[floral dagger]

where did the other 3 come from

[shy vigil]

add the corresponding coefficients

[floral dagger]

3i

3j

[shy vigil]

-3i

-3j

[floral dagger]

yeah

[proper kelp]

what why adding i thought dotprod

[shy vigil]

that's the sum

[shy vigil]

sorry, we had first done the addition part which seemed to have a little error. So just correcting it now 😅

[floral dagger]

can you just do the question im getting lost by the minute

like do it in the same format

like the first cuz I don't get it

we got a+b

[shy vigil]

Did you understand the first part?

[floral dagger]

but how do we multiple

multiply

[shy vigil]

Okay

[floral dagger]

a i did

21.9a

[shy vigil]

a = 5i + j
a + b = -3i - 3j

now just take corresponding coefficients and multiply them

[floral dagger]

-15i + -3j

[shy vigil]

without i and j

just numbers

[floral dagger]

-15+(-3)

[shy vigil]

yep

which is?

[floral dagger]

wait

answer is supposed to be

12

not -12

[shy vigil]

oh lmao

[floral dagger]

😭

[shy vigil]

so like

[shy vigil]

the sum a + b is actually 3i - 3j instead of -3i - 3j

now again

5i + j
3i - 3j

[floral dagger]

i feel so bad😭

15+ (-3)

12

[shy vigil]

YEAH!

[floral dagger]

YEEEEEEE

[shy vigil]

🎉

now please try part c

[glad jetty]

noice

[floral dagger]

wait i need to write this down

[shy vigil]

yes

[floral dagger]

wait now im confused

why did we need a+b if we could do the question w/o it

[shy vigil]

how could we do it without it

[floral dagger]

aa wait nvm

sooo c

3i+(-3j)*3i-3j?

-3-3 = 0

3+3 = 6

but where is the multiplying part......

if we already reached our answer by just subtracting

[shy vigil]

https://media.discordapp.net/attachments/903430332324405288/1123244470050230372/912969666295627787.png

c part is

(a+b).(a-b)

so first find a+b, then a-b, then multiply both

@floral dagger ^

[floral dagger]

33 and 3-9

ugh

3 * 3 n 3 * -3

6 and -9

do i add

but it still doesn't get us to 6

[shy vigil]

but where are u getting those numbers from

lets do one step at a time

how do we add 2 vectors?

[floral dagger]

by adding the same coefficients

3i + (-3) is a+b

7i + 5j

is a-b?

10*2 is 20 😭

[shy vigil]

yeah!

so we have 3i - 3j
7i + 5j

now we just need to multiply these

steps to multiply:

• multiply corresponding coefficients
• add them

[floral dagger]

21-15!!!!!!!!!!!!!

6

[shy vigil]

pls just make sure u remember the steps

write them down, and attempt similar questions

remember - this is just one type of vector product, called "dot product"
cross product is second type of multiplication... which you might study later on

[floral dagger]

yeah we only had dot product for a lesson

the rest were about lines, circles and parabolas which I don't get either but im taking it 1 step at a time 😭

r u still up for helping me w/ a different type of question...

[shy vigil]

Yes. You can request a channel here and ping me separately if you wish

or just here right now if that suits

[floral dagger]

yeah sec

"Check if the vectors are perpendicular to each other."

@shy vigil

[shy vigil]

in this case u need the cross product. 2 vectors are parallel if their cross product is zero

to understand cross product I'd advice an online video, since it'll be easier that way

[floral dagger]

aaa okeei

i'll have a go at it later

[shy vigil]

I had vectors and 3d geometry over 2 years back, and I didn't study it nicely since covid => no schools...
i have to look at them again

[floral dagger]

aa it's fine really

I just wanted to ask since you were already helping me

[shy vigil]

but really, if you get the concepts, then its interesting

[shy vigil]

of course!

[floral dagger]

math is fun when you get it

but I never get it 😭

[shy vigil]

and tbh, it's the same with all topics

[floral dagger]

yet I'm in an advanced math lane

yeahh I guess

[shy vigil]

haha nothing too bad
just a bit more consistency, and it'll be great

[floral dagger]

yeahh

[shy vigil]

do you plan on doing any other topic at the moment?

