help-13

[past pond]

only

[tepid trout]

what you need help with this?

[past pond]

just checking

[tepid trout]

k

a) should be 52/112 not 57

[past pond]

OH

thankss

[tepid trout]

um I got class right now so I’ll have to wait to help with the rest, hopefully someone helps you but if I comeback and they didn’t then I’ll message you

[past pond]

np

thanks

[cedar kiln]

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[crimson sedge]

\begin{align*}
x^{\frac32} + y^{\frac23} &= 1\
\left(y^{\frac23}\right)^{\frac32} &= \left(1 - x^{\frac32}\right)^{\frac32}\
\end{align*}

[crimson sedge]

how do i continue

[wraith dagger]

mahmooz

[crimson sedge]

gotta derivate this

uh nvm i dont think i need to open the parenthases

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[hazy palm]

A rod of mass M= 3.1 kg and length L = 1 m is free to rotate around a fixed horizontal axis passing through point P. The system is in the Earth's gravity field, g=9.8m/s2. The bar is initially at rest in the position shown in the figure. The bar is released and the end comes to hit a body of mass m = 1.6 kg located below the axis as described in the figure below. The bar comes to a standstill just after impact.

[hazy palm]

determine the magnitude of the velocity of the body after impact

will I use conservation of energy to solve that question?

[devout radish]

yeah definitely

[hazy palm]

ok, so I simply equal mgh=mv^2/2 ?

the first been the potential of the rod and the second the cinetical of the body?

the problem is that I won't use the lenght of the rod

oh wait

the height will be the lenght?

[devout radish]

the height is half the length, bc you do it from the center of mass

theres do steps too tho, you use CoE at first and then conversation of angular momentum for the second part. or could you use conservation of linear 🤔 bc the rod stops and the mass will move linearly

[hazy palm]

I don't understand that

on the image it says L is the height

here

how do the center of mass affect it?

[devout radish]

gravity acts on the center of mass

so the force pulling it down acts on L/2 from the pivot

this is the best i could do rn lol

[hazy palm]

oh, bc the other extreme is fixed on P ?

but will I use half of the mass or not?

[devout radish]

well it doesn't matter where the pivot is in terms of where gravity will act its always on the CM, and no everything else will be the same because its still the same mass gravity is pulling

[hazy palm]

ok, got it

thx dude

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[unique cloak]

how do i go from left to right

[dire geode]

expand and factor out 100 from all terms. then when you take the square root of 100, it becomes 10

$\sqrt{100 \cdot (...)} = \sqrt{100}\sqrt{(...)} = 10 \sqrt{(...)}$

[wraith dagger]

riemann

[unique cloak]

but my equation doesn't look that way after i expand it

i got $\sqrt{1600+80t+t^2+100^2-6000t+90t^2}$

[wraith dagger]

tacku

[unique cloak]

from $(a*b)^2 = a^2+2ab+b^2$

[wraith dagger]

tacku

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[dire geode]

you're missing t^2 multiplying 1600

this is multiplication, not addition

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[tulip harness]

Hi! How can I find possible different values for two numbers a and b such that ab - 5 | 4a

[dire geode]

did you use the definition yet?

[tulip harness]

It doesn't matter if there are infinite many, I'm going to intersect that with something else

[tulip harness]

(ab - 5)*k = 4a?

I don't know how to operate from that

[dire geode]

so k takes on at least one* positive integer here

[tulip harness]

yeah

I can for example rewrite that as a(bk - 4) = 5k

But I don't get how does that tell me what are possible values for a and b

[dire geode]

show the original question where you're asked to do this

[tulip harness]

Wait a sec I'll send a pic of it

thr

[dire geode]

can't really read this handwriting

is that a 5?

is that a+b on the left?

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[celest badger]

Please help me, I don't know how to find the side length of the triangle if the diameter of the circle inscribed within it is 10. Atleast, I got 16.77 but when solving it appears that is wrong so someone please help.

[cedar kiln]

@celest badger Has your question been resolved?

[boreal horizon]

so the radius is 5, and since it's a equilateral triangle, all the internal angles are 60 degrees, that means the line connecting the center of the circle and any vertice would be 10

[celest badger]

well that makes sense and is what i got

but

when i try to solve the rest of the answer it doesn't work out

if what u are saying is correct

then the height of the triangle is 15

and the side length is radical(15^2 + 7.5^)

but when finding the surface area using all that it doesn't result in any of the possible answers

[boreal horizon]

side length is 2 * 5 * sqrt(3)

half of the side length is not 7.5

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[median grove]

A car drives to and from work on weekdays, calculate the fuel economy per month.

MPG = 42
Gallons = 12.2
Distance from work and home= 15.2
Gas = 3.75```

_I used Distance Driven divided by MPG times 3.75 and got 29.85 however I believe you can come to the same answer through a different equation idk

[somber belfry]

Uhm what us MPG!

