help-33

[urban bobcat]

so do we solve this?

[hallow crown]

Yea pretty much

[urban bobcat]

e^t = 2 or e^t = 1?

[hallow crown]

You can if youre allowed to use hyperbolic cosine and hyperbolic sine

[urban bobcat]

hmm i am not sure if i know what that is

[hallow crown]

cosh and sinh

If you dont know its ok you can solve with a quadratic equation

[urban bobcat]

yes okay i can try it with sin and cos after i have solve it later!

[hallow crown]

sinh and cosh btw not sin and cos

[urban bobcat]

ohh sorry

so t can be either 0 or ln(2)?

[hallow crown]

Yeah but 0 is impossible

[urban bobcat]

oh yeah okay

so i can put in ln(2) into r´ now?

[hallow crown]

Yep

[urban bobcat]

(3/2, 5/2, 23/4)?

its this right?

[hallow crown]

r(A) + tr'(A)

Remember, tangent at point A

[urban bobcat]

okay so this?

[hallow crown]

ln(2)?

[urban bobcat]

is that not our t?

[hallow crown]

No in a linear approximation your t varies

[urban bobcat]

ohh okay so that but with just t instead?

[hallow crown]

Yea pretty much

[urban bobcat]

okay i get this then

[hallow crown]

@urban bobcat i think it would be (3/2, 5/2, 17/2) on the right

[urban bobcat]

ohh yeah you are right

[urban bobcat]

okay this but with 17t/4 instead

[hallow crown]

17t/2 but yea

[urban bobcat]

oops yes

okay so that is all?

[hallow crown]

Regarding this last question you can study the max and min by calculating the partial derivatives and setting them to 0 to get a system of equations

Extremums*

Which would get you also the range of the function

In both cases i guess youd get what the question asked for

[urban bobcat]

Yes okay thank you for the help!

[hallow crown]

No problem best of luck

[urban bobcat]

Thank you!

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[supple umbra]

can someone explain this one?

[neat oak]

is that the cross product?

yeah then its trivial

cross product of parallel vectors is 0

|a x b| = |a||b|sin(x)

if a and b are parallel then x=0

[supple umbra]

thx, that is so obvious

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[fossil grail]

If given lateral surface area of a cone, how do I find the cones with maximum volume?

[fossil grail]

or rather, how do I make a function for V (for the volume) to find the global maxima

I've gotten 1/3pi r^2 sqrt(S^2/(pi r)^2-r^2)

but by observing, there's no global maxima

for r -> inf, S - some fixed number, S->inf, r - some fixed number

Lateral surface area being S= pi r l

[novel juniper]

wait, so given lateral surface area, you want to maximise volume

that's it, right?

[fossil grail]

yes

[novel juniper]

ok, so let the semi-vertical angle of a cone be \theta

and the radius of the base be r

express lateral surface area in terms of r and $\theta$

[elfin berry]

ƒ(Why am. I here)=I don't know

[novel juniper]

and volume too

you can do it from here, me thinks

[fossil grail]

uhh, semi-vertical angle?

[novel juniper]

alpha is the semivertical angle

here

[fossil grail]

how do I express the lateral surface area via alpha

or theta

[novel juniper]

what's teh formula for lateral surface area

[fossil grail]

S = pi r l

[novel juniper]

what's l in terms of r

and theta

[fossil grail]

sine

I think?

[novel juniper]

ok, I kind of have to go now, sorry

[fossil grail]

alright

[marsh citrus]

@fossil grail Has your question been resolved?

[marsh citrus]

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[frank eagle]

Consider $n$ sets $B_1,\ldots,B_n$, not necessarily disjoint that partition another set $X$, meaning their union is $X$. Consider the collection $\mathcal{C}$ of sets of the form $C_1\cap \ldots\cap C_n$, where $C_i$ equals $B_i$ or $B_i^c$. There are $2^n$ such sets, though they may not all be distinct (many of them may be the empty set). One can also show that the distinct sets from this collection are pairwise disjoint. \

However, something I can not understand is why the union of the sets in $\mathcal{C}$ where $B_1$ occurs equals $B_1$. How do you show this?

[elfin berry]

Philip

[frank eagle]

The complements are taken in X, so X is the entire space, the universe.

[devout mauve]

well the union is clearly a subset of B_1. because all of the sets are subsets of B_1 because they are of the form B_1 intersect (something)

so you only need to show that every element b in B_1 is also in the union

[frank eagle]

indeed, good point

[devout mauve]

and for every element b you can tell whether its in B_i or B_i^C

[frank eagle]

ok, I think I understand, thank you👍

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[vestal mist]

how would i do this?

[red nimbus]

Use this

[vestal mist]

thanks

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[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[cyan dove]

this is wrong somehow

[hazy lion]

,w Solve[11x^2 - 15x + 10 == 0, x]

[elfin berry]

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Requested by janniku

[hazy lion]

i believe they want it to be =0

not =y

would be my guess

[cyan dove]

oh yeah fr

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[gentle remnant]

Any tip on this stats question?

[gentle remnant]

My attempt

[hallow crown]

@gentle remnant for Y = -1 it would not be 1/6

[gentle remnant]

what would it be?

2/3?

[hallow crown]

Yep

Makes sense since the sum of probabilities needs to be equal to 1

[gentle remnant]

ye

but

like

is my apporach correct?