[floral dagger]

I mean I can ss the course subjects

[shy vigil]

or continue with vectors (which I prefer, since you've started it
but it also depends upon what you want to do)

[floral dagger]

hence the dates

I'm like super late on hw

im at chapter 21

[shy vigil]

interesting

[floral dagger]

so i have a lot ahead of me

plan to finish all of this by 10.8

achievable but it'll be extremely slow w/ the rate I'm progressing through exercises

[shy vigil]

it'll be okay

what's 10.8?

[floral dagger]

10th of august

[shy vigil]

ahhh

not too bad

[floral dagger]

I mean yeahh but I also have chemistry

that I'm behind on

I got burnt out towards the end of the academic year so i just skipped the exams

for these two courses

and didn't end up doing the hw I planned on saving as "practice" for the exam

if ygm

[shy vigil]

I see

[floral dagger]

but yeah I don't wanna take up too much time

thank you for the help

[shy vigil]

you can start slope of a line. it's interesting

[floral dagger]

all I remember is that it has a weird formula 😭

[shy vigil]

y = mx + b

lets forget that

[floral dagger]

aa not that one

[shy vigil]

ax + by + c = 0

or (y2 - y1) / (x2 - x1)

[floral dagger]

sec

lmao

do u wanna see the first question for that sector

[shy vigil]

sure

[floral dagger]

i google translated it for the sake of this

wait

[shy vigil]

its easy really

[floral dagger]

X_X

[shy vigil]

• beautiful graph

[shy vigil]

@lucid surgewalker3302 knows when he sees one

oh fuck

sorry whoever got pinged

[floral dagger]

lmaoo

[shy vigil]

so do you know what's slope?

[floral dagger]

no.....

[shy vigil]

perfect

[floral dagger]

the line

LMFAOO

[shy vigil]

imagine you're buying apples

and one apple costs 2 bucks

then what does it mean

what can you say about the total cost of all apples you buy?

say you buy 7 apples

what'll be the cost?

[floral dagger]

14?

[shy vigil]

yeah, you got it by

multiplying the cost of one apple by the number of apples

[floral dagger]

jepp

[shy vigil]

to put it in other words, the apple situation, we can say that

with each successive apple that you buy, the increase in total cost will be 2 bucks.

that my friend, is rate of change of the cost of apples. otherwise known as slope of the cost line

is it good so far?

[floral dagger]

aaa

yeah it is

[shy vigil]

alright. to put it into some maths

[floral dagger]

nooooooo

jp

[shy vigil]

if we say the cost is C, and the number of apples is n

then we can say C = 2n, can we not?

[floral dagger]

yeah

2 apples?

[shy vigil]

no, n apples

[floral dagger]

aa

[shy vigil]

2 is the cost of 1 apple

[floral dagger]

whats the 2 for

aaaa

okei

[shy vigil]

now, suppose cost of apples is twice the number of apples, PLUS 5 bucks

then the cost equation will be
C = 2n + 5

We just added 5 in the earlier equation

[floral dagger]

wait i don't get it

why can't you just say C = 2

[shy vigil]

Because cost is not 2 in all cases. If you increase the number, cost will increase

[floral dagger]

aa specifically for apples it's 2

so 2n

[shy vigil]

Yes

[floral dagger]

okk

[shy vigil]

and we're also adding 5 bucks anyway, in all cases

thus 2n + 5

which means
1 apple costs 7 bucks
2 apples cost 9 bucks
3 apples cost 11 bucks

and so on

See the increase in each step?

[floral dagger]

yeahh

[shy vigil]

the increase in cost

its still 2.. the same as earlier

[floral dagger]

yeah but + 5

[shy vigil]

hence the slope still remains the same

and in our equation
C = 2n + 5

2 is said to be the slope
5 is said to be the intercept

Each equation has 2 things in common. The C and the n.
Mostly you'll have y instead of C, and x instead of n.

