help-28

[harsh pilot]

Nice - I do the same

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[upbeat vapor]

can someone help me solve this, not sure what to do next

i know how to do it when C's height is 0

should i cancell out all masses?

[tired zephyr]

ah just conserve the energies , C's height is 25m

[upbeat vapor]

what

[tired zephyr]

total energy at A = total energy at B = total energy at C

[upbeat vapor]

idk what that means

im trying to find the speed at point c

[tired zephyr]

total energy = potential energy + kinetic energy

and kinetic energy would help you find out the speed at C, the m (mass) get cancelled out in the equation

[upbeat vapor]

so ill get rid of all the masses

[tired zephyr]

it gets cancelled out so yeah you don't need to know the mass

[upbeat vapor]

i did this before but there was a dead end

[tired zephyr]

your approach is correct, note that the speed at A is zero since it's released from rest

[upbeat vapor]

oh so i can get rid of that whole term?

[tired zephyr]

correct

[upbeat vapor]

ohh thats where i was stuck ty

.close

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[tired zephyr]

np have a good day/night

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[slow tangle]

(arcsinx+arccosx)^2

[slow tangle]

Is it 1-2arcsinx arccosx or π/2-2 arcsinx arccosx

[spice knot]

arcsin x and arccos x are not the same thing as sin x and cos x

so you cant simplify (arcsin x)^2 + (arccos x)^2 to be 1 or to be π/2

are you trying to take the derivative of (arcsin x + arccos x)^2

[spice knot]

an identity you use instead is that arcsin x + arccos x = π/2
so (arcsin x + arccos x)^2 is π^2/4

[slow tangle]

function is given f(u)= u^3+(π/2-u)^3......u is arcsinx

I have to find minimum value of it@spice knot

F'(u)=3u^2-3(π/2-u)^2

[spice knot]

you mean $f(u)=u^3+\qty(\frac\pi2-u)^3$?

[glossy valve]

matt07734

[slow tangle]

Yes

[spice knot]

now whats the derivative of (π/2 - u)^3?

[slow tangle]

-3(π/2-u)^2

[spice knot]

are you sure? isnt the derivative of u^3 equal to 3u**^2**?

[slow tangle]

Ohh

Forget the power

F'(u)=3u^2-3(π/2-u)^2=0

[spice knot]

and remember this is $\dv{f}{u}$

[glossy valve]

matt07734

[spice knot]

so now that you have 3u^2 - 3(π/2 - u)^2 = 0, how would you solve for u

[slow tangle]

I got u=π/4

So it will be the critical point

[spice knot]

thats the one critical point of f(u)

[slow tangle]

Yes

What should I do next?

[spice knot]

you put this u back into f(u)

wait

[spice knot]

unfortunately this is still wrong

when you d/dx (π/2 - u)^3, you use chain rule

[slow tangle]

Damn why?

[spice knot]

so thats 3(π/2 - u)^2 (-1)

the derivative of π/2 - u is -1

[slow tangle]

Yes it is and we have -1

[slow tangle]

See

Are you there?

[spice knot]

the - was already there

the -1 changes it to a +

so it should be f'(u) = 3u^2 + 3(π/2 - u)^2 (= 0)

[slow tangle]

Nope

• was not there it came from derivative of -u

[spice knot]

bruh

you fixed a typo in the original problem but didnt tell me

[slow tangle]

We are here let me do it again

[spice knot]

anyways for f(u) = u^3 + (π/2 - u)^3, we got that f'(u) = 3u^2 - 3(π/2 - u)^2

[slow tangle]

Nope i only fixed u to x?

[spice knot]

and for f'(u) = 0 we got u = π/4

so the x-coordinate of the critical point is π/4
now you do f(π/4) to find the y-coordinate of the critical point

[spice knot]

theres so many things wrong with this

first of all, we calculated u = π/4, not x = π/4

so that means f(u) = u^3 + (π/2 - u)^3

so f(π/4) = (π/4)^3 + (π/2 - π/4)^3

so what is f(π/4)

[slow tangle]

Ahhh

[spice knot]

you know it would help if you screenshotted the original problem

[slow tangle]

But i have solved it

I was solving with max minima

So f(π/4) is π^3/32

[spice knot]

yes

[slow tangle]

Ohh i got it now

[spice knot]

do they intend for you to use calculus for this problem

[slow tangle]

A option

[spice knot]

yes its option A

[slow tangle]

Nope

[spice knot]

do you want to see the proper way

[slow tangle]

In my exam we just need to save time

The quickest method

[spice knot]

this method isnt quick

[slow tangle]

Even unorganised methods via options

Could you explain any short?