[hallow crown]

Yea

Looks correct

[gentle remnant]

but then how does it differ for the sum for -1?

[hallow crown]

Just some error in calculations

Just recalculate it briefly and youre sure to find 2/3

[gentle remnant]

ye nvm

I c i t now

thank uuuuu

[hallow crown]

No problem

[gentle remnant]

started at wrong index

[hallow crown]

Yea happens

[gentle remnant]

.close

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[gentle remnant]

lifesaver

ppl champion

[hallow crown]

Nahhhhh

Best of luck thi

Tho*

[gentle remnant]

.reopen

[marsh citrus]

✅

[gentle remnant]

Can sm1 explain what the approach is to this question?

[valid cape]

!1q

[marsh citrus]

It is suggested that you limit yourself to one question per help channel, opening a new one once your original question is answered and your original channel has been closed. This is to make your channel easier to follow for potential helpers and can bring attention to the fact that your question has changed.

[gentle remnant]

oops my bad

ok

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[wooden elbow]

Wouldn't P(Y=-1) just be the complement of P(Y=1)?

[gentle remnant]

Ye solved it

Indexing issue

so that's why I thought my approach for finding P(Y=1) was wrong

but was correct

[marsh citrus]

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[cyan dove]

-b so b = -2

b^2 -4ac

so 4+4c = 4

or 4 - (-4c) = 0

[eternal sand]

I suggest to use sum and product

[late geode]

it's incorrect to set D = 4

[cyan dove]

D?

[eternal sand]

sum of root and product of root

I suggest

[late geode]

discriminant, the b^2 - 4ac

[cyan dove]

oh is it supposed to be sqrt16

im dum

[late geode]

yeh, 4 = sqrt(16)

[cyan dove]

c=3

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[bright badger]

Can I get help with this? Find the product of the positive divisors of 6480 that are multiples of 12

[bright badger]

You have to use Product notation and I am confused.

[marsh citrus]

@bright badger Has your question been resolved?

[marsh citrus]

@bright badger Has your question been resolved?

[bright badger]

Can I get help with this? Find the product of the positive divisors of 6480 that are multiples of 12

You have to use Product notation and I am confused

[marsh citrus]

@bright badger Has your question been resolved?

[marsh citrus]

@bright badger Has your question been resolved?

[bright badger]

Can I get help with this? Find the product of the positive divisors of 6480 that are multiples of 12

You have to use Product notation and I am confused

[normal sky]

hint 6480 is already divisible by 12

and as for product notation, you'll have $\prod_{d\mid6480, 12\mid d} d$

[elfin berry]

chebyshev's infinite pee norm

[normal sky]

you can simplify with the fact that 12 divides 6480

[marsh citrus]

@bright badger Has your question been resolved?

[marsh citrus]

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[lucid flare]

am i doing guassian elimination wrong

[blazing rampart]

where did you get stuck at

[lucid flare]

the results i get are wrong

i dont get stuck anywhere

i did elimination twice and got the same wrong solution twice

so maybe i misunderstand the algo

[hard gull]

nvm

[lucid flare]

the z I get is wrong

[hard gull]

whats the correct answer?

@lucid flare is z = -1/4?

[lucid flare]

yeah

[lucid flare]

i said 6/-24 = -4

instead of -0.25

xd

thanks

[hard gull]

i checked ur entire work and the mistake was really simple

np

[lucid flare]

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[flint reef]

Determine a vector equation of the plane using lines: r = (4, -3, 5) +t(2, 0, -3), tER. and r = (4, -3, 5) +s(5, 1, -1), sER

[flint reef]

Would the answer be (4, -3, 5) +s(2, 0, -3) + t(5, 1, -1), s,tER?

[tight furnace]

that's the plane that contains both of the lines yes

[flint reef]

Ok that's helpful thank you, but if both lines did not contain (4, -3, 5) and they contained different points, how do ik which one is correct?

Unless I can choose one or the other?

[tight furnace]

You can't choose either

If the lines don't share a point and aren't parallel then there won't be a plane

[flint reef]

Wait but doesn't the direction vector determine if it's parallel

Not (4, -3, 5)?

[tight furnace]

Yes

I'm saying there's no general way for the starting points aren't the same because there's no guarantee there will be a plane at all

[flint reef]

Ok, so the lines above have direction vectors that aren't scalar multiples of each other meaning they aren't parallel

[tight furnace]

If the lines intersect, move the starting point to the point of intersection

[flint reef]

So would they not create a plane at all?

[tight furnace]

No, they do create a plane

Because they intersect

[flint reef]

Oh hmm ok

[tight furnace]

And if they intersect you can write both of them as

point of intersection + t*(direction vector)

here it's already written like that

[flint reef]

Ok I understand what ur saying

[tight furnace]

otherwise if they're parallel you can choose either point as your starting point and the second direction vector will be the vector between the starting pointa

actually as long as the lines intersect you can use any starting point you want

[flint reef]

So if lets say if their points were (4, -3, 5) and idk (7, 2, -9) respectively there would be no plane?