Okay im not American so I'm not into freedom units but here is my take

First calculated the distance you want to travel, he drives each day to and from work

Lets assume we have 20 work days a week

Now we got our distance D

[fair hare]

miles per gallon

[somber belfry]

Then D/MPG

[median grove]

Dam the same answer I got

Isn't there another way to calculate it?

I got the same answer doing (Total distance / mpg) * gas price

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[lyric leaf]

where have you gotten so far? or are you confused on how to start

[civic flicker]

im confused on how to start

yall got that $$ bot that show the numbers?

cause thats helpful if you could use it

[lyric leaf]

yeah, though i dont really know how to use it

[civic flicker]

alr thats fine

[lyric leaf]

simplifying square root expressions usually have a couple tricks to them, one of them is just the basic rules you might've already learnt, sqrt(a) * sqrt(b) = sqrt(ab)

and the other is looking for square numbers to take out of the square root

for example, sqrt(32)=sqrt(16 * 2) = sqrt(16) * sqrt(2) = 4 * sqrt(2)

you should try and look for square roots with numbers that can be divided by square numbers like 4, 16 or 25, etc.

3 and 7 evidently arent divided by any square numbers (they're prime, for starters)

but try and take a look at sqrt(175)

can you see a square number you can take out of that?

[civic flicker]

7*25

[lyric leaf]

yeah

so you can write it as sqrt(7) * sqrt(25) = 5*sqrt(7)

now the question might be a lot easier

[civic flicker]

how would it be written out now

like the problem

wait lemme try sum

$sqrt(4)$

[wraith dagger]

ariki

[civic flicker]

wtf

how i use this

[lyric leaf]

sqrt(3) * ( sqrt(7) - 5*sqrt(7))

oh idk what you actually use to get square roots, i'm just abreviating it

$\sqrt{7}-5\sqrt{7}$

[wraith dagger]

Az

[lyric leaf]

thats the one

$\sqrt{3}\left(\sqrt{7}-\5\sqrt{7}\right)$

beh, nevermind

you get what i mean with the first one, though?

[civic flicker]

yes but plug it in the paranthesis

?

[lyric leaf]

yeah

you can also simplify it to -4sqrt(7)

and then just do that multiplied by sqrt(3)

[civic flicker]

theres like an imaginary 1 in front of sqrt 7 kinda

thats how its -4 now

?

[lyric leaf]

yeah because sqrt(7) is pretty much just 1 * sqrt(7)

we just dont really write the 1

[civic flicker]

ohh

[lyric leaf]

then you just have sqrt(3) multiplied by 1 term so its a bit easier

[civic flicker]

so js sqrt3(-4sqrt7)

mb i aint feel like putting brackets

so just multiply the radicands?

so liek distrubute

so its -4_/21

@lyric leaf

[lyric leaf]

yeah thats right

[civic flicker]

W

is it distrubute or do i just multiply the radicands

[lyric leaf]

just remember with qs like these to multiply out and check for square numbers

[civic flicker]

alr

[lyric leaf]

id write it as -4 sqrt(21), its simpler multiplied out

like you wouldnt write 4 * 8

hope that helped tho

[civic flicker]

it did thanks

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[cold isle]

i am having trouble finding sin on largewr numbers of the unit circle. My question is to find sin(17pi/6)

[cold isle]

sorry, i dont know how to write pi on my keyboard itself

my thinking is that its passed 2 pi because 17 passes 6 twice to then im left with 12pi/6 + 5pi/6

2pi + 5pi/6*

am i correct so far?

[pulsar skiff]

yes

17pi/6 = 2pi + 5pi/6

@cold isle

[cold isle]

actualy sorry, i am finding cosine

[pulsar skiff]

ok

[cold isle]

so i have cos(2pi + 5pi/6) and i know that its 150

i enter sqrt3/2 but i am getting it wrong

[pulsar skiff]

cos(150 deg) is not positive

if you have 150 degree rotation from center axis, then the x-coordinate is negative

so cos(150) must therefore be negative

[cold isle]

ahhh and cos = x

[pulsar skiff]

yes x = r cos(theta)

on a unit circle r = 1

and in this case theta = 150 deg

so x = cos(150)

[cold isle]

is there a way that it is easier to memorize the other quadrants of the unit circle? I have a question today to find the sin(285) buit when the number is so large, i have an issue

its easier if 285 was on the unit circle but it isnt so i am a little lost

[pulsar skiff]

285 degrees is on unit circle

unit circle is from 0 to 360

285 degrees is quadrant 4

[cold isle]

this is the unit circle i pulled online and its also the one that was given to me in class

[pulsar skiff]

oh you mean specifically labeled

sin(x) = - sin(360 - x) yes?

because sin(x) = - sin(-x) and sin(x) = sin(x + 360)

so sin(285) = -sin(360 - 285) = -sin(75)

[cold isle]

its a question from my sample final.

