help-10

[gritty grotto]

ahhh iccc

[mystic dagger]

solve it?

@gritty grotto

[gritty grotto]

doing it right now

slowly

brain slow

i got 4k^3 -12k +8

for the last bit

[mystic dagger]

oh i might be false

left hand?

[gritty grotto]

because i did (2k+16/k^2) which is eq 3

minus

eq 2 6 x (2+4/k)/2k

[mystic dagger]

i mean eq3's third term

[gritty grotto]

that k-4/k

[mystic dagger]

so when this is 0, that type of function will be not exist.

right?

@gritty grotto ok?

[gritty grotto]

wdym?

isnt there supposed to be values?

[mystic dagger]

this is from image that you sent me

if h=0 and i not 0

[mystic dagger]

there will be not exist solution

[gritty grotto]

oh yup

[mystic dagger]

i mean another image

[mystic dagger]

yes

[gritty grotto]

h=0 i=/0

[mystic dagger]

no.2

y

[gritty grotto]

yeah what does that mean

[gritty grotto]

OHHH

SO

when

the

3rd value

in the 3rd row

[mystic dagger]

this mean ( k-4/k)*c = ...

[gritty grotto]

=0

[mystic dagger]

if k-4/k=0

c not exist

right?

[gritty grotto]

yeah?

so what does this mean for the answer?

[mystic dagger]

if k is -2 or 2, c is not exist and one type of function is not exist.

got it?

[gritty grotto]

wait how id u get k = +-2

sorry i still dont get it

so isnt k-4/k = 0

since k-4/k = h

[mystic dagger]

no

it is not k-4/k = h

(k-4/k) * c = h

[gritty grotto]

why times c?

[mystic dagger]

and k-4/k = 0 means that c is not exist

this is gaussian elimnation

wait

Ax+By=m
Cx+Dy=n

[gritty grotto]

yeah?

[mystic dagger]

Ax+By = m
(D-BC/A)y = n - mC/A

right?

@gritty grotto

[gritty grotto]

wait what

[mystic dagger]

this is simple one

simplest e.g. of gaussian

if D-BC/A=0, y is?

??

[gritty grotto]

sorry i am so confused rn

i learnt this today

[mystic dagger]

i c

think more

[gritty grotto]

did you minus them

[mystic dagger]

yes

[mystic dagger]

no eliminate x

eq2 - eq1*C/A

ok?

this is principle of gaussian elimination

???

why dont you answer

@gritty grotto

[gritty grotto]

I’m doing it in my book

So

Dy-Bcy/A

[mystic dagger]

yeah

and we can get y from this one

ok?

[gritty grotto]

Ok

But how does this relate to the q?

[mystic dagger]

but if D-BC/A=0, you can get y?

[mystic dagger]

answer my q

if D-BC/A=0, you can get y?

[gritty grotto]

Y=0?

[mystic dagger]

no

right hand is not 0

0*y = ...

[gritty grotto]

But if D-BC/A=0

[mystic dagger]

you can get y?

[gritty grotto]

then

You can’t get y

cuz y time 0 is 0

[mystic dagger]

right

yes

you finally got it

so this means if D-BC/A = 0

[gritty grotto]

wait but how do I put this into context

[mystic dagger]

there is not appropriate solution

right?

[gritty grotto]

yes

and so how did u get 2 or -2

😭 sorry

[mystic dagger]

so if k-2/k=0, you can get c.

right?

[gritty grotto]

I thought it was k-4/k

[mystic dagger]

oh, yeah

k-4/k = (k-2)(k+2)/k

right?

[gritty grotto]

Yup

[mystic dagger]

so if k is -2 or 2, you can't get c

[gritty grotto]

Ahhh ic

[mystic dagger]

so there is not solution

right?

[gritty grotto]

ohh

okk

[mystic dagger]

gaussian elimination will be useful when number of variables goes higher

[gritty grotto]

Damn that’s hard

Thanks so much

[mystic dagger]

x,y,z...

[gritty grotto]

I get it now

thanks so much for your time and effort

I really appreciate it

[mystic dagger]

you are welcome

if you get any question DM me!

[gritty grotto]

TYSMMM

I SEE

I GET IT

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[obtuse pebble]

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[obtuse pebble]

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[next widget]

if f(x) = x^2 -4x +5 what is the largest possible positive domain for which f^-1 (x) has an inverse func

[brisk matrix]

is this a multiple choice, seems like there could be multiple answers

ah ok

[next widget]

i edited my question

[brisk matrix]

what have you tried

[next widget]

i got the range of f(x) and made it the domain for f^-1 (x)

and got x>= 1

which i wanna know if its right

[brisk matrix]

it is not

you should be looking to make f(x) pass the horizontal line test

if you draw a parabola, you'll see that that means you want only 1 side of the axis of symmetry

which side you want is given in the question

[next widget]

doesnt f(x) already pass line test

or did u mean f^-1 (x)

[brisk matrix]

the vertical line test yes

the horizontal line test no

[next widget]

ohh

[next widget]

doesnt it only pass the horizontal line test at the axis of symmetry

which is 2

so x=< 2 which is invalidated because the qus asking for largest possible domain

or x>= 2

@brisk matrix

[brisk matrix]

yes either of those

but your question says positive

so which should you pick

[next widget]

this

@brisk matrix does a function only have an inverse if it monotonically increasing or decreasing?