[tight furnace]

Assuming those lines don't intersect somewhere

yeah

If they intersect you can use either starting point as a starting point or the point of intersection or whatever else you want

just has to be any point on the plane

[flint reef]

To determine that, all I have to do is see if it intersects is putting point (4, -3, 5) into the parametric eqn of the other line to see if it matches up

[tight furnace]

no, just because it intersects doesn't mean it will intersect at (4,-3,5)

[flint reef]

Wait ya ur right

[tight furnace]

You have to set the equations for the lines equal to each other and solve

[flint reef]

Yes I remember this

[tight furnace]

using different variables for t, like t_1 and t_2

or s and t

[flint reef]

Ok I think ik what to do ur were helpful thank you

[tight furnace]

np

[flint reef]

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[polar magnet]

this is what I have done

[placid hound]

Did you try applying C2 = C2 - C1 and then C3 = C3- C1 and then developping along the last row?

(i didn't do it yet, but seems like a good start)

[old cradle]

this question is from the topic differentiability....Can sm1 help me understand this question ..and help me solve this question.....

[placid hound]

Try using an availlable Channel ;)

[marsh citrus]

@polar magnet Has your question been resolved?

[polar magnet]

no

I thought it would complicate it

lemme try

ahh
got it
answer is 0

thank you

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[knotty trellis]

.open

[still temple]

Is it possible that after integrating something we get a negative value
If yes so why. Because AREA under curve would be always positive no?

[still temple]

Area is alwys positive right

[desert dirge]

Area is positive as a measure, integral is not always so

integrals do 'signed areas'

if its below the x axis itll come out negative, if above then positive

[still temple]

If velocity is sinx
So integral is -cosx
So displacement negative?

[desert dirge]

not necessarily, no

depends over what interval youre doing it

if you do it from 0 to 2pi, displacement is 0 for example

but distance travelled would be 4

from pi/2 to 3pi/2 would be positive displacement

oh ive misread that

velocity is sin(x)

the principle applies though

[still temple]

Can u give me example of a velocity function by displacement can be positive and negative

[desert dirge]

y=x
y=cos(x)
y=sin(x)
y=x^3
y=x^2-12

anything that is on both sides of the axis at some point

v(x)=:
-1, x<=0
1, x>0

anything

[still temple]

No I mean such functions where first velocity and increasing then decreses and then increases

Basically like sin wave yea

[desert dirge]

v=sin(x)

[still temple]

So how displacement-time graph would look which is derived from it?

[desert dirge]

displacement will be s=-cos(t)+c
assuming its 0 at t=0 then s=-cos(t)+1

[still temple]

Cos or -cos?

[still temple]

Oh sorry I didn't see "-"

[still temple]

I have a doubt
In velocity ( sinx) initially velocity was increasing so why in -cosx graph initially displacement is in negative?

[still temple]

@desert dirge

[desert dirge]

because we integrated sin(t) to -cos(t)+c

and weve just assumed c=0
which means at t=0 s=-1
it then increased precisely because sin(t) is positive

[still temple]

That's math
Now let's imagine a velocity of car is sinx
So initially it's velocity is positive so displacement won't EVER be negative no?

[desert dirge]

are you assuming displacement is 0 when starting?

if so, then yes, itll never be negative, just fluctuate between 2 and 0

[still temple]

Yep

[still temple]

Soo... How the graph would look like...

[still temple]

Which graph is this?

[desert dirge]

displacement

[still temple]

-cosx?

[desert dirge]

-cos(x)+1

[still temple]

Ohh.. Cool cool
So one more doubt
Magnitude of positive and negative integral of sinx is equal?

[desert dirge]

you realise when you ask these things, youre not saying enough right

over what interval

0 to 3pi/2? no its not

0 to 2pi? yes it is

[still temple]

As in?

[desert dirge]

it would be 2+(-1)=1

[still temple]

Ohh. Okok
So again same question
Is there any function which has increasing and decreasing velocities through which we can see positive as well as negative displacement in graph

[desert dirge]

v=cos(x), if you want one that has s=0 at t=0
s=sin(x)

can make most functions into one if you choose the right starting point

[still temple]

Oh damn
Thanks a lot man
Thanks very very much

You freaking made me day good

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[twin ether]

I messed up, but I don't know where

[stoic slate]

Before last step

You had a minus sign with a parenthesis

When you eliminated the parenthesis you didn’t eliminate correctly

@twin ether

[twin ether]

ahhhhh

these are always my mistakes

so the last cos should be negative

$\frac{\cos{x} - \sin{x}\sqrt{3} - \sin{x}\sqrt{3}-\cos{x}}{2}$

[elfin berry]

rynite

[twin ether]

like that?

[stoic slate]

Yes, you can simplify much more

[twin ether]

alr sec

[twin ether]

$\frac{-2\sqrt{3}(\sin{x})}{2}$

[elfin berry]

rynite

[stoic slate]

Doesnt look correct

[twin ether]

let me check again

wait I forgot the cos

[stoic slate]

Forget about it

No it was my dyslexia

Just simplify more

[twin ether]

alr

[stoic slate]

So u get

-sinxsqrt3

Right?

[twin ether]

$-\sqrt{3}(\sin{x})$

yeah I think so

[stoic slate]

Negative

-2/2 = -1

[elfin berry]

rynite

[twin ether]

alright

[stoic slate]

To me thats correct, lets check

,w is cos(x+pi/3)-sin(x+pi/6)=-sinxsqrt3?

[elfin berry]

Results provided by WolframAlpha

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Requested by _arkimond_

[stoic slate]

Yes

[twin ether]

awesome! thank you very much

I'll close this.