If you have

[spice knot]

actually its option C because it wants you to find when the equation has no solutions

u^3 + (π/2 - u)^3
u^3 + (π/2)^3 - 3(π/2)^2 u + 3(π/2) u^2 - u^3
(π/2)^3 - 3(π/2)^2 u + 3(π/2) u^2
π/2 ((π/2)^2 - 3(π/2)u + 3u^2)
π/2 (π^2/4 - 3π/2 u + 3u^2)
3π/2 (π^2/12 - π/2 u + u^2)
3π/2 (u^2 - π/2 u + π^2/12)
complete the square
3π/2 ((u - π/4)^2 - π^2/16 + π^2/12)
3π/2 ((u - π/4)^2 + π^2/48)
3π/2 (u - π/4)^2 + π^3/32
this is a parabola in vertex form
the vertex is (π/4, π^3/32) and the parabola opens up because 3π/2 is positive
so the minimum is 1/32 π^3 and you get C

the calculus way is faster if you knew how to do the calculus

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@slow tangle Has your question been resolved?

[slow tangle]

How can we know the method we solved previous that it will he max or min?

[slow tangle]

Here we got the value

.close

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[winter island]

let a = {1,2,3,4,5,6,7,8,9,10} and b ={0,1,2,3,4} the number of elements in the relation R 2(a-b)^2 + 3(a-b) is ___

[strong crystal]

this isnt a question

[winter island]

oopsies, wait lemme just send the image.

[strong crystal]

do you have any idea where to start?

or have you started?

if so, what have you done so far?

[winter island]

I do not know where to start

how do I make/count those pairs

[strong crystal]

my idea (not sure how great it is, but its something) was to start by letting x=a-b

then solve 2x^2+3x=0,1,2,3,4

then find the 10 compliments of x and youre done

[winter island]

oh ic, lemme try that rq.

[winter island]

so it's supposed to be in a-b=0 and a-b=-2? but this way I cannot get that.

cuz I get something weird as the root

[strong crystal]

lets start with 0

2x+3=0, then x=-3, and x=0

so, we have all pairs (a,a) because a-a=0

then, x=-3

so, a-b=-3, a=b-3

so, how many pairs is that?

[strong crystal]

once you figure this one out, note that pairs (a,a+3) and pairs (a,a) are distinct, so we can sum them up, and the rest are going to be easy because - is closed on the integers

[full forum]

@winter island Has your question been resolved?

[winter island]

1 right?

[winter island]

yep so 10?

[strong crystal]

10 pairs of (a,a)

But not one pair of (a,a-3)

Think about it this way
(1,-2)
(2,-1)
(3,0)
(4,1)
(5,2)
(6,3)
(7,4)
(8,5)
(9,6)
(10,7)

But (1,-2) and (2,-1) and (3,0) arent in A

How many do we have left?

[winter island]

uh, could you elaborate. I'm prolly getting confused here.

[winter island]

oh ic

[winter island]

7

[strong crystal]

Yes

So thats 0

Now we have 1,2,3,4 but these are easier

2x^2+3x=1 ==> x is not an integer

And since - is closed on the integers, there is no two integers we could subtract from one another and get a non integer

So, for 1 from B, there are no relations

Carry this logic on, and how many total elements are in R?

[winter island]

I don't get why we check for x is not an int.?

cuz it can't be?

[strong crystal]

Because x=a-b

And if (a,b) are in R, then a,b are in A

But A is a subset of the integers

So, if x is not an integer, then by closure of - on integers, a,b must not be integers

Then a,b must not be in A

[strong crystal]

By contrapositive of closure, i should say

[winter island]

Wait while counting the pairs, I'm getting it wrong, It should be 18.

[strong crystal]

Wait

I messed up

2x+3=0 ==> x=-1.5

My bad

[winter island]

yes?

[strong crystal]

So by the same logic that we ignored the others, this one is ignored

Back to the 10 pairs of (a,a)

2x^2+3x=2, has a solution at x=-2

Then all the others will be non integer x, so we ignore those

So, you may already know, but how many integer pairs (a,a-2) are on the interval [1,10]

That is, how many pairs (a,a-2) are in A

[winter island]

how'd we get this?

[strong crystal]

I realized i overlooked this sol

We were solving
2x^2+3x=0,1,2,3,4

So, enumerate

Solve 2x^2+3x=0, then 1, and so on

Ignore non integer solutions

They only integer solutions were x=0 for 2x^2+3x=0 and x=-2 for 2x^2+3x=2

There may be a faster way than examining every element, but it is fast enough to do it this way, and itll work

[winter island]

for 0, it's -1.5?

[strong crystal]

Right, and 0

Quadratics can have two solutions

[winter island]

-1/4 and -5/4?

[strong crystal]

Wdym

[winter island]

for 2x^2+3x=0?

[strong crystal]

No, x=0 or x=-1.5, if 2x^2+3x=0

,w 2*(-1.5)^2-1.5*3

[glossy valve]

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Requested by Cycadellic#0025

[strong crystal]

Yeah

Oops

🤦‍♂️

My bad

[winter island]

,w 2(-0.25)^2-0.25*3

[glossy valve]

Results provided by WolframAlpha

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Requested by jprcoder

[strong crystal]

Either way, we have those solutions, with x=0 and x=2

And only those two

[winter island]

whoops, I mistook b^2 to be 4 while it is 9 while solving bcz I wrote it haphazardly

it is 0 and -1.5

[strong crystal]

Yeah

And since - is closed under Z, we may ignore x=-1.5

[winter island]

Got it, 10 pairs for x=0 and 8 pairs for x=-2

Thanks!