[brisk matrix]

im not sure how to answer this

[next widget]

ohk

[brisk matrix]

yes under some other conditions

in general, no

[next widget]

oh

ok

[next widget]

so how do u determine under what domain a function has an inverse

@brisk matrix

[brisk matrix]

im not sure how to answer that either

what class are you in

[next widget]

year 11

australia

[brisk matrix]

i dont know what that means

[next widget]

ohk

[brisk matrix]

do you know what "injective" and "surjective" are

or "one to one"

and "onto"

[next widget]

ik 1 to 1

whats onto tho

[timid silo]

surjective

[next widget]

dunno what that is too

[timid silo]

1 to 1: no two elements map to the same element

surjective: for every element in the second set, there is an element in the first set that maps onto it

[dense imp]

is (-infinity, 2) a larger interval than (2, infinity)? 🤔 how does infinity work..

[brisk matrix]

1 to 1 is like saying every horizontal line touches at most 1 element of the function
onto is like saying every horizontal line touches at least 1 element of the function

if both of those are satisfied, the function has an inverse

[next widget]

okh

whats the big difference between 1 to 1 and onto @brisk matrix

[brisk matrix]

i suggest you look at some graphs and apply the definitions i gave you

you will gain some intuition for the difference

[next widget]

alr

thx

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[dense imp]

if you think of a function as some kinda machinery with input and output, imagine a vending machine where you press some button and you get out a soda of some kind. say the machine is stocked with 5 different kinds of soda. if our function is ONTO then there will be some button for every kind of soda. If it is not onto, then you try to press all buttons, but turns out you can't get some kind of soda out whatever you press.
for one to one: if it's one to one it means every button pressed will give you a different kind of soda (no repeats, no two buttons will give you the same type of soda)

x = which button you press, y = which soda you get out

[obtuse pebble]

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[trim cloud]

need help with row reduction of a matrix

[trim cloud]

6 6 -24
4 5 -15

the right most column is past the = sign

first i divide 1st row by 6

1 1 -4
4 5 -15

then row2 - 4*row1

1 1 -4
0 1 -31

then row2 - row1

1 0 -35
0 1 -31

but answer should be -5 and 1

[gilded needle]

this is wrong

-4 times -4 is +16, not -16

[trim cloud]

oh

whoops

thanks

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[graceful marten]

Uhhh

[graceful marten]

How do I solve for y, values

Ik how to do for theta range

WITHOUT THE circle diagram

also srry, i didnt see u write

at the begining? or we typed at same time

[solar hill]

I forgot the double angle formula, but this looks straightforward

2 sin(2y) sin(y) = 3 cos(2y) cos(y)

[graceful marten]

Can i just plug my theta vlaues

into y

?

Somehow

idk

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[gray ore]

I need help with derivatives. This is the task and what i've ended up with:

[gray ore]

How do i get to the same solution? In our lecture tutorial the tutor just skipped all the steps

[primal wadi]

wait is it (log theta)^t or log(theta^t)

or log(theta^t(1-theta)^h)

[gray ore]

oh i'm stuck at the first task without the log

the first derivative without log

[primal wadi]

💀 i just kinda need to know what the goal is

[gray ore]

this is the goal

ive only got this far

[primal wadi]

so your equation is
$$t\theta^{t-1}(1-\theta)^h+h\theta^t(1-\theta)^{h-1}$$

[warm shale]

qwertytrewq

[gray ore]

yes

[primal wadi]

note that they have a common factor of $\theta^{t-1}(1-\theta)^{h-1}$

[warm shale]

qwertytrewq

[primal wadi]

try to take out the common factor and simplify

on the other hand

[gray ore]

but they have -1 in the exponent

[primal wadi]

on your (1-theta)^h derivative you forgot a -1

[gray ore]

how do i deal with them having / not having -1 in the exponent

[primal wadi]

cuz d (1-theta)/dtheta=-1

[primal wadi]

i dont quite get what you mean by -1 in the exponent

[gray ore]

This part doesnt have $\theta^{h-1}$ only $\theta^h$ $$t\theta^{t-1}(1-\theta)^h$$

[primal wadi]

note this it the same as $$t(1-\theta)\cdot \theta^{t-1}(1-\theta)^{h-1}$$

[warm shale]

Yannick

qwertytrewq

[primal wadi]

thats why i say they have a common factor of $\theta^{t-1}(1-\theta)^{h-1}$

[warm shale]

qwertytrewq

[gray ore]

oooooohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

[primal wadi]

also you should fix your original solution up to that step you forgot to multiply by -1

$g'(\theta)=-h(1-\theta)^{h-1}$

[warm shale]

qwertytrewq

[primal wadi]

not $h(1-\theta)^{h-1}$

[warm shale]

qwertytrewq

[gray ore]

oh i get it

thank you very much

[primal wadi]

np

[gray ore]

🙂

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[gray ore]

.reopen

[obtuse pebble]

✅

[gray ore]

also it worked without the $g'(\theta)=-h(1-\theta)^{h-1}$

[warm shale]

Yannick

[gray ore]

without the - was fine too $g'(\theta)=h(1-\theta)^{h-1}$

[warm shale]

Yannick

[gray ore]

came to the same result as the solution

is this an oversight on my part, due to how the - would be eliminated?

ahhh probably doesn matter

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[primal wadi]

the -would not be eliminated and what you got is incorrect without the - sign. You probably made another mistake and double negated it?