What will vary in each case, is the slope and intercept

---

In simpler words, intercept is nothing but the value of C when n = 0. So the cost of zero apples in this case

the starting point in some sense

that's about it for the introduction. You can take up a video tutorial, they're all pretty nice. I hope you'll enjoy it. Please feel free to tag me if you run into any doubt 🙂

[floral dagger]

that made a lot of sense 😭

thank youu I will <3

[shy vigil]

You're very welcome

[obtuse pebble]

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[tawdry meteor]

is this formula for computing Ej correct?

i use qj instead of Kj because it reads bettre on paper

[obtuse pebble]

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[undone sentinel]

i dont understand what the question is asking

if its parallel that means the slope is the same right

and a negative x axis

so isnt the answer B?

[timid silo]

I think you should look for an equivalence for the equation, which option is the same as the equation above?

[undone sentinel]

B

C and D cant be true cause of the slope

A has a positive y intercept

[timid silo]

i thinking b can't

because the 3/2 is multiplying the parentheses

and the b option, the 3/2 is only multiplying the x

[undone sentinel]

A?

it might be right but i dont want to make vague equivalencies to find an answer

[timid silo]

A nop

try make the multiplication for each option if you find the answer

[undone sentinel]

the answer is supposed to be A

its just the explanation says "we multiplied -8 by -1 and there's our answer"

[timid silo]

ohh

i dont see the c and d were reversed T_T

[obtuse pebble]

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[modest pine]

tf is this??

[modest pine]

how am i plugging a+h to sqrt(5

[kind hawk]

take every x that appears and replace it with a+h

[modest pine]

no x's in sqrt(5

[kind hawk]

yes

[modest pine]

so is

[kind hawk]

so you dont have to replace anything

[modest pine]

is that the answer

bnruh

stupid teacher

[kind hawk]

no, not stupid

[modest pine]

how about this

[teal turret]

$(a+h)^2 \neq a^2 + h^2$

[warm shale]

northsteve

[modest pine]

oh shit

2a+2h

?

[teal turret]

what

no

[modest pine]

i mean the equation u posted

not my problem

[teal turret]

(a+h)^2 ?

[modest pine]

wia thwat

OHH

its

a^2+2ah+h^2

times 3

[teal turret]

times -3

[modest pine]

gottem

help

i think i figured it out

but how does it make sense to have 2

is it cause it's not including -5?

ok\

[bitter topaz]

Exactly, the denominator can't be 0 so you are solving equation x+5=0 first. From this you will get x=-5, so this number can't be in the domain range. It could also be rewritten as R-(-5), which means all the real numbers except -5

[modest pine]

what about root squares

can the number inside be 0

so 0+ is valid

[bitter topaz]

Do you have an example, so I can visualize it?

[severe hornet]

I'm sorry for disrespecting you with my previous message, I attempted to do this and realized I do not know how to do it either hehe!

[modest pine]

😎

[severe hornet]

Just jazzing you cuz!

[bitter topaz]

But it all depends on the question. If you have more equations in the denominator, then you need to solve all of them, and the domain is only the numbers all of them have in common. If there is just 1 equation and that is a square root, then yes.
In this case, it can't be, and all the numbers inside the root must be positive and bigger than 2 (because smaller than 2 will give you the negative number in the root which would lead to a complex number where you don't want to go and 2 also can't be because the denominator would be 0, which can't be)

[modest pine]

thanks

[obtuse pebble]

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[pastel lark]

Hey all

[pastel lark]

.close

[obtuse pebble]

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[brave bramble]

hey

[obtuse pebble]

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[placid smelt]

nvm

lol

.close

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[astral plover]

how do i evaluate this indeterminate limit

[thin pawn]

can you use L'Hopital's rule?

[astral plover]

we have to use trigonometric limit laws apparently

[thin pawn]

So like, use trig identities and manipulate the limit into something that's not 0/0

?

[astral plover]

yeah

[thin pawn]

uhmm, well then there aren't many identities for tan^2, so maye transform into sec^2 -1 ?

[astral plover]

do u thikn sin^2 10x/cos^2 10x would be better

[thin pawn]

maybe. It's hard to see those first try. Usually you experiment a bit and find the solution afterwards

so the limit would look like $\newline \frac{\sin^2x}{3x\cos^2x}$

[warm shale]

imtyp0

[thin pawn]

which is still 0/0

we can factor a 1/3 out though

for what that's worth

and those are functions of 10x my mistake

So maybe use the fact taht 10x = 2*5x and use double angle identities?