.close

[marsh citrus]

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[light swift]

can someone explain why one is negative

[small berry]

I mean there’s a whole derivation. Have you went through it? https://tutorial.math.lamar.edu/classes/de/VariationofParameters.aspx

[light swift]

ah ill take a look at tha tty

does which ones negative matter though

eh ill read it ty

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[dreamy hull]

Hi, I need help with some basic vector exercises in R².

[void elm]

[dreamy hull]

It asks me to determine the components of the vector represented in each case

[void elm]

right, I think you need to shift the origin to the base of each vector

[dreamy hull]

that would be the easiest thing to do, but I would have to do it analytically or justify it

in the first one I was thinking of doing something like this:

Vector AB would be OA + OB right? I am not sure

[final skiff]

that would be <0,2>

[final skiff]

You can justify it by the distance between the head points and tail points

[marsh citrus]

@dreamy hull Has your question been resolved?

[dreamy hull]

is what I said correct? the vector AB should be (4, -2)

[final skiff]

yeah

[dreamy hull]

wouldn't it be OB - OA?

[final skiff]

yeah that is basically what I said earlier

That is a better way of saying it though

[marsh citrus]

@dreamy hull Has your question been resolved?

[marsh citrus]

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[obtuse quest]

g(x) = x^2-3x-4 , f(x) = -x^2 + 1
C and A are the intersecting points between the two functions
B is the other intersecting point of one of the functions with the X axis
Find the area of Triangle ABC
I need help with this ^

[errant garnet]

you can calculate the x values using the functions

so we can find B by plugging in 0=-x^2 + 1

so x=1

[obtuse quest]

oh I already found B, A and C

I'm just not sure how to calculate the area of the triangle

[errant garnet]

it would be (1/2)bh no?

[obtuse quest]

what would h be

[errant garnet]

theres probably a more elegant way but you could just apply the distance formula

[obtuse quest]

i tried that and the result only corresponded to the one in the answers listed in the paper if rounded up

[errant garnet]

wdym? wouldn't that make it correct?

[obtuse quest]

I guess, but this isn't what we were taught so I don't think it's the correct way

[errant garnet]

what do you normally do for this?

i would normally just brute force this with the distance formula but like i said there are probably some geometric properties to better solve it

are you meant to integrate this?

[obtuse quest]

either Pythagorean or Just the AB*h/2

I could skip this one part, as the others on this question don't require it to solve them

[errant garnet]

if you know how to integrate you could come up with the equation for the line between B and C and A and C, then integrate over A to B

[obtuse quest]

what do you mean by integrate? if this is a math term, i am not familiar with those in english, I had to translate the question I got

[errant garnet]

its a calculus term which involves taking the area under a curve, or between two curves

i noticed you had an integral written in the background

[obtuse quest]

ah, that's just some jumble written in the background

obviously this is in hebrew, so I can't translate it

maybe I didn't translate all of the info correctly

[errant garnet]

sorry but I'm honestly not sure of any other way to do it

I only know how to sovle it using calculus or with the 1/2 * b * h equation

[obtuse quest]

I know the answer is supposed to be 5.25 squared units but they don't show how they got there

I'll skip (א) (finding the area of the triangle) and do the other ones and maybe come back to it later

thanks!

[errant garnet]

alright, good luck!

[obtuse quest]

how do i close this

[errant garnet]

you would do .close

also this may help later: https://www.youtube.com/watch?v=nW0mNaG3Z2o

[obtuse quest]

.close

[marsh citrus]

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[marsh citrus]

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[glad burrow]

^^

[fierce field]

!

[glad burrow]

ok please continue ur explanation >_>

[fierce field]

Alright so lets say i have the equation: a = b + c

and i want to solve for b

do you know what i can do ?

i want b to be all by itself

[glad burrow]

isolate it?

[fierce field]

yeah, and how can i do that

[glad burrow]

i think you multiply? im not sure.

[fierce field]

well what happens if i subtract c from both sides

[glad burrow]

then it disappears right ?

[fierce field]

what disappears ?

[fleet plover]

has your teacher ever introduced the idea of "whatever you do to the left side you do to the right side"

[glad burrow]

yes i think tat was like 2 months ago

[fierce field]

a - c = b + c - c

[glad burrow]

i think i can pull up an example question he show'd us

[fierce field]

gives you a - c = b

because the c's "cancel out"

[fleet plover]

no need its just the same principle here.

[glad burrow]

ohh ok i remember now

[fierce field]

so you notice i had a = b + c and then i subtracted c. Because subtraction is the opposite of addition.

now if you have a = b/c

you can do a similar thing

but with just a different operator

as we divide b by c

we now have to multiply both sides of the equation by c

$a \times c = \frac{b}{c} \times c = a \times c = \frac{bc}{c} = a \times c = b$

do you understand this process?

[elfin berry]

SollyPolly

[glad burrow]

a lil bit ?

[fierce field]

which part dont you quite grasp?

its okay

[glad burrow]

kinda like the whole equation u just put ^

[fierce field]

thats okay

so we have $a = \frac{b}{c}$ and we want to isolate b, or in proper terms, we want to make b the subect. we do this by multiplying both sides of the equal sign by $c$. This is because multiplication is the opposite of division. just like subtraction is the opposite of addition. so when we do this we get $a \times c = \frac{b}{c} \times c$

[elfin berry]

SollyPolly

[fierce field]

we'll go step by step, do you understand this part?

its completety okay if not, just let me know

[glad burrow]

this is a lil bit more clear then the equation u put before lol

[fierce field]

yeah ik :(

so you understand this step

multiplying both sides by c

[glad burrow]

yes i understand it now

[fierce field]

okay so now we can simplify this right hand side

are you able to do that?

remember $\frac{b}{c} \times c = \frac{b \times c}{c}$

[elfin berry]

SollyPolly

[glad burrow]

like which part do I simplify?