[strong crystal]

Np

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[brazen falcon]

Could you help me with circle theorems?

[brazen falcon]

I don't understand what to do

[gentle totem]

are you supposed to prove it
that's just a statement

[brazen falcon]

I think I'm supposed to prove it

[gentle totem]

take the angle as some "a" with the tangent and ||start by finding angle BCA||

[brazen falcon]

.close

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[urban coral]

Hi for question c how did they get sin60?

[hot wadi]

That's the general formula for area of a triangle

If you have 2 sides of a triangle, let's say of length a and b, and the angle between them is theta

[feral ermine]

Sin60 represents h/a

[hot wadi]

U can say area is 1/2absin(theta)

[feral ermine]

Where h is the height of the triangle

[urban coral]

wait but how do they get 60?

[feral ermine]

They are trying to do 1/2baseheight

[urban coral]

im really confused about that

[hot wadi]

They assumed the triangle is equilateral, so all sides are equal and all angles are 60°

[feral ermine]

Do one thing join height from the top vertex to the base

[urban coral]

OHHH

TYSMM

that makes sense :)))

tyyy

[hot wadi]

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[torn jolt]

Hello

[torn jolt]

In questions 6 – 9, state the solutions for the quadratic equation depicted in the graph, and explain how you know

6

7

8

9

This is my question

[rough tundra]

the solution for a quadratic is where it passes the x-axis

[torn jolt]

can you exlpain it a little more

[rough tundra]

for some quadratic ax^2+bx+c we want where it equals 0

and the graph equals 0 where it intersects the x-axis

[torn jolt]

the solution for this question would be (-2,1)?

[torn jolt]

what about question 7

[rough tundra]

nope, not quite

[torn jolt]

?

[rough tundra]

where is crosses the x-axis

the point you said is the vertex of the parabola

[torn jolt]

-3 and -1 ?

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@torn jolt Has your question been resolved?

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[terse dove]

Can someone help me to finish this?

It’s moivre formula

I don’t know what to do after this step

Is this correct?

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[hardy plinth]

if these are 2 different equations how can i multiply the bottom one and then subtract it from the top one and get an answer, can someone try to explain?

[plush egret]

well theyre saying x-2y=5 right

so 2(x-2y)=2*5

[hardy plinth]

.close

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[solemn garden]

Hello I unfortunately need help

[solemn garden]

I am trying to solve this komplex number

And while calculating the degree I got a number like -0.98

How am I supposed to work with that really, when it does not have pi in it

[deep mauve]

t(cos(-0.98)+isin(-0.98))

Or what exactly are you trying to find?

[solemn garden]

Oh I am solving Re (z1/z2)

And since -2+3i is not is cartesian form I need to convert it to be

I mean I can insert this number to the e-function, but idk

[deep mauve]

≈ 3.59×e^-i0.98

[solemn garden]

But -0.98 is probably wrong

[deep mauve]

why

[solemn garden]

Not probably it is

[deep mauve]

why

it is true

as i see

what is the problem

[solemn garden]

Yeah

[deep mauve]

(5/3.59)×(e^iπ/2/e^-i0.98)

[solemn garden]

I need to + pi so that I have it positive

[deep mauve]

Re of this is (5/3.59)×cos(π/2+0.98)

[deep mauve]

why

[deep mauve]

okay very good

what is the problem

[solemn garden]

pi / 3 = 1.047

And 1 pi is 180 degrees

[deep mauve]

okay

[solemn garden]

Meaning with 0.98 it would be less than 60 degrees

[solemn garden]

Does this mean I can just work normally with the number even when I don't have a pi in it?

[deep mauve]

tanx = -tan(180-x)

tanx = tan(x+180°)

so it will be the same

[solemn garden]

Yeah

Because previous task had a pi in it

So I thought every solution has to have a pi

[deep mauve]

why would it be like that

[solemn garden]

I don't know

Maybe I just falsely thought every solution has to have a pi

So it's also fine, when it is 2.159?

[deep mauve]

@solemn garden This time the imaginary and reel seems to be wrong

[quiet stirrup]

Hint: cos(-atan(x)) =cos(atan(x)) = 1/sqrt(1+x^2))

[solemn garden]

I don't understand that

[solemn garden]

This is how we learned it

Oh hmm

[deep mauve]

yes @solemn garden

I think

[solemn garden]

I think somewhere I went wrong

[deep mauve]

180+098

rather

π+0.98

[solemn garden]

But the 0.98 is negative

[solemn garden]

What that would do, would be

The orange arrow

So both π + 0.98 and -0.98 + π, we would get to the same spot

Since when you add them together you get 2π

Which is 180° + 180° = 360°

So both our ways are correct

I went somewhere wrong

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@solemn garden Has your question been resolved?

[solemn garden]

<@&286206848099549185>

.close

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[torn wave]

Good day! How to factorize x^2-2x+1 so that we can cancel out x+1 in the denominator

[torn wave]

It's a limit problem

And if we just factor the numerator normally, the results will be (x-1)^2 which doesn't cancel the (x+1)

[dull seal]

Can you show the original problem?