[gray ore]

i still dont know why there should be a minus, but if i made the same mistake for f'(0) then it would cancel out right?

[primal wadi]

because by differentiating (1-theta)^h you are implicitly using a chain rule with 1-theta

and the derivative of 1-theta is -1

you forgot to multiply it by the derivative of 1-theta

[gray ore]

oh

yes

i understand

ur right, in my result there was a plus and in the solution there was a minus

i didnt notice that

[obtuse pebble]

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[fiery salmon]

for this limit problem, even though it's not possible I reason it by taking out the x^2 on its own which results in x/2x = 1/2 as lim x approaches -infninity. My question is why the answer is -1/2 and not just 1/2, I also know that my logic to arriving at 1/2 is incorrect so help explaining the proper way to do it would be great

[primal wadi]

because as x-> -infinity square root of x^2-9 is similar to |x| which is -x when x is negative so -x/2x-3 is approximately -1/2 not 1/2

[fiery salmon]

why do i apply an absolute value and how does negative infinity squared - 9 equal a negative variable

this is my string of logic but I'm definitly missing something here

[primal wadi]

you probably should not do operation on infinity if you want rigor.

[primal wadi]

bottom fraction should be -2infinity cuz we are approching -infty

[fiery salmon]

i-

ya i really did just miss that

OKAY THANKS

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[stiff locust]

Suppose, you are to jump and touch a wall 10m away. However, you can only jump halfway to the goal in every jump. Can you find how many jumps needed to get to the wall in this way?

[sweet cobalt]

What have you tried?

[stiff locust]

i've halfed the distance covered by the jumps till the 7th jump

there's still 0.078125 m left

[sweet cobalt]

Okay cool yeah

[stiff locust]

im unsure if i should just leave it there or continue

[sweet cobalt]

It's a bit of an odd question to be honest

Imagine the final jump that makes you reach the wall

[stiff locust]

its very

absurd

imo

its supposed to have a continuous answer right?

[sweet cobalt]

You would have to have jumped the whole distance to the wall

[signal tendon]

I'm trying to see if the question is tricking you. It's saying "halfway to the goal'. The goal is 10m

Doesn't that mean you jump 5m every time?

[stiff locust]

the thing is

the first jump is 10/2

the second is 5/2

the third 2.5/2

it goes on like that

[signal tendon]

No, I get that. But, the goal is 10m. Does your goal change?

[stiff locust]

no

[sweet cobalt]

You will never reach the wall in this case

[signal tendon]

The goal is still 10m. So what if it means you jump 5m in both jumps

[strong ermine]

The goal doesn’t change but your distance relative to the goal does after every jump

[stiff locust]

yes

[signal tendon]

This is what I mean yeah

So you'd still jump 5m each time no?

[stiff locust]

idk im pretty confused

[strong ermine]

I meant it as in you reassess your position after every jump and see the distance to the goal

[signal tendon]

oh

[strong ermine]

Then hop over half of that

[signal tendon]

Then doesn't that mean it's continuous

[strong ermine]

This a weird question ong

[signal tendon]

I think you don't have to reassess you position after every jump

You're just going to jump 2 times. 5m 5m

[strong ermine]

It does say “halfway to the goal in every jump”

[signal tendon]

But, that doesn't mean your distance relative to 10m becomes your new goal does it?

[strong ermine]

Yes, and you jump half of that

You know what @stiff locust it just depends on how you view the question

If you see it Kai’s way then the answer is 2

If you see it my way then the answer is infinite

[signal tendon]

I think what it's saying is that

You jump in discrete values

[strong ermine]

Depends on how you interpolate it tbf, the semantics is really crucial hete

Here

[signal tendon]

You can't jump 0.623872 meters, but you can jump 1 or 2.5 meters

Is what I'm guessing it's assuming

It needs to be discrete numbers

[stiff locust]

the thing is

i think the question is looking for an infinite answer?

its not really a trick question

the question is trying to see how far i can calculate without fail?

thanks tho

[signal tendon]

I doubt

[signal tendon]

Probably not

They're either looking for the answer 2, or using discrete numbers (I'm guessing 5 jumps)

Can you confirm with the teacher / professor what the question is asking for? I'd love to know the answer

[stiff locust]

I'll ask him tomorrow and let you know :)

[timid silo]

the answer is infinity

the seq. certainly can never end

[obtuse pebble]

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[slow prawn]

if ln0 is DNE why isnt this DNE

[timid silo]

10^x is continuous on [0,1] and therefore exists

[timid silo]

*its integral between 0 and 1 exists

[warm canopy]

Show your work for the antiderivative

[slow prawn]

antiderivative of 10^x

is

10^x ln10

right?

thats what i got

[warm canopy]

What happens if you differentiate your answer?