[fierce field]

well we can actaully remove c right now

but can you realise why

what if they were numbers instead? lets say $\frac{7 \times 3}{3}$

[elfin berry]

SollyPolly

[fierce field]

what does this equal to?

[glad burrow]

7?

[fierce field]

yes!

which is weird because thats the first number...

[glad burrow]

true...

[fierce field]

this will happen, the 3 on the top and the 3 on the bottom will cancel out

as we can technically rewrite it as $\frac{7 \times 3}{3} = 7 \times \frac{3}{3}$

[elfin berry]

SollyPolly

[fierce field]

and how many 3's go in 3

just 1!

now back on track

$\frac{b \times c}{c} = b$

[elfin berry]

SollyPolly

[fierce field]

so we have simplifed this right hand side of the equals

now lets remeber what the left hand side was

$a \times c = \frac{b \times c}{c}$

[elfin berry]

SollyPolly

[fierce field]

but that is now just $a \times c = b$

[elfin berry]

SollyPolly

[fierce field]

so we managed to go from $a = \frac{b}{c}$ to $a \times c = b$

[elfin berry]

SollyPolly

[fierce field]

by multiplying both sides by c

Now the origonal question here: https://cdn.discordapp.com/attachments/907022703230345237/1229222143187877980/image.png?ex=662ee554&is=661c7054&hm=f4a84887da114db01ecca75c8bf9be7c21455f92f7edd9e6aa0dea8c21eb59dd&

$D = \frac{m}{v}$ we've moved $v$ to one side, now we'd like to make sure its alone. So now we divide both sides by D.

[elfin berry]

SollyPolly

[glad burrow]

by any chance can you use numbers as an example like u did before?

[fierce field]

sure

so we have move v to the side with D

as v = c and D = a

in the example i used earlier

so we have $D \times v = m$ and we want to divide both sides by $D$ now. $\frac{D \times v}{D} = \frac{m}{D}$

[elfin berry]

SollyPolly

[fierce field]

now can you notice something similar from before?

$\frac{D \times v}{D}$ looks suspiciously similar to a previous example

[elfin berry]

SollyPolly

[glad burrow]

it does .

the one with 7 before?

[fierce field]

yeah!

so in this case

what does this simplify to?

[glad burrow]

m / d?

[fierce field]

focus on just this fraction for now

$\frac{D \times v}{D}$

[elfin berry]

SollyPolly

[fierce field]

before we had $\frac{b \times c}{c}$

[elfin berry]

SollyPolly

[fierce field]

and we were just left with b in the end

because the top c and the bottom c, cancelled each other out.

[glad burrow]

oh is it like now d/d then?

bec v gets cancelled out right?

[fierce field]

not quite but almost

in the b,c we cancelled out the c.

leaving just b

so in this case we cant cancel out the v because there is just 1.

so what can we cancel out?

[glad burrow]

this one right?

[fierce field]

yes

[glad burrow]

can we cancel out d because there is 2 of them?

[fierce field]

yes!

Leaving us just with v.

on the left hand side of the equals

and on the right hand side

well. Whatever we do to one side

we have to do it to the other.

so on the left, we had D * v

and the divided by D

so we also have to do that on the right side.

If u can remember the right side was just m.

so can you write down what the right side is now?

[glad burrow]

m/d ? because u gotta divide them right?

[fierce field]

yes!

[glad burrow]

m was by itself so u added d under it

[fierce field]

yes

so we done the first question

we had D alone but v wasnt.

so we first

multiplied both sides by v.

leaving us with D*v=m

and then divided both sides by D

leaving us with v = m/D

are you happy with this?

[glad burrow]

yes :)

[fierce field]

lovely

Do you think you could have a go at (b)?

[glad burrow]

maybe , also does it always involve multiplying both sides or other factors?

like dividing

is ther ever a problem were you gotta divide or is it only multiplying

[fierce field]

there can be

lets say for example

[glad burrow]

how would you know?

[fierce field]

well actually in the first question

we did both

[glad burrow]

ohh right

[fierce field]

we multiplied both sides by v

and then divided by D

[glad burrow]

would you ever need to divide both sides ?

lik yk what i mean?

[fierce field]

do a division twice?

[glad burrow]

yes

[fierce field]

well lets see

$D \times P = \frac{m}{v}$

[elfin berry]

SollyPolly

[fierce field]

now i would multiply both sides by v

to get $D \times P \times v = m$

[elfin berry]

SollyPolly

[fierce field]

and now you have two options

you can either divide twice

first by D

then again by P

or you can just divide it once but you divide by D * P. in one big step

There are probably further examples which have tougher questions

Also if the letters are written right next to eachother like DP

it means $D \times P$

[elfin berry]

SollyPolly

[fierce field]

just so u can read the answers yourself

[glad burrow]

okk i think i can solve b) myself , can u assist me incase i do anythign wrong ?

[fierce field]

of course

[glad burrow]

so like for
d = m/v you multiply both sides

[fierce field]

by what tho :O????