[torn wave]

Sure wait I'll ss

As x approaches -1

The upload is slow 😭

[dull seal]

Xd

Just type the limit

I have 5 minutes so better be quick xd

[torn wave]

Lim as x approaches -1

(x^2-2x+1)/x+1

[dull seal]

You can’t factor here

Is it asking for one side limit or two sided?

You can split into the product of two limits

[torn wave]

I don't know 😭 the only thing that I know is if you plug in -1 the answer must not be 0

[dull seal]

One will be 1/(x+1)

The other will just be x^2-2x+1

I have to go now, but you can continue from here, just check these two and you will find what happens when u try from the left and from the right

[torn wave]

Ok 🙏 thanks

[limpid furnace]

If both limits match, the limit exists

If they do not match, or one of them doesn’t exist

Then the limit is not defined

[torn wave]

We weren't taught both limits yet, only plugging in first 🥹 but from what ur saying the limit is not defined in this problem?

[limpid furnace]

We can always approximate how the limit is going to look, right?

[torn wave]

Ohh the -0.9999

[limpid furnace]

So let’s say if x is -0.99

Then we have a positive denominator

And a positive numerator

When x is -1.01
We have a negative denominator and a positive numerator

If you want to be be sure, sub in some more values

But for this question, the left limit approaches negative infinity

And the right limit approaches positive infinity

Are you still with me?

[torn wave]

Ohh I see, i forgot about approximation but now I remember

[torn wave]

Yep just letting you write it first, thanks 🙏

[limpid furnace]

You are welcome.

[torn wave]

.close

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[near mirage]

Are all values of $sin(x)$ unique? Where $x \in \mathbb{Z}$.

[glossy valve]

Oğuzhan

[near mirage]

This is a random thought I've just thought and I think it is, but just couldn't figure out a straight proof of this

Maybe because its period is irrational

But how does that affect it 🤔

[stiff musk]

that's not enough.. consider that cos(1) = cos(-1)

[near mirage]

No not cosine, my question is about sine

[stiff musk]

no, i'm saying that the period being irrational isn't enough
after all, cosine has irrational period and it's not true for cosine

[near mirage]

Oh yeah that's a great example

https://brilliant.org/wiki/expansions-of-certain-trigonometric-functions/

Just remembered this formula, gonna try it

Not sure where to go with this one either

[stiff musk]

if sin(m) = sin(n), then cos(n) is either +cos(m) or -cos(m)
(just by looking at the unit circle)

[near mirage]

Yeah

[stiff musk]

if cos(n) = cos(m) then:
exp(in) = cos(n) + isin(n)
= cos(m) + isin(m)
= exp(im)
so m and n differ by a multiple of 2pi
they can't both be integers

you can reason similarly if cos(n) = -cos(m)

[near mirage]

Oh that's really clever

Thanks!

[stiff musk]

sure, gl!

[near mirage]

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[calm trail]

Is the second last inequality correct in this proof

[gritty rose]

Why do you think it's wrong

[calm trail]

So when I was writing it, my idea was like "oh |z| should be a least something, but it should be greater than 0, so like I get |z| < |w| - \delta" and then the way i got that formula was "solve 2\delta/|w|-\delta = \epsilon or |w| - \delta should be a number above zero

The thing I don't like is that im possibly dividing by 0

But I don't know what the right way to fix it is: like is it a *2 or /2 on the |w| term

Ah wait

Im dumb.

It's so obviously divide by 2

But hmm; does that fix the issue

|z| >= |w| - \delta is always true though

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[calm trail]

idk does my little rant above make any mathematical sense; ie that delta should be max{|w|/2, the other thing}?

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[echo wind]

Linear Algebra: Matrices

Hey I'm working on a question where I need to construct a 3x4 matrix augmented with a 3x1 matrix where the solution should be the vector: [3, -2 , 1. 0].

I'm just wondering if there is a general way to approach these types of questions

[echo wind]

so like I'm given control on making an augmented matrix; however, the solution should be [3, -2, 1, 0]

I'm just wondering if there is a formula or method of some kind to do this (I've been using guess and check and it's painful)

[spiral vigil]

well you can come up with a very simple set of equations

like x1 = 3
x2 = -2
x3 = 1
x4 = 0

but if you only have 3 equations then you won't be able to uniquely specify a 4-vector

[echo wind]

yea thats what I want, but

I have this so far, but then x4 would be free and would be allowed to be anything which means that x4 != 0

[spiral vigil]

yeah

[echo wind]

😦

[spiral vigil]

with 3 equations (rows) you can't uniquely specify a 4-vector

[echo wind]

wdym? like are u saying it's not possible?

[spiral vigil]

yeah

you'll always have a free variable

[echo wind]

.

oh yea that makes sense, bec we can only have 3 leading 1's; therefore, the last column will always be free...

oh shoot maybe I misinterpreted the question:
"give an example of a 3x4 matrix A and a 3x1 vector b such that the solution set of the augmented matrix [A|b] is a line which has a direction vector [3, -2, 1, 0]"

not sure what a direction vector is tho

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[buoyant saffron]

Could someone tell me if i am on the right path?