[slow prawn]

isnt it already differentiated?

[warm canopy]

You do not differentiate to find an antiderivative

[slow prawn]

this

oh

[warm canopy]

Your antiderivative should give you back your original expression when you differentiate it

[slow prawn]

mb i differentiated instead of integrate

wait how do you integrate

10^x

is it

tbh idk

[warm canopy]

Try use the fact that $a = e^{\ln(a)}$ for any a non zero

[warm shale]

ΣAC

[obtuse pebble]

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[feral acorn]

could someone help me understand 1. what is a singular solution? ( i dont understand the definition) 2. why y = 0 is a singular solution to the given DE?

if ye have any more examples of singular solutions of DEs thatd be useful thanks

[obtuse pebble]

@feral acorn Has your question been resolved?

[feral acorn]

<@&286206848099549185>

[obtuse pebble]

@feral acorn Has your question been resolved?

[brazen viper]

I'm not quite sure exactly how to explain it better than the text of the definition. So instead let's look at the example.

Let's first verify that it's a solution to the DE

y = 0, so 4y = 0

and y' = 0 so (y')^2 = 0

The left hand side is equal to the right hand side (LHS = RHS)

so this is indeed a solution to the DE.

Next, let's look at the solution given, y = (x + c)^2

This expands to y = x^2 + 2cx + c^2

Now for whatever value of c we specify, we will still be left with a parabola.

we cannot generate a constant function.

We can check that this satisfies the DE by taking the derivative and verifying the identity holds.

y' = 2x + 2c

(y')^2 = (2x + 2c)^2 = 4(x+c)^2

and (x + c)^2 is y, so the identity holds.

What went wrong here? Well, typically this occurs because as part of the solution you wind up accidentally assuming something about the solution. In this case, to solve this problem you would divide through by y, and then integrate both sides of the equation.

\begin{align*}
(y')^2 &= 4y \
y' &= \pm2 \sqrt{y} \
\frac{y'}{\sqrt{y}} &= \pm 2 && \text{Assumes $y \ne 0$} \
\int \frac{y'}{\sqrt{y}} , \dd{x} &= \pm \int 2 , \dd{x} \
2 \sqrt{y} &= \pm 2x + c \
y &= \pm (x + c)^2 && \text{Implicitly renaming c/2 to c}
\end{align*}

[warm shale]

全能の存在

[brazen viper]

Because in step 3 we assumed that $y \ne 0$ we must check the case $y = 0$ separately.

[warm shale]

全能の存在

[brazen viper]

@feral acorn

[obtuse pebble]

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[versed stratus]

other than actually writing down all such numbers, is there any way to solve this?

[last pilot]

i mean

you could see

the permutations of numbers

i mean combinations

that are equal to 24

[versed stratus]

yeah, just thought of that, realised how stupid a question I asked

sorry

[last pilot]

lol

xD

is fine i'd probably struggle too

am dumber than u

[versed stratus]

still have to list them out though, no way to avoid that I suppose?

like 888

789

ah, the only possible cominations are the permutations of 7,8,9 and 888

I think

[last pilot]

yes-

lol

well not really i mean

6, 9, 9 as well

[versed stratus]

ah, ok

yeah

[versed stratus]

so that's 6 combinations

oh

[last pilot]

yep

[versed stratus]

6/2!

[last pilot]

wait 6?

[versed stratus]

3!/2!

right

that's just 3

[last pilot]

ye

[versed stratus]

so 3+6+1

so that's 10/900

which is 1/90

which checks out with the answer

tysm!

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[steel ivy]

Somebody help me

[steel ivy]

I am dying

Q5 goofy aahhhahahh shit

HOW MAN

[thin stone]

you're pretty close

[indigo blade]

V/CSA

[thin stone]

forget about the numbers, try to express V in terms of CSA or vice versa

[steel ivy]

Hello

I am here

[steel ivy]

HUh

How does that work

[steel ivy]

Help

[thin stone]

yeah

[steel ivy]

I’m dying

I thought it was the simultaneous shit

But it’s not

[thin stone]

so you have $V=\pi r^2 h$ and $S=2 \pi r h$

[warm shale]

anonymemes

[steel ivy]

Yes

[thin stone]

okay

they look kinda similar right?

[steel ivy]

Oh I gotta do that

[thin stone]

how can you make S look like V or V look like S

[steel ivy]

Cancel out pi or something

R square = 2 r

Ohhhhhh

WAIT

How do I get the answer

I’m dying again

[thin stone]

thats if you want them to be equal, i mean i want you to multiply V or S by something to get the other

like what happens if you multiply S by r

[steel ivy]

462 x r = 132?