[normal oracle]

Use the squeeze theorem.

[obtuse pebble]

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[cunning locust]

Alright, here's my question.

I play two dice games.

In Game 1, I take a handful of n 10-sided dice, and roll them. Any die that comes up 8 or higher is a "hit", and any die that comes up as a 1 subtracts a hit.
(ie a roll of 🔟9️⃣7️⃣6️⃣1️⃣ scores 1 hit.)

In Game 2, I take the same n dice, and roll them again. In this game, any die 8 or higher is a "hit". Ones don't subtract hits this time, but whenever a 10 is rolled, it "explodes", so an extra 10 sided die is rolled, and if that rolls 8+ it's a hit, and if that rolls a 10, that explodes, until such a time as no more tens are rolled.
(Ie the same roll of 🔟9️⃣7️⃣6️⃣1️⃣ scores two hits, but an extra d10 is rolled, say 8️⃣, making the total number of hits three.)

My question is this - what's the probability of rolling 5 or more "hits" in game 2 for n dice, and how many hits would equate to the same probability in game 1?

I've already tried looking at equations, but I keep getting mired in summations and such that it's too late in the day to wrap my head around.

(Context: I'm a rookie DM looking to use some material from another sourcebook that uses a different dice roll system, and want to convert the mechanics into my native system so they're usable without me having to carve out a specific exception for it.)

I found this, earlier which could work for game 2. Just unsure of a simple way to reduce this down and plug game 1 into this.
https://math.stackexchange.com/questions/1644794/exploding-a-k-a-open-ended-dice-pool

[obtuse pebble]

@cunning locust Has your question been resolved?

[somber spear]

I started writing out an expression for the probability but it is getting dummie long

[obtuse pebble]

@cunning locust Has your question been resolved?

[somber spear]

ok so this is my exhaustive expression for getting 5+ hits in game 2

although now that I'm thinking about it, it probably overcounts in some ways I can't handle

There is probably a better way

[obtuse pebble]

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[grizzled shore]

,calc 3(1/4)^2 + 8(1/4) - 3(1-1/4)

[warm shale]

Result:

-0.0625

[grizzled shore]

please

do

not

use

mixed fractions

[jagged fossil]

oh i should use decimals

[royal basin]

no

you should use improper fractions

[grizzled shore]

use improper fractions

[jagged fossil]

OH

[royal basin]

mixed fractions are hell to calculate with

[jagged fossil]

that's a much better idea

[royal basin]

you should only ever use them to report a final answer, and that ONLY if you're explicitly asked for mixed fractions

[grizzled shore]

or in cooking it seems

mixed fractions for cooking is fine

but we are doing maths here not cooking

[jagged fossil]

,calc 3(1/4)^2 + 8(1/4)

[warm shale]

Result:

2.1875

[grizzled shore]

also

$3\left(\frac{1}{4}\right)^2\neq \frac{3}{2}$

[warm shale]

frosst

[jagged fossil]

i thought i resolved the exponent before multiplying?

WAITTT

1^2 is not 2

[grizzled shore]

$3\left(\frac{1}{4}\right)^2=3\left(\frac{1^2}{4^2}\right)$

[warm shale]

frosst

[jagged fossil]

oh

shit

so that’s 3/16?

[grizzled shore]

yes

[jagged fossil]

,calc 3(1-1/4)

[warm shale]

Result:

2.25

[jagged fossil]

this should be right

,calc 35/16

[warm shale]

Result:

2.1875

[glossy saddle]

Doesn't my friend look lovely

[jagged fossil]

i'm not 100% these are correct, if someone would check that'd be epic

[violet kindle]

You’re having some sign trouble

Remember (-3)^2 = 9

And -7(-3) = 21

I think you’ll end up getting that it’s not a solution anyways, but partial credit could be docked

The second one also has some issues

(1/2)^2 is not 2/4

And the last equality you give is not right

[jagged fossil]

shit

[jagged fossil]

i thought it was (-3^2)

if that's wrong then rip me

but i do understand at least

[jagged fossil]

ah this was silly

[jagged fossil]

DAMNIT

[violet kindle]

If you plug in -3 to w^2, you get (-3)^2

[jagged fossil]

tyty

.close

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[analog fjord]

need help with this (calculas 2)

[azure anchor]

!status

[obtuse pebble]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[analog fjord]

1

i already sketched it

but lost from there

[azure anchor]

do you know what to do next

well, we are looking for the region bounded by the curves and boundaries.