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[buoyant saffron]

<@&286206848099549185>

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[dark ember]

I don't understand why I keep getting this error code

[dark ember]

the first code displays the sequences an and gn

[gritty rose]

This is a math server

Go find a python server

[dark ember]

the second code works out the first value where the difference between the two sequences is less than 0.001

i went to a python server but they said they don't understand the maths

[lilac flume]

👀

[gritty rose]

Then explain it to them

[dark ember]

Do you not know python?

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[strong crystal]

his point was that this isnt particularly math related

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[strong crystal]

this is debugging python code

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[cobalt bear]

i don’t see where i went wrong

[novel crest]

heyy could someone help me with this question

[cobalt bear]

wtf twin

this mine

😭

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[novel crest]

loll

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[high basalt]

Hi I tried to solve this limit

[full forum]

Please don't occupy multiple help channels.

[high basalt]

I substitute cos(2x) and multiple it whit sin/cos of the tan(X)

I can't solve the limit, I even tried the sandwich theorem

[gritty rose]

Multiply top and bottom by 1+cos(2x)

[high basalt]

I am really struggling with it

By the conjugated?

And then I should re write 1-cos ^2 (X)?

<@&286206848099549185>

[gritty rose]

what are you stuck on

[high basalt]

I have

[torn jolt]

1-cos^2 (2x) is ?

like think

[high basalt]

The denominator

[torn jolt]

yeah

[high basalt]

I can write it in other ways

[torn jolt]

ok so which form, show here

[high basalt]

Well, cos ^2 (2x) = cos^4 - 2cos^ 2 • sin ^ 2 + Sen ^ 4

[torn jolt]

2 sin^2x is also posibel

[torn jolt]

okay we might go somewhere and waste time

can you expand cos(2x) and try?

it will be 2sin^2 (x)

which is whole deniminator

[high basalt]

Cos ^2 - sin ^2

[torn jolt]

1 - of that u said

[high basalt]

And the ^2?

[torn jolt]

wdym?

please do it

you will get like x/sin(2x)

[torn jolt]

1-c^2(x)+s^2(x) = 2* s^2(x)

I hopw u got it

[high basalt]

I got this

I even can cancel the sin (X)

[torn jolt]

ok so 2 c (t) s(t) as you know obviouisly = sin(2t)

[high basalt]

The limit is still indeterminate

[torn jolt]

nope

you need to find limit of x/sin(2x)

which you know very well using sandwich theorem or any other rule like L'hospital rule

I don;t know why its taking long time. I hope u got it.,
Bye!

[high basalt]

Thanks you very much

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[remote steppe]

Can anyone help me find d?

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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

[remote steppe]

1

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[remote steppe]

.reopen

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✅

[robust slate]

Hint: Consider what is shared between both triangles (it’s ||side CD||)

[remote steppe]

oh ok

do I need to use the Pythagorean theorem

yeah but what about the width

of like the shaded rectangle

how do go around that?

[robust slate]

That doesn’t rlly matter

I was referring to triangles CBD and CAD

[remote steppe]

oh

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[urban creek]

So, how do I write the function f

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[urban creek]

<@&286206848099549185>

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@urban creek Has your question been resolved?

[urban creek]

<@&286206848099549185> I help everybody but nobody helps me, sus

[polar valve]

its given that the graph consists of a parabola, a circumfence and a hyperbola (in this order), you have 3 formulas one for a parabola, one for a circle and one for a hyperbola. you can look for special points on the graph e.g. (0;1) or (1;0) so it should be easy to determine a,b,c.

[urban creek]

oh ok thank you!

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[full forum]

Please don't occupy multiple help channels.

[torn jolt]

How do i change the origin point

y = 4

but i want to generally start this equation on an (x,y) coordinate that i want

here exactly

how do i freely move that function as a whole

[spiral vigil]

replace y with (y-k) and replace x with (x-h)

[torn jolt]

i dont really get this

can u explain in detail

[spiral vigil]

i mean take ur equation

and whenever it says y

replace that with (y-k)

and do the same for x

but with (x-h)

[torn jolt]

okay!, i got it

also

I am writing a mathematics assesment so, how do i explain the additional letters as

[spiral vigil]

wdym you're writing a mathematics assessment like you're creating a test for people to take?

[torn jolt]

It's like a personal assesment, I graph using functions

however, I need to clearly explain all of the things that I have done for each part

within the context of mathematics

How do I mention the addition of h and k

inside the y=mx+c function

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[spiral vigil]

you probably need to review your lessons...

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[blissful tangle]

Help

[blissful tangle]

You draw a 5 card power hand and are thrilled to discover it is a straight flush (all the cards are the same suit in the same order). How many ways can you be delt a straight flush? Do not include a royal flush (10, J, Q, K, A of the same suit.

I need help starting the problem

I have my test in 20 mins i need help fast

[summer echo]

How many cards can your straight start with?

the starting card determines the entire straight flush

[blissful tangle]

Um

[summer echo]

You should do this more time before the test btw, doing it now mostly gives stress and you won't learn that much more

[blissful tangle]

Alr

[blissful tangle]

Idk

All of them

[summer echo]

J?

[blissful tangle]

The answer is 36

[blissful tangle]

5 cards

It can start with 8 of them

?