[thin stone]

forget about the numbers for a sec

the expressions

[thin stone]

what happens if you multiply this S by r

[steel ivy]

We get volume

[thin stone]

no we dont

[steel ivy]

But s needs to become h

[thin stone]

take this step by step

just take the expression for S and multiply it by r, what do you get. i dont want it in terms of volume or anything yet

[steel ivy]

We get something but we don’t know the radius so how

[thin stone]

what do you get

[steel ivy]

132 x r

[thin stone]

dont deal with the numbers

$S \cdot r= 2 \pi r h \cdot r = ?$

[warm shale]

anonymemes

[steel ivy]

😭

[thin stone]

lol its okay

[steel ivy]

I don’t understand

[thin stone]

okay here

gimme a sec lemme lay it out for you

Forget about the numbers for a lil bit. I want to express my surface area in terms of the volume

i know that $V=\pi r^2 h$ and $S=2 \pi r h$

[warm shale]

anonymemes

[thin stone]

with me so far?

if not thats okay tell me which part is confusing :)

[steel ivy]

Ok one sec

Yes

Ik that

[thin stone]

okay thats great

you can see that the expressions are pretty similar, the expression for $V$ is missing an extra factor of $2$ and has an extra factor of $r$

[warm shale]

anonymemes

[thin stone]

for it to be like S

is that part a lil confusing?

[steel ivy]

No I got it

[thin stone]

okay

[steel ivy]

R^2 for volume and x2

For surface area

[thin stone]

so you can see that $\frac{S \cdot r}{2}= V$

[warm shale]

anonymemes

[thin stone]

does that make sense now?

[steel ivy]

No what did you do

[thin stone]

okay

[steel ivy]

You cancel the pi and the height?

[thin stone]

im not doing anything with the height

here look

$S=2\pi r h$. So, $S \cdot r = 2\pi r h \cdot r= 2\pi r^2 h$

[warm shale]

anonymemes

[steel ivy]

Yea because you multiply by r there you can just make it r square

[thin stone]

From here you can see that if you divide $S \cdot r$ by $2$ you get $\frac{S \cdot r}{2} = \pi r^2 h = V$

[warm shale]

anonymemes

[steel ivy]

Yes

[thin stone]

great :)

now you can rearrange for $r$, what do you get?

[warm shale]

anonymemes

[thin stone]

remember that you know $S$ and $V$, so the purpose of doing this manipulation was to express $r$ in terms of $S$ and $V$

[warm shale]

anonymemes

[steel ivy]

132 x r = 924

Radius 7

So diameter 14cm

[thin stone]

I meant for you to see that $r=\frac{2V}{S}$, but this is correct :)

[warm shale]

anonymemes

[thin stone]

yup

[steel ivy]

@thin stone

[thin stone]

you can find the length pretty easily now

or h in this case

[steel ivy]

Yea one sec

@thin stone it’s basically 3cm right

I used 3.14

With 22/7 yea

Ok

[thin stone]

yeah 3.0012... cm

[steel ivy]

Ok ty

[thin stone]

yw

[steel ivy]

.close

[obtuse pebble]

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[obtuse pebble]

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[stable hollow]

Nah man I don't know how to do log graphs, here is the question:

The relationship between two variables, x and y, is of the form y = ab^x, where a and b are constants. An experiment to text this relationship was performed and provided the data that when log(e)y is plotted against x, the curve passes through the points (0, 2) and (2, 8). Find the value of a and b.

[stable hollow]

I know that I need to turn it to Log(e) = Log(ab^x)

[versed stratus]

not really IMO

first write $2=ab^0$

[warm shale]

Why am. I here

[versed stratus]

that will give you a

then find b by comparing indices

[stable hollow]

.close

[obtuse pebble]

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[obtuse pebble]

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[blissful bane]

Classify the isolated singularity of $\tan z$

[warm shale]

Dubs

[blissful bane]

I think the pole happens when cosz =0 implies z=(2n+1)pi/2

how do I find the order of the pole?

[obtuse pebble]

@blissful bane Has your question been resolved?

[obtuse pebble]

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[random olive]

"(The old man in the box) A cubic box has a corner at the origin and edges OA, OB
and OC with others in the points A = (2, −2, 1), B = (1, 2, 2) and C =
(−2, −1, 2). A dot-shaped old man is at the point with coordinates
(0.9, −0.9, 4.5). Is the old man in the drawer?
Tip: Transfer to a suitable new base. Rephrase the question 'lie in the drawer' in ¨
algebraic terms. What finesse does the change to a new ON base offer?"

[random olive]

could someone help me solve this?

[surreal forge]

do you understand the significance of the tip?

particularly the “Transfer to a suitable new base” part

[random olive]

No not really

Is it to make it easier to calculate?

[obtuse pebble]

@random olive Has your question been resolved?

[random olive]

<@&286206848099549185>

[mystic arch]

i cant do this

[random olive]

okay?

[mystic arch]

stonks

[random olive]

.close

[obtuse pebble]

Channel closed

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[obtuse pebble]

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[green field]

cont := 0
for i := 1 to n do
for j := 1 to i - 1 do
if a[i] < a[j] then
cont := cont + 1
endif
endfor
endfor

Hello folks, I need help calculating average number of times that cont: := cont + 1 executes in this algorithm. The part I'm struggling with is the probability of a[i] being less than a[j]. This is my result, I would appreciate if someone with more proficency on maths than me can tell me if I'm right or I messed up.

[drifting wraith]

needs more context

[obtuse pebble]

@green field Has your question been resolved?