[glad burrow]

by v

[fierce field]

yes!

[glad burrow]

since its at the bottom

[fierce field]

yup

[glad burrow]

so then

would dv = m/d?

[fierce field]

why is m/d here?

wait

are you doing

[glad burrow]

because of the d at the first part

[fierce field]

(a) or (b)?

[glad burrow]

b

[fierce field]

Have a look, you just want to have m by itself.

and you dont do anything to oneside without doing it to the other.

[glad burrow]

does the v at the bottom cancel out?

like after you x both sides

[fierce field]

yes it cancels out when you multiply by v

[glad burrow]

so its dv = m?

bec v is cancelled out?

[fierce field]

yup!

and thats the final answer!

[glad burrow]

also for a ) when the v cancelles out do you replace it with d? since its D = m /v?

[fierce field]

what do you mean by replace?

the v's on the right cancel out

but remember on the left you are just multiplying by v

so its Dv = m

but for (a) we wanted to have v by itself.

[glad burrow]

no no on a ) were the final answer was v = m/d

[fierce field]

whereas in (b) we only wanted m by itself

yeah so in (a) we cancel out the v's

[glad burrow]

do we replace the cancelled out v with d? since D = m/v?

[fierce field]

im not sure what you are writing, is it meant to be bold?

[glad burrow]

no like i mean since its D we replace the cancelled out v with d?

[fierce field]

no? we have Dv = m

and want to isolate v

so now we can divide both sides by D

[glad burrow]

ohh ok

is it the same concept for this?

[fierce field]

kinda of

yeah

[glad burrow]

better ss

[fierce field]

yes you can see they have divided by \pi and h on both sides

things you should know

addition and subtraction are opposites

multiplaction and divison are oposites

[glad burrow]

yep

[fierce field]

squaring anumber and square rooting are opposites

you remember

[glad burrow]

mhm

[fierce field]

if you have a*b = c and you want to isolate a

you have to do the opposite

of whats happening to a

so you have to divide by b

maybe this video might help

https://www.youtube.com/watch?v=8U9u_itcs7k

its getting late for me and i have to be up earlier tomorrow

[glad burrow]

ok !!

thats fine!

thanks 4 ur help i understand the concept now ^_^

[fierce field]

best of luck and i hope the video helps

it also includes harder examples

like square roots

[glad burrow]

thamk u i have a quiz tmmr </3

ok !! thank u

[fierce field]

bye and good luck!

[glad burrow]

.close

[marsh citrus]

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[marsh citrus]

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[hollow pelican]

I need to find the greatest distance these two graphs are apart in the closed area

How do i do that

[silk sable]

Is there a constraint on the points you can choose or can you pick any point from each function.

[marsh citrus]

@hollow pelican Has your question been resolved?

[marsh citrus]

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[marsh citrus]

Please don't occupy multiple help channels.

[sonic pivot]

Can someone help me with question 1 and 2?

[marsh citrus]

@sonic pivot Has your question been resolved?

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[wooden grotto]

Someone solve this problem

[thorn epoch]

What is the question?Send full question

[eternal schooner]

Which angle of d? ebd, eda, adb?

[wooden grotto]

what do you mean?

[eternal schooner]

Isn’t it asking for the degrees in d?

[final skiff]

no

[final skiff]

Do u k ow what deg() does

[wooden grotto]

the degree of the vertex labeled 'd' in the graph.

[final skiff]

Ok do you know what the degree of a vertex is

[wooden grotto]

tbh i forgot, isnt it the number of edges connected to it

[final skiff]

Yes

So just count

[wooden grotto]

or vertices

[final skiff]

Edges

[wooden grotto]

so 4

for deg(d)

[final skiff]

Ye

[wooden grotto]

and for the sum of the vertex degrees, isn't it just 2 * deg(d)?

wait no

yeah im lost, what would the sum of the vertex degrees be?

anyone?

[final skiff]

deg(a)+deg(b)+…+deg(e)

Or alternatively 2*#no if edges

[wooden grotto]

so it would be 8?

[wooden grotto]

Am I thinking about it wrong?

[final skiff]

there are 8 edges so 2 times that is 16

Seems like you might be overthinking the problem. It is really just a bunch of counting.

[wooden grotto]

I thought there was 4?

[wooden grotto]

we said there was 4 here for deg(d), how is there 8 edges now?

[final skiff]

We r looking at all vertices and the entire graph for last problem

[wooden grotto]

Could we look at one last problem?

[final skiff]

yeah

[wooden grotto]

Do you understand this problem?

[final skiff]

is the box cutting anything off

[wooden grotto]

23wcno

[wooden grotto]

no

other then "is"

[final skiff]

Do you know the definitions for each of those

[wooden grotto]

yes

[wooden grotto]

for 1, I figured that its a closed walk that is not a trail.

[final skiff]

A path is not a trail if there are repeated edges. Do you see any edges that are repeated

[wooden grotto]

a is shown twice

[final skiff]

Yeah but that is just a vertex

an edge is a pair of vertices

trails can have repeated vertices but can't repeat edges

[wooden grotto]

so this is a closed trail that is not a cycle?

[final skiff]

Why is it not a cycle

[marsh citrus]

@wooden grotto Has your question been resolved?