Or no

?

[summer echo]

Almost, you can start with an ace I think

[blissful tangle]

So 9

?

[summer echo]

Yeah, 9 for each suit

[blissful tangle]

Okay

Then the next possible u can have 8 cards

?

[summer echo]

What I meant was, if you have a straight flush what possible cards can it start with. Those are A23456789, so 9 possibilities for just one suit

[blissful tangle]

Mhm

[summer echo]

So how many are there for all 4 suits?

[blissful tangle]

36

Cus 9 times 4

[summer echo]

Yes, and that's the answer

[blissful tangle]

But why would you only take into account the starting number

[summer echo]

It uniquely defines a straight flush, the other 4 cards just follow

If I have a straight flush that starts with 5 of spades, then 6, 7, 8, 9 must come after (also spades)

[blissful tangle]

Why isnt it this

And then multiply each one for the number of possible

[summer echo]

Since order doesn't matter here, you can assume your hand is in increasing order

[summer echo]

What do you mean with this?

[blissful tangle]

So you basically just multiply the total number of starting possibilities by each hand?

[blissful tangle]

Cus for the first slot there could be 9 cards then next would be 8 then so on

[summer echo]

You add them, in this case it works because each starting card determines the rest of the hand

[summer echo]

You are confusing this with factorials

If there are 9 possibilities for the first card, there is just 1 for the next one in each of those 9 cases

[blissful tangle]

Ohhh

Cus they would follow in order

[summer echo]

Yes

[blissful tangle]

Ok I see

What does it mean in the problem when it talks about a royal flush

[summer echo]

It means that 10, J, Q, K, A doesn't count as a straight flush, because it is a royal flush

[blissful tangle]

And the 9 starting possible eliminates that option right

Cus we cant start with 10

[summer echo]

Yes

otherwise it would have been 10 starting options

[blissful tangle]

Alright ty sm bro

Have a great day

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[boreal cedar]

How many positive integers can be formed from the numbers 1 through 9, such that the sum of the numbers is odd?

[boreal cedar]

1×2×3×4×5×6×7×8×9 ? Is this answer right?

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[rigid iron]

9! would be all possible arrangement of digits 1-9 but your problem has a constraint

[boreal cedar]

What is the solution

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[gusty carbon]

how do i simplify

[fickle finch]

change to exponent of 1/3 then distribute

[gusty carbon]

so( -27c^19 x^12)^1/3

[balmy rose]

$\sqrt[a]{b}^c=b^{c/a}$

[glossy valve]

Martin

[balmy rose]

👍

[gusty carbon]

then what do i do

[balmy rose]

$(a\cdot b)^{c/d}=a^{c/d}\cdot b^{c/d}$

[glossy valve]

Martin

[balmy rose]

that is what Anson meant by distributing the 1/3

kinda like the distributive rule a(b+c)=ab+ac

[gusty carbon]

image arent loading for me

[balmy rose]

(a * b)^c=a^c * b^c

and then, we wanna use:
(a^b)^c=a^(bc)

[gusty carbon]

-27c^19/3 * x^4?

[balmy rose]

dont forget to also apply it to the 27

[gusty carbon]

-9c^19/3?

[balmy rose]

now dont forget the x 🙂

[gusty carbon]

-9c^19/3 * 1/3x^4

[balmy rose]

if 27^(1/3)=9, then 9^3=27
which is not true

however 3^3=27

also idk how you got that 1/3 infront of the x^4

you already did x^12/3=x^4

[gusty carbon]

so its -3c^19/3 * x^4

[balmy rose]

👍

HOWEVER

as a side note

this now simplified form is not exactly the same as before

before, we had a root

and by simplifying them, we often loose information

we also have to distribute this 1/3 to the minus sign, so we get a
(-1)^(1/3)
Now, we might say that this is just -1

but that is not the full picture

here, we can see that if we are dealing with complex numbers, then we actually get three results

[gusty carbon]

I cannot see

[balmy rose]

hmm

[gusty carbon]

school wifi

[balmy rose]

thats ok, you are not supposed to consider these two other solutions

just thought i'd mention them

so for your case, your result is correct 👍

[gusty carbon]

yayy

can you help me with a few other algebra problems

[balmy rose]

sure thing

[gusty carbon]

I think I have solved at least very similar problems but for this one I keep getting it wrong

[balmy rose]

we can start by simplifying the denominator

we have x^a * x^b = x^(a+b)

like how x^1 * x^1 = x^2

[gusty carbon]

x^-9/14 y^1/4

[balmy rose]

yep

now, we can use this property:
1/x = x^-1

that also implies that for example
1/( x^(-9/14) )=x^(9/14)

in words:
we can swap factors from the numerator do the denominator and vice versa
doing so, we have to change the sign of the exponent (which is the same as raising it the power of -1)

[gusty carbon]

would i do x^-9/14 -(-8/7) for x?

[balmy rose]

we wanna move everything to the numerator, not the denominator
so for x, we would get -8/7 + 9/14

[gusty carbon]

oh ok

-7/14 = -1/2?