[analog fjord]

i know the limits of integration are 2 and 8

[azure anchor]

have you tried finding the intersections of y1 and y2?

how does your graph look

[analog fjord]

no, I'm not sure how to do that

ill send the graph

[azure anchor]

so looks like all we need are the 2 and 8

no intersection, only bounds

so you're right about the limits of integration

[analog fjord]

i used desmos since we are allowed in my class

[azure anchor]

so now, the basics of this are, when integrating with respect to x, the formula goes

$\int_{a}^{b}top\left(x\right)-bottom\left(x\right)dx$

[warm shale]

b0ngl0rd

[analog fjord]

the top is 10-1/2x and the bottoms is the y = -3/8x(x-8)?

[azure anchor]

looks like it

do you understand why the formula is that way?

[analog fjord]

it is it because we are trying the find the region between those 2 graphs?

[azure anchor]

yes, but why we would subtract the top curve from the bottom?

[analog fjord]

im not too sure

[azure anchor]

when we integrate, we find the area under the curve.

[analog fjord]

ahhh, is it also because we wnat a postivie value?

[azure anchor]

when we subtract in an integral, we are subtracting one area from another.

[analog fjord]

makes more sense now

Is this how the integral is setup?

[azure anchor]

so we are "cutting a hole" in the top curve's area the size of the bottom curve.

yeah try to integrate

helps to shade in the area you are calculating too

[analog fjord]

oh shit, I didn't know about that feature

also can i seperate the functions?

so like the integral for 10-1/2 minus the integral for -3/8x(x-8))?

[azure anchor]

absolutely. that's a property of integration.

you can "distribute" the integration operator

if you think about it, that's exactly what we are doing anyway. subtracting one integral from another, or one area from another. to find an area between the two.

[analog fjord]

sorry for the bad handwritting

[azure anchor]

it looks like you wrote the same thing

ohhh i see

keep going

actually

if it helps. you can simplify the integrand

10 - 1/2x - ( -(3x/8)(x-8) )

simplify

[analog fjord]

now we plug in the limits of integration to each one of them?

[azure anchor]

looks good

[analog fjord]

I got 43 for the first equation

and then 27 for the second one

which i subtracted and got 16 which is wrong since the final answer is 18

[azure anchor]

to avoid mistakes, it might be better to simplify the integrand

so instead of separating it into separate integrals, simplify it and then split it up however you like

[analog fjord]

I did it right, I just need to plug it in neatly

[azure anchor]

did you get the right answer?

[analog fjord]

yes

[azure anchor]

for the future

you could do it like this

simplify and then integrate

[analog fjord]

that looked so much more easy

[azure anchor]

it wont always be possible, but its good to check if you can simplify

[analog fjord]

Is this still valid to do?

[azure anchor]

yes. there's almost always multiple ways to integrate

as long as you get the same final answer you did it right

[analog fjord]

Okay thank you so much!

[azure anchor]

👍

[analog fjord]

!done

.done

.close

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[obtuse pebble]

Please don't occupy multiple help channels.

[nocturne minnow]

You already have #help-13 don't open multiple channels

.close

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[long oyster]

this is coming from partial fractions btw*
My confusion is, the lecture said that since the coefficients are 10 on the left side and 2A on the right side we can assume that 10 = 2A. How does that work?

[long oyster]

how can we just assume that 10 = 2A?

[wooden cipher]

We want to make sure the value on the left and right vary at similar rates depending on x

So you match coefficients to ensure this happenz

[long oyster]

mm still dont get it

how can you assume that 10 = 2A?

ohhhh

because for the equality to be an equality the two values would need to increase equally whenever the value of x changes hence 10x = 2Ax

[wooden cipher]

Indeed

[long oyster]

ohhh now GarlicBredFries words make much more sense

i didnt understand the exact same words before

😭🙈

ty @wooden cipher @versed cave and @timid silo

got it now!