[marsh citrus]

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[tidal kindle]

Someone pls help

[tidal kindle]

.close

[marsh citrus]

Channel closed

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[boreal rose]

I can help 😭

[tidal kindle]

NO

NOT YOU AGAIN

😭😭

[tidal kindle]

YOU MADE ME MORE COOKED

[boreal rose]

HOW

[tidal kindle]

YOU BARELY HELPED

👹

[boreal rose]

bro 😭

[marsh citrus]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[urban lodge]

I am really stuck with this question
I was thinking
9C1+9C4(2)=261
Since umpire is 9C1
9C4(2) is the total team possible?

[urban lodge]

But the answer said it's 315

Really stuck on what I did wrong, I think something is wrong on 9C4(2) but can't say why

[topaz hull]

for a or for b?

[urban lodge]

a

Though haven't done b that is

[topaz hull]

i havent done this topic in a while but dont u also have to find the arrangement

so the permutation

youve only found the combination

or

i might be wrong

let me think

[urban lodge]

Because if I put 9P1+9P4 it would be 3033

Which is not what I would expect from a question like this

[topaz hull]

oh wait

yes

oh hang on

i think i got it

so basically

uhh wait let me think how to word this

first you choose the umpire - as you know its 9C1

then you form the teams since youve already chosen the umpire- we’venow got 8 players left to make into two teams of 4. If we pick any 4 players for the first team, the other 4 automatically become the second team. hence this is just 8C4 which is 70

HOWEVER picking the first team and the second team counts each pair of teams twice. so you have to divide it by 2 - which is 35

and now the total number of combinations is 9*35. the mistake you did in the first one is adding it- just remember if it says and it usually wants you to multiply

does that make sense T-T

@urban lodge

[urban lodge]

Yeah I think it does, just let me read it first lol

[urban lodge]

Ah I see. So what I did wrong was I did the teams part first, making the range 9 instead of 8. I also thought I have to multiple it first lol

Yeah that makes so much more lol

[topaz hull]

yayaa

this is good practice for me its in my exams 😭

[urban lodge]

Just following a bit up if you don't mind
For b I was thinking it may need two senarios, given the two persons are A and B

1. A is in one of the 4-man queue, so B has 5/9 slots remaining
2. A is the umpire, so B has a 8/9 slots remaining?

[urban lodge]

True, got one up this early May. I am spending so much time just to train myself to these applied math questions. IDK why I could do pure maths so much faster than applied

[topaz hull]

hm yes i think that would be right

[topaz hull]

yes me too early may as well!

[urban lodge]

Don't tell me you're Aussie as well lol

[topaz hull]

LMAO

yeah

[urban lodge]

Good grief, NSW?

[topaz hull]

vce

vic

have u guys done vectors yet its really confusing me D:

[urban lodge]

Ah fair enough, I am NSW lol, probably the same year. I am so cooked lol, I got 1st in year in advanced math but I am stuck here with 0 ideas and my egos are shattered

[urban lodge]

Nope

[topaz hull]

u should join my server i have some atar resources on there ill send you the invite if youd like

[topaz hull]

HAHAHAH

[urban lodge]

We did functions and Permutations-Combinatroics

I am yr11 though

[topaz hull]

with counting you start to get the hang of it once you do lots of practices

im y11 too

[urban lodge]

YES PLEASE, I am so stuck on resources lol

[topaz hull]

i gotchu dw

[urban lodge]

I am basically scavaging on past papers lol

[topaz hull]

goodluck you got the grind this year 🙏

lmk if u need more help

i will also try T-T

[urban lodge]

Not really helped me being an ESL decided to take English Advanced

And I am aiming for ATAR 98

So I really need Math extensions

[topaz hull]

im hoping for 99.95 HAHAH

[urban lodge]

And so I am having a mid-life crisis over math extension and my own abilities to get 98 because I am literal garbage in applied math

[topaz hull]

but if you look at the tiscwebsite you should aim to average at least 85 to get 98 i think

[topaz hull]

DW U GOT THIS

[urban lodge]

Yeah hopefully lol. I also got chemistry exam following up my math so... yeah

[topaz hull]

i havenet gotten my exam schedule yet..

i actually dont really like chemistry the marking key is so strict!?!?!

[urban lodge]

Our chemistry teacher is borderline useless

[topaz hull]

LMAOO

[urban lodge]

Legit played IG shorts on a chemistry lesson

[topaz hull]

aint no way

[urban lodge]

I am so ended

[topaz hull]

what other subjects are u doing?

[urban lodge]

Like I am revising for it tomorrow the entire day
So I have to done my math revision for cambridge today, I am still 40 questions behind. Took me 2 hours to grind 10...

[urban lodge]

Modern Hist
PDHPE
Bio

I hate PDHPE but it's my own fault I went this way lol

[topaz hull]

ohh do u use the cambridge textbook? the sadler one is really good i can send you pdf it has really good counting questions in there too

[urban lodge]

Our school use Focus, I use Cambridge (The hardest development questions) for my own revision

Sadler? Is it a Vic textbook?

[topaz hull]

its western australia i think

[topaz hull]

ohh i see

[urban lodge]

Interesting, yeah I would appreciate that albeit idk if I got time. I need to grind past papers too

[topaz hull]

DW U GOT THIS

okay im gonna go lock in school starts tmr for me D:

[urban lodge]

I am one day into my break and I am having a mid-life crisis over the subject I am supposed to be good at (Asian math man 💀)

[topaz hull]

AHHAHAHA my parents are the same- theyre always like "samantha lee you should be good at this"

[urban lodge]

My parents are not really strict, it's more like my own expectations. I come from Hong Kong, I come from a "good" school in HK, I can't fail my background

It's the dutiness for me 💀

[topaz hull]

ohh i see

u just feel obligated to

if u dont mind do u have ig?