[balmy rose]

yep

[gusty carbon]

do i also do that with y

[balmy rose]

yep

[gusty carbon]

1/24-1/4 = -5/24

[balmy rose]

👍

[gusty carbon]

so x^-1/2 and y ^-5/24

what do I do with the 2 numbers now

[balmy rose]

we can leave them like this
some people might prefer them to be expressed with roots, but this is as simplified as it gets

some people would also prefer these two numbers to instead be in the denominator, because then the exponent will be positive

[gusty carbon]

would it be like 1/x^1/2 * y^24

[balmy rose]

yeo

yep

[gusty carbon]

ok thakn you

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[olive magnet]

need help with this proof

[olive magnet]

proof by definition

[viral jasper]

By definition of limit? As in, you must use only the epsilon limit definition of sequences?

[olive magnet]

yes

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[glad pine]

I had trouble with this problem

[glad pine]

I tried to find 'm' in y=mx+b by using the point slope formula

I got a slope of 0, so I made my equation y=0x+19/2

I don't know how else to move forward because I think that's incorrect

[steel mist]

have u seen derivatives?

[glad pine]

Yes, but we haven't used them yet to solve problems

[steel mist]

well, they exist to be used

[glad pine]

sorry I mean my class has not taught them yet

Is there some other way to solve this problem or can it only be done through derivatives?

[steel mist]

limits? xd

[glad pine]

yes thats what I mean sorry 😆

[steel mist]

a derivative is a limit in the end

[glad pine]

I assume my math must be off thne

[steel mist]

so idk

[glad pine]

ah ok

with derivatives, are we supposed to find the derivative of f(x)

[steel mist]

$y-f(a) = f'(a)(x-a)$

[glossy valve]

Aza

[steel mist]

is this formula familiar to u?

[glad pine]

let me check my notes

no I dont think I am

[grim skiff]

Maybe they have to use the limit definition of the derivative rather than the shortcuts

[steel mist]

idk what he can use

dont u have other examples on ur notes?

[glad pine]

I had an earlier question I solved that was similar

Let me get it

[steel mist]

yes pls

||because if u dont know about derivatives, only way is u thinking and guessing what a derivative is xD||

but show what u have done

[glad pine]

f(x)= -x^2 +4x

Find the slope m and an equation for the line L tangent to the graph of f at the point (−1, −5)

[steel mist]

and how did u solve it?

[glad pine]

I did

[steel mist]

how

[glad pine]

lim x towards 1

sorry im writing it out

[steel mist]

1 or -1?

[glad pine]

lim x towards -1 (-(-x-5)(x+1))/(x+1)

sorry towards -1

I plugged in the numbers to the formula (f(x)-f(a))/(x-a) = L

to find L

and then I used L as m in y=mx+b

[steel mist]

yeah

exactly that

is the definition of derivative

[glad pine]

wait really

[steel mist]

the derivative of a function at a point a is the slope of that function and that point

and m is indeed the slope

[steel mist]

so by doing this, u get the slope of ur function f(x) at point a

[glad pine]

ohhhhh

[steel mist]

and then u use that to build ur tangent

[glad pine]

okay

[glad pine]

I am struggling still, I tried to doing something similar for the problem and I was getting it wrong

[steel mist]

since for building a line u need at least 2 points

what u do to build the tangent is picking 2 points, one at x and the other at x+h

then u make that h super small

so those point are rlly rlly close

as close as u want

[glad pine]

cant you build a tangent using a point and the original function?

[steel mist]

resulting in a line that intersects f only on that point

u can

wait, holdon

[glad pine]

ok

[steel mist]

U have ur f(x), and ur point a

a+h is another point, so now u can build a line that passes through both points, right? the orange one

if u make, with a limit, h close to 0

[glad pine]

and h is just any constant?

[steel mist]

both points will be super close

h is the only variable

a is a number, u know it

-1, 8, 5

whatever

on that draw we can say a=1

and a+h whatever

the point is, u make the limit as h goes to 0, so both points get closer and closer

almost the same point, so u can build the tangent

[glad pine]

ohhh

[steel mist]

if u have this triangle, what is the slope?

y/x, no?

[glad pine]

yes

rise/run

[steel mist]

sorry for my bad draw xd

[glad pine]

I did ((4x^2+1/2)-(19/2))/(x-(-3/2) for my problem

[steel mist]

do u understand the picture?

[glad pine]

Yes

what is the mark next to a+h and a

[glad pine]

oh its f

[steel mist]

yeah xd it is an f

so u can build a rect triangle

with base (a+h-a) = h

and height f(a+h)-f(a)

no?

[glad pine]

I guess so

I am just confused why I cannot use the formula I used earlier for this problem again

[steel mist]

so that is the slope

u can, im explaining u where it comes from

[glad pine]

ohh

do u mind if I save this picture?

[steel mist]

i guess children nowadays dont care about where things come from and they only wanna be machines blindly applying formulas

so w/e, im not wasting my time more. Good luck

[glad pine]

no I find it interesting

I just didn't understand what you were getting at

[steel mist]

do u want me to continue?

[glad pine]

yes

[steel mist]

so on that picture, i have a point a, and another point a+h, where h is a real number

like a slider

and offset

so far so good?