[obtuse pebble]

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[verbal bane]

Need help

[verbal bane]

I know to find area from 0 to 30s but cant seem to figure out right values

[timid silo]

subdivide tbe thing into 3 rectangles? show what u did anyways

[verbal bane]

1st rectangle area= 45*15

2nd rectangle = 10*15

[tardy epoch]

that height is not 45

[verbal bane]

Would it not be 55-10?

yeah

I already tried that

it was 55*15 = 825

• area of the second rectangle = 15*10

[tardy epoch]

,calc 825+150

[warm shale]

Result:

975

[verbal bane]

yes

[tardy epoch]

Show the error

[tardy epoch]

Why are you putting units

Oh it tells you to

[verbal bane]

Yeah

[tardy epoch]

You're putting the wrong units

Read the problem

[verbal bane]

If I enter any other units for ex J/s or kW/s it says incorrect units

So joules is the only correct unit

"The unit in your answer is not correct" 😦

[knotty crow]

kW * s -> kJ

[verbal bane]

ohh

Got it thanks lmaoo

.close

[obtuse pebble]

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[obtuse pebble]

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[vagrant nacelle]

For functions such as
f(x)=(x-1)/(x-2)
how would i describe the transformations? I understand that in functions like 2/x its a vertical stretching by a factor of 2, but I dont know what to write when its something like that

[knotty crow]

do division first

and then it will be simpler

[warm shale]

Modus

[vagrant nacelle]

OH

so for like

2x+1/x-2

hold on

[knotty crow]

now start from 1/x

[vagrant nacelle]

2+5/x-2

OH

okay

tysm

.close

[obtuse pebble]

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[obtuse pebble]

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[coarse night]

Can someone help on this I don’t know where to start

[fathom flicker]

Do not close and reopen a channel just to post the same question

Be patient

What are the points for, is this a test?

[coarse night]

I assumed no one was going to help by looking at old channels, I’m not going to sit and wait for nothing to happen

[coarse night]

Homework, this is all the information I’m given about the question

Question before wants you to understand why vectors a x b = 0

[fathom flicker]

Be patient like everyone else, or don’t use the help channels, you’re not supposed to close and reopen them just because no one has answered yet

[coarse night]

No worries let me wait 5 hours like another guy who wasn’t answered

.close

[obtuse pebble]

Channel closed

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[tardy epoch]

!vol

[obtuse pebble]

Helpers are just people volunteering their time to help you. Be polite and patient.

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[odd harbor]

How do I calculate the flux from the curved surface here,

[odd harbor]

specifically, how to take an elemental area (da) along the curved surface?

s5

[obtuse pebble]

@odd harbor Has your question been resolved?

[obtuse pebble]

@odd harbor Has your question been resolved?

[odd harbor]

<@&286206848099549185>

[obtuse pebble]

@odd harbor Has your question been resolved?

[obtuse pebble]

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[coarse pelican]

what happens to the numerator? sorry to whoever i just closed i actually didn’t mess up the question

[coarse pelican]

2-2-x

isnt 2-x a hole?

can i just substitute both of them with one in the numerator and the denominator?

[obtuse pebble]

@coarse pelican Has your question been resolved?

[short wing]

Your Last step is incorrect

[obtuse pebble]

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[civic socket]

hai

[civic socket]

I need help with this question

!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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[civic socket]

1

[latent walrus]

very efficient of you

try finding the derivative first

[civic socket]

okie

[versed stratus]

I'd probably differentiate it and the solve an inequality

I encourage you to think what that inequality would be

hint ||dy/dx>0|| (spoilers)

[civic socket]

yea I did that and got $\frac{ad-bd}{(cx+d)^2}$ sorry for slow latex

[warm shale]

MSC2020 11N36 (Akira)

[civic socket]

idk what's after

[versed stratus]

,w differentiate (ax+b)/(cx+d)

[warm shale]

Results provided by WolframAlpha

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

Requested by physicsrocks

[latent walrus]

lovely

[versed stratus]

$\frac{ad-bd}{(cx+d)^2}>0$

[warm shale]

Why am. I here

[versed stratus]

why is this?

[civic socket]

wdym

[twilit meteor]

ad-bc>0

[versed stratus]

why should dy/dx>0

[civic socket]

no idea

[twilit meteor]

coz the slope is positive

[latent walrus]

why are you answering my man?

[civic socket]

yeah

[twilit meteor]

k nvm

u ans

[civic socket]

okay so I don't get why it should be dy/dx>0 but its just the slope?

if we had a slope was negative we can say it's dy/dx<0 right

alright got it thank you

.close

[obtuse pebble]

Channel closed

Closed by @civic socket

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[timid silo]

look into mobius transformations @civic socket

[versed stratus]

pretty sure this is a problem involving elementary calculus

[obtuse pebble]

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[solar roost]

The table summarizes results from pedestrian deaths that were caused by automobile accidents.

Pedestrian Deaths
Driver
Intoxicated? Pedestrian Intoxicated?
Yes No
Yes 56 68
No 248 605
If two different pedestrian deaths are randomly selected, find the probability that they both involved pedestrians that were not intoxicated.
Report the answer rounded to four decimal place accuracy.