.close

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[versed cave]

I didn't do anything
Still happy you got it though, it's such a cool method to solve these kinds of stuff

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[patent zealot]

Let G be a connected graph that is not complete. We need to prove that there exist vertices a, b, and c in G such that ab and ac are edges in G, but bc is not an edge in G.

[patent zealot]

Assume the opposite, that for any vertices a, b, and c in G, if ab and ac are edges in G, then bc must also be an edge in G. Since G is not complete, there must exist at least one pair of vertices that are not directly connected by an edge, say b and c. Since G is connected, there must exist a path between vertices b and c through some intermediate vertices, including vertex a.

So that means that ac and ab are edges, but bc isn't => but because we assumed the opposite, ab, ac and bc must all be edges, is that right

So that is the contradiction

[wooden cipher]

Sounds good to me

[obtuse pebble]

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[runic void]

Can you help understand why we write Miyi ≡ 1 (modmi) ?

i understood that if (a,m)=1, then ax≡ b (modm) has a unique solution

but not sure, why we choose b=1 here

[obtuse pebble]

@runic void Has your question been resolved?

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[slow phoenix]

need help with qn7b! (calculus of vector - valued function)

[slow phoenix]

the ans just in case required.

[pseudo swift]

you got any idea or no ? @slow phoenix

[slow phoenix]

er no tbh

[pseudo swift]

here's the basic situation

your plane is in orange, your parameterized curve is in red

[slow phoenix]

mhm

[pseudo swift]

the curve intersects the plane at the big red dots I put

[slow phoenix]

mhm

[pseudo swift]

now your plane has a normal vector, and your curve has a tangent vector at every point

[slow phoenix]

mhm

and we needa find the angle between those 2 rite?

[pseudo swift]

the curve also has a tangent vector at the points where it intersects the curve

and yeah we want to find the angle between those two

[slow phoenix]

mhm but i not sure how to do the mathematical solving part oml :"> like as in the workings

[slow phoenix]

i get until this part

[pseudo swift]

can you compute the tangent vector or no ?

since you're doing a calculus of vector - valued function class, I'd expect you to be able to do that

[slow phoenix]

let me try it on a piece of paper brb in few mins

@golden glacier

oh @pseudo swift ignore that ping hehe :>

just pinging a fren dun take it wrong :>

,rotate

[slow phoenix]

idk if what im doing is exactly right but er i got dis 0v0

[pseudo swift]

sounds alright

still need to evaluate your derivative at t_0

if we're talking about the linear approx of the curve around t=t_0

otherwise it's fine

anyway we won't need the approx, just the tangent vector r'

[slow phoenix]

oooh

[pseudo swift]

now we're interested in the tangent vector at 2 specific times

when they intersect the plane

[slow phoenix]

mhm

and t = 0 at these points right?

[pseudo swift]

idk

you found the intersection points already

you certainly know at what times they happen

[golden glacier]

The intersection points are <1,1,-3> and <-2,4,6>

[slow phoenix]

mhm

[slow phoenix]

Oh so u sub it in dis?

o.0 issit?

[pseudo swift]

yeah

[pseudo swift]

^

[slow phoenix]

oki oki lemme do the subbing

okay what is sup to be done after subbing btw?

[pseudo swift]

well do you know how to compute the angle between two vectors ?

if you can do that you're pretty much done

[slow phoenix]

er issit like cos inverse smth?

[pseudo swift]

you'll need it

you prolly need to know the cosine beforehand if you want to do that

[slow phoenix]

wait this is the tangent right? what about the normal?

[golden glacier]

I can't tell if I'm right

[slow phoenix]

er we have dis atm?

[golden glacier]

I need a . b cos theta to get the degree right?

[slow phoenix]

@pseudo swift btw the subbing issit correct?

[pseudo swift]

you have 2 accts or something @slow phoenix

[pseudo swift]

the linear approx is irrelevant