[glad pine]

ye

yes

[steel mist]

and the images is just f(a) and f(a+h)

no? on y-axe

so u can build a triangle, what i paint in blue

and the length of the base is end-start

[glad pine]

so the triangle has width h and height f(h)?

[steel mist]

end = a+h cuz it is the furthest point

no

it has width h

but height

is f(a+h)-f(a)

[glad pine]

ohhh

[steel mist]

look on the y-axe

[glad pine]

thats not the same

roger

[steel mist]

it is

this is like a zoom

[glad pine]

f(a+h) - f(a) = f(h)? or no?

[steel mist]

no???

hell no XD

[glad pine]

ok thats what I said

[steel mist]

not even close

so we have that triangle

[glad pine]

yes

[steel mist]

now, whats the slope of that triangle?

it is y offset divided by x offset

[glad pine]

(f(a+h)-f(h))/h

right?

that is sthe slope?

[steel mist]

hence $\frac{f(a+h)-f(a)}{h}$

[glossy valve]

Aza

[steel mist]

no?

[glad pine]

yesss

[steel mist]

so if u make h super small

which in mathematics means a limit

u have the slope on point a

of that line

this is indeed the derivative

[glad pine]

ok

[steel mist]

also the slope of the line u are looking for

[glad pine]

the derivative is the slope of a line based on a single point?

[steel mist]

$y-f(a) = f'(a)(x-a)$

[glossy valve]

Aza

[glad pine]

sorry I mean based on two points

[steel mist]

and this is one of the multiple ways of describing a line

actually, very usefull for ur problem

[steel mist]

no. a limit is special

u dont care what happens at x=a

u care what happens NEAR a

so those points are never the same

they are super close

[glad pine]

as the line approaches the limit

[steel mist]

as close as u want

but never the same

so u have 2 different points

they can be super close, but they are different

and with 2 points, u can build a line

u have a=1 and a+h=1.00000000000000000000000000001

they are suuuuuper close

but they arent the same point

so u can build the line

[glad pine]

doesnt this often times make the question harder if you have such a small number?

[steel mist]

no

again, u are not a machine plugin numbers and solving

[glad pine]

I suppose I am a human

[steel mist]

$f(x)= -x^2 +4x$ at (-1,-5)

[glossy valve]

Aza

[steel mist]

so we need to find the slope first

[glad pine]

yes

[steel mist]

$y-f(a) = f'(a)(x-a)$ because on this formula for a line, we need f'(a), which is the slope

[glossy valve]

Aza

[steel mist]

so just do the thing

[glad pine]

yes the slope is 6

and from there we plug back into the original equation

[steel mist]

$\lim_{x\to-1} \frac{-(-1+h)^2+4(-1+h)}{-1}$

[glad pine]

and then get b

[steel mist]

damn, sec, i forgot the syntax

[glad pine]

ah

[steel mist]

no, wait

$\lim_{h\to0} \frac{-(x+h)^2+4(x+h) - (-x^2+4x)}{h}$

[glossy valve]

Aza

[steel mist]

do u see that $f(x+h) = -(x+h)^2+4(x+h)$?

[glossy valve]

Aza

[glad pine]

yes

[steel mist]

okey, and $f(x) = -x^2+4x$

[glossy valve]

Aza

[glad pine]

and the other part at the numerator is the original equation

this is the TRIANGLE

wow

y/x

[steel mist]

so u have to solve that limit

and the limit will depend on x, of course

and then, that will be f'(x)

then u do f'(-1)

[glad pine]

I understand that h is a variable

[steel mist]

and u get the slope

no, i dont mean that

okey imma help u with this

but u do the other alone

[glad pine]

how can I solve when I have two variables?

[steel mist]

We have this: $\lim_{h\to0} \frac{-(x+h)^2+4(x+h) - (-x^2+4x)}{h}$

[glossy valve]

Aza

[steel mist]

so lets do some algebra (imma use paint now)

[glad pine]

ok

[steel mist]

correct me if im wrong with any sign. I am doing algebra

[glad pine]

ok

I am confused how do you solve both for x and h

oh wait

you make h 0

nevermind I understand

[steel mist]

yeah

h tends to 0

[steel mist]

yeah, i messed up on a sign

that was a +, not a -

[next hawk]

it was left on purpose

[steel mist]

but anyway, what gert did

[next hawk]

but for another time

[steel mist]

finally u end up with -2x + 4

[glad pine]

so you get -2x+4

[steel mist]

this is f'(x)

[glad pine]

this is the derivative?

[steel mist]

and now u want the slope on x=-1???

[next hawk]

don't remove the (, ) to early

[steel mist]

u plug -1 there

cuz exercise telling u on point (-1, -5)

so x = -1

[glad pine]

yes

[steel mist]

so 2 * (-1) + 4

[glad pine]

so does that give us a slope of -2?

[steel mist]

2

possitive

[glad pine]

2 lol whoops

[steel mist]

so now u go back to

$y-f(a) = f'(a)(x-a)$

[glossy valve]

Aza

[steel mist]

and u have all u need

y is y. As the y-axe on a line