[gilded needle]

!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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[solar roost]

I am trying to figure out rn, and I think I might have it give me a moment

okay sorry I figured it out, I reailsed i was doing something wrong

.close

[obtuse pebble]

Channel closed

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[obtuse pebble]

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[sharp ether]

I got this wrong, could someone help me out?

[sharp ether]

plz

[tardy kindle]

!f 10

[sharp ether]

huh?

what?

hello?

[obtuse pebble]

@sharp ether Has your question been resolved?

[timid silo]

No lol I got it tho

[sharp ether]

bro...

I was willing to wait lol

[timid silo]

If I can help should I not?

[sharp ether]

no, plz help me~

https://tenor.com/view/vergonha-gif-1631438418695329354

[timid silo]

Lets set up a relationship where gear 1 (the larger) has T1 and R1 where T1 is the teeth, and R1 is the rpm

T1 * R1

Then for the other gear (smaller), lets have T2 and R2

so T1 * R1 = T2 * R2

Now you can fill in the information, and solve it out algebraically, solving for X where x is the teeth of the smaller gear

[sharp ether]

huh?

ok ok ok give me a sec

[timid silo]

$T1R1=T2R2$

[warm shale]

spectral

[sharp ether]

so 50times110=xtimes275 right?

[timid silo]

yeah

[sharp ether]

ok give me one sec I am really dumb

got it!

its 20!

[timid silo]

Nice, don't say you're dumb tho, we all have our moments

[sharp ether]

says the person who majoring in computer something~

[timid silo]

Says the person majoring in law

[timid silo]

But this discussion isn't for this channel

[sharp ether]

lol

see ya~

[timid silo]

I was getting there austin

[sharp ether]

.close

[obtuse pebble]

Channel closed

Closed by @sharp ether

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[obtuse pebble]

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[hexed gull]

what's your approach so far

[daring vessel]

nope.

it's too hard

[hexed gull]

the wording "sensible" :D

[daring vessel]

it's too sensible

[hexed gull]

Attempt to sketch it

It doesn't have to be numerically acurrate

just so you can envision the task

and then you should know which formula among sin/cos/tan you need to get the height

[daring vessel]

yes I know

[hexed gull]

k

[daring vessel]

sin50 = h/6

h= 6sin50

right?

[hexed gull]

ys

[daring vessel]

why is math so east

easy*

[hexed gull]

bc you wear a crown I suppose

[daring vessel]

yes

wait please don't leave me

[daring vessel]

halp

halp

no

I regret it

pls help

PLEASE

[hexed gull]

I perceive it to be a timed test so no direct answers

[daring vessel]

ok

wait what

it's not test >:[

this is base -> bottom

[hexed gull]

mind proving? Someone else just stated the same

regarding the same task

[daring vessel]

a practice book :]

[hexed gull]

kk :]

[daring vessel]

Noice

now pls help

how long is the tope

rope*

[hexed gull]

You know the difference on the ground is 6m

[daring vessel]

yes

[hexed gull]

And you know the angle from wall to rope

so you can apply sin

[daring vessel]

is this cos19 = 6/R

[hexed gull]

ys

[daring vessel]

omg this too easy

wait

what about window

[hexed gull]

and lastly window height is the same

but cos ofc

since you seek the side next to the angle

[daring vessel]

hmm

so is it

idk

HALP

[hexed gull]

if yk sin then you should get cos as well

[daring vessel]

how

[hexed gull]

sin = OPPOSITE / HYPOTHENUSE

[daring vessel]

K

[hexed gull]

what's cos

[daring vessel]

cos =hyp / opp

[hexed gull]

nop

adjacent / hypothenuse

[daring vessel]

oh

ok

[hexed gull]

and tan = opposite / adjacent

from here on you'll get it

[daring vessel]

adjacent side and hypotenuse

ok

How do I get now

[hexed gull]

yop

[daring vessel]

pls help

[hexed gull]

huh you know all variables you need

[daring vessel]

yes

[hexed gull]

cos(angle) = adjacent / hypothenuse

hypothenuse is the rope

adjacent is the wall

[daring vessel]

cos45 =

what is the adj

is it 6

[hexed gull]

it's the window height ofc

15.4 even visualizes it :D

you get the same explanation there which you get here

[daring vessel]

wait I know how

6m * sin(19°)

why does math exist

it's toooo easy

alright peace

.close

[obtuse pebble]

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[hexed gull]

K

[obtuse pebble]

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[tawdry path]

Hey guys, I was working on a problem that was essentially a carbon copy of the one here https://math.stackexchange.com/questions/263229/how-do-we-prove-that-fx-has-no-integer-roots-if-fx-is-a-polynomial-with. The official solution in my book was the same as the first reply and I actually understood most of it except for the part where it said a-b must be a factor of f(a)-f(b). I've just never learned it so can someone prove it to me/point me to a proof I'd really appreciate it, thanks

[twin sandal]

this is a consequence of the fact that if a and b are integers and n is any power, (a-b) | (a^n - b^n)

f in this case is a polynomial, so f(b) - f(a) will be a_n (b^n - a^n) + a_(n-1) (b^(n-1) - a^(n-1)) + ...