[urban lodge]

Yeah, the East Asian mindset is stuck with me from all those indoctrination lol

[urban lodge]

Yup

Dmed if you wanna follow so

[topaz hull]

OKAYY YAYY

alright im actually going to go off i have to study maths

[urban lodge]

Tbh I really seldom give it out since I come to Austrlia. No one in my school knows my contact

[urban lodge]

TRUE
I am stuck in 22b still so...

[topaz hull]

U GOT THIS

i need to finish 3 more chapters before i go back tmr.... i have a test first thing arghh

[urban lodge]

Good luck

Seems realy tough to me

I am just gonna close this lol

.close

[marsh citrus]

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[marsh citrus]

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[true girder]

How to solve Number 7?

[true girder]

And how to find the value of <2

[topaz hull]

correct me if im wrong but isnt it 135 because opposite angles in a rhombus are equal?

[true girder]

I think so

Thanks

.close

[marsh citrus]

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[marsh citrus]

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[real helm]

how to solve these or rather to find the last digit of anything by breaking it up into modular arithmetic
https://cdn.discordapp.com/attachments/1167513725125795964/1229291189908279296/image.png?ex=662f25a2&is=661cb0a2&hm=2ed75dc597b7743e3090c2297a3f1d99e293321db52780861cde58ed9e886db7&

[real helm]

dont know how exactly im supposed to break it up

[marsh citrus]

@real helm Has your question been resolved?

[lofty gyro]

so ehhhh

usually we just write the first few down

and spot a pattern

of course we can proof it, but not necessarily for these type of questions i guess?

[lofty gyro]

you can first try with
3
9
27
81
243
729
...
and see if you can find a pattern in the unit digits

[bold lichen]

So 7 next?

3^0 = 1

So 1397,1397?

[lofty gyro]

yea if you start from 0
1,3,9,7 is the pattern

[restive sun]

pull a few out from the exponent until you get something congruent to +/-1 mod 10 (i.e. try to get 9 as your base)

that should save on a bit of trial and error

also i don't think the last digit is 3?

is this a true/false question

[bold lichen]

I need to open my number theory book lolololol

[real helm]

Oh i see

Ty ty

I also dont understand this

[restive sun]

I'm guessing the notation on the left means the digits concatenated and read in base 7

then it's basically the same proof as the divisibility by 9 rule (for base 10)

[marsh citrus]

@real helm Has your question been resolved?

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[patent vine]

my teacher has a bad handwriting so ....

could someone clarify what these are written?

are they + signs? or what...

[fervent rampart]

they are + and - signs:
[ \lim_{x \to 0^+} f(x), \quad \lim_{x \to 1^-} f(x)]

[elfin berry]

pnoןɔ

[fervent rampart]

they mean the directional limits (+ meaning from the right, - meaning from the left)

[proud prism]

but I haven’t seen this one tho

[fervent rampart]

i assume that is a typo

[proud prism]

right

[marsh citrus]

@patent vine Has your question been resolved?

[patent vine]

tysm!

[patent vine]

tysm!

.close

[marsh citrus]

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[still temple]

Hi

[still temple]

18. The product of the two numbers is (-40). The sum of the two numbers is 3. What are the two integers?

[past dust]

what is your doubt?

[still temple]

I dont know how to do it

i doubt im gonna solve it

[past dust]

can you set up the equations?

[still temple]

i dont know where to start

[past dust]

start by reading the first sentence

the product of two numbers is -40

[still temple]

2x20

[past dust]

do you know what product means?

[still temple]

5x8

10x4

[past dust]

do you know the numbers?

[still temple]

yea

[past dust]

no

you know that the product is -40

[still temple]

okay

[past dust]

but you dont know exactly which two numbers

you dokt know if its 4 * 10 or 5 * 8 etc

[still temple]

nvm i got it

[past dust]

so you are gomma call those numbers x and y

[still temple]

ty

-5 and +8

.close

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

@simple thicket Has your question been resolved?

[marsh citrus]

@simple thicket Has your question been resolved?

[marsh citrus]

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[alpine gulch]

?

[still temple]

ok so

redoing what we did

you know the steps

[alpine gulch]

what stepss

[still temple]

bruh

[still temple]

right angle

the triangle

in the rhombus

[alpine gulch]

ohhhh yhh so we're finding base?

[still temple]

yeh

base of the green triangle or smth

[alpine gulch]

4 is hypotenuse right

[still temple]

no

[alpine gulch]

3 is perpendicular?

ulta?

[still temple]

the black side is

[alpine gulch]

oh shit opposite of that?

black side is hypotenuse?

[still temple]

yep

[alpine gulch]

whats 3 and 4 then😭

[still temple]

sides

do pythagoreom theorem

[alpine gulch]

ohhh

okk im so dumb

5?

[amber birch]

Ye

[still temple]

ye

[alpine gulch]

ohhh okk

uhh now height right

[still temple]

ye

i think pythag too right?

[alpine gulch]

yess

2.5?

for base

[still temple]

ye