[tardy epoch]

couple more proofs here:
https://imomath.com/index.cgi?page=polynomialsWithIntegerCoefficients

[tawdry path]

Ok I think I get it

Thanks for the help

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[drifting mortar]

How to do 1/4 roation coutner clock wise (topic is trig like terminal angles smth like that)

[hexed gull]

via matrix?

[drifting mortar]

No

like

Lesson 9.2 Angles and Radian Measure
this is the name of the topic

[hexed gull]

in what way do you perform the rotation?

coordinates to new coordinates & rotation around origin?

[drifting mortar]

The queston is Given the roation of an angle, sketch the angle in standard position and give its measure in degrees

ie

a 1/4 roation CCW(counterclock wise) would be 90 but how do i show that?

[hexed gull]

well a full rotation is 360°

1/4 * 360° = 90°

[drifting mortar]

omgaull I am dumb

one sec

that what it is

thanks

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[tiny temple]

Let V = The set of all n × n skew hermitian matrices. check whether V is a real (or
complex) vector space under usual addition and scalar multiplication of matrices

[tiny temple]

I am a bit confused by what field I am supposed to be taking here

[haughty coyote]

hermitian -> complex coefficients

[tiny temple]

cuz the vectorspace itself will obviously be Complex, cuz a skew hermitian matrix can have complex entries

[tiny temple]

yeah,
but what about the field?

[haughty coyote]

So C-vector space

[tiny temple]

i want it to satisfy closure under scalar mulitplication

[haughty coyote]

Wait

[tiny temple]

that is fine
but what field are we talking?

[haughty coyote]

It asks you to check for both

"whether V is a real (or complex) VS"

[tiny temple]

it can't be for real anyways though
cuz skew hermitian can have complex entries, so no way the set of all nxn such matrices will be real

[haughty coyote]

Real vector space

C is a R-VS

This is fine

[tiny temple]

huh?

[tiny temple]

how?

[haughty coyote]

C ~ R^2

[tiny temple]

can you explain with words
i am new to this language of symbols

[haughty coyote]

C is a R-VS isomorphic to R^2

[tiny temple]

you mean its similar to R^2?

[haughty coyote]

I mean the precise definition of vector space isomorphism

[tiny temple]

yeah so they are similar you are saying
but how can you prove that?

also, why is that even required

[haughty coyote]

I'm simply arguing to you that C is a R-VS

[tiny temple]

why

[haughty coyote]

Exercise

[tiny temple]

wdym?
explian urself

[haughty coyote]

It's a much easier and far more fundamental exercise than this so you should start with it

[tiny temple]

what excercise are you even talking about?
can you simply just clear my doubts instead

[haughty coyote]

This is a fact
It may as such be proven
Proving it is an exercise

[tiny temple]

then show me how to prove it

[haughty coyote]

No

Just check all VS axioms

[tiny temple]

you could have said so in the beginning

[haughty coyote]

I was never I was gonna help you prove it
I take it as obvious so I leave it to you as a practice exercise
If you can't do it you clearly don't know your theory well enough

[tiny temple]

i would appreciate if you could stop with the assumptions now if ur not help
i'll wait for 5 more min and then i'll ping helpers for someone else to clear my doubts
thank you and have a nice day

[tardy epoch]

don't be entitled

[fathom flicker]

Do you know that C is a real vector space

[obtuse pebble]

@tiny temple Has your question been resolved?

[fathom flicker]

@tiny temple what do you expect help with?

[tiny temple]

i was
they said they didn't want to show me how

[tiny temple]

i simply wanted to know
what field will we be taking for proving it to be a vectorspace

[tiny temple]

i am not
that was a request

[fathom flicker]

It says, real or complex

So you’d check over both fields

[tiny temple]

yeah
so, with both real and complex
it will be a vectorspace if we take it over the real field
however over the complex field, i was having a doubt that it won't be a vectorspace

[fathom flicker]

Why is that?

[gilded needle]

your doubt is warranted

[fathom flicker]

Also, the question doesn’t say it has to be, it basically wants you to check if it is or is not

[tiny temple]

cuz when I take the conjugate over the complex scalar
and then multiply it with the complex entries of my skew hermitian matrix
i am not getting -A as I need to for satisfying the critera of skew hermitian matrices

[tiny temple]

no no
see
that's the thing
the question is asking us about the vectorspace being real or complex
it does not specify which field are we taking it over

[fathom flicker]

It is asking you to check both

[gilded needle]

"real vector space" means "vector space over the real scalar field"

[fathom flicker]

It is asking you, is it a real vector space, yes or no? And also, is it a complex vector space, yes or no?

[tiny temple]

ahh

oh shi
i see now

hm
so, over the real field
lets say R is my scalar
[0 Ra-Rbi ]
[ Rc-Rdi 0 ]
for the matrix
[ 0 c +di]
[ a+bi 0]
I'd say for the real scalar field, its fine and is a vector space
cuz its closed under mulitiplication
but for complex field
R will change the matrix, so i think its only a vectorspace over the real field and not the complex one

[tiny temple]

@fathom flicker @gilded needle please crosscheck

.close

[obtuse pebble]

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[kind hawk]

cus what is closed under multiplication

also you are was too unspecific about what exactly changes when you allow R to be complex