[pulsar skiff]

actually -sin(75) does have a nice form

[cold isle]

i tried to simplify it to be multiples of the unit circle meaning sin(45)+sin(30)

[pulsar skiff]

no

sin(a + b) =/= sin(a) + sin(b)

[cold isle]

i dont think i have been taught -sin yet

like as in the professor has never gave us an example or pulled it up in class

[pulsar skiff]

• sin is -1 times sin

[pulsar skiff]

instead, you use sin(a + b) = sin(a)cos(b) + sin(b)cos(a)

so sin(30 + 45) = sin(30) cos(45) + sin(45) cos(30)

= sqrt(2)/4 + sqrt(6)/4

[cold isle]

he just showed that today actually. that formula

i am not familiar with it just yet

thank you @pulsar skiff

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[cedar kiln]

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[mystic abyss]

is there any reason why you write it out the second way instead of the first?

[mystic abyss]

is there a certain rule?

[hushed veldt]

not really
just more "normal" format

[mystic abyss]

ok ty

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[cedar kiln]

@atomic seal Has your question been resolved?

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[silk sail]

How Will You Solve 48?

[modern compass]

do you know about equivalence classes?

[silk sail]

Can You elaborate?

[modern compass]

two numbers in an equivalence class of 5 if they have the same remainder when divided by 5

or, in other words, a and b are in the same equivalence class of 5 if a-b is divisible by 5

[silk sail]

Oh So modular.

Go Ahead.

[modern compass]

if you know the sums of every 6 out of 7 of these are divisible by 5, you can show they must all be equivalent mod 5

[silk sail]

Oh Wait.

Ah Okay I Misread The Question a bit.

ThANk You.

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[wintry heart]

If $X_1, \dots ,X_n, Y_1, \dots , Y_m$ are two random samples from a $N(\mu _1, \sigma ^2)$ and $N(\mu _2, \sigma ^2)$ respectively, how can I rewrite the following expression as a function of a statistic $Z$ which has a well known distribution?

[wraith dagger]

Eduardo291299

[wintry heart]

$\frac{\sum_{1}^{n} (x_i-\mu_1)^2 + \sum_{1}^{m} (y_i-\mu_2)^2}{ \sum_{1}^{n} x_i^2 + \sum_{1}^{m} y_i^2 }$

[wraith dagger]

Eduardo291299

[solid mango]

Checking my understanding of Logarithms . let f(x) = log3(3x-2). let f(3) = log3(3*1.5-2) = 1.19 (2 decimal places)

[cedar kiln]

@wintry heart Has your question been resolved?

[solid mango]

All the examples i went though are not showing any decimal places so, log3(3) as 1 and not 1.4313.

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@wintry heart Has your question been resolved?

[wintry heart]

<@&286206848099549185>

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[rotund sluice]

What does it mean when a correlation matrix isn't symmetric?

[finite garden]

Most likely that something has gone wrong.

A correlation matrix has to be symmetric; if it isn't, then cov (X, Y) != cov (Y, X) for some pair of random variables X and Y.

[rotund sluice]

I thought as much

thank you

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[feral isle]

in which direction should i think

or am i missing some concept ? idk how to solve it

[finite garden]

Linear transformations are, well, linear.

[feral isle]

like is there any way i can get aT(1) + bT(1) = (1,0) so it not possilbe, therefore not possible ?

[finite garden]

You're thinking in the right way.

But the linear transformation takes input in R2, not R.

Here's a really big hint: see if there's any way you could express T(1, 1) as a linear combination of the last two expressions T(2, 3) and T(3, 2).

[feral isle]

yeah like we can have 1/3(1,2) + 2/3(2,1)

like that?

[finite garden]

I'm not following.

(1, 2) and (2, 1) are points in the image.

[feral isle]

okay let me try some more

;-;

[finite garden]

Okay, hold on. I should probably let you try some more, but I'll give you another really big hint, which should give you the idea.

What is T(2, 3)+ T(3, 2) equal to?

[feral isle]

T(5,5)?

[finite garden]

Correct!

And from the information given, what is this equal to?

(We have the values of T(2, 3) and T(3, 2), remember.)

[feral isle]

ummm

(3,3)?

[finite garden]

Correct again.

Now, one final step of logic. What is T(5, 5) in terms of T(1, 1)?

[feral isle]

5T(1,1)

[finite garden]

Correct.

And what is 5T(1, 1) from the information given?

[feral isle]

5(1,0)

[finite garden]

Which is?

[feral isle]

not equal

[finite garden]

Which means that?

[feral isle]

they are not linear transformation

[finite garden]

There you go.

[feral isle]

yah

we did it

thanks :3

[finite garden]

Glad if it made sense to you.

[feral isle]

thanks i will ask more on linear combinations, i dont really get it so get stuck often ;-;

thanks for now

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[final elbow]

can someone help me in grouped data mean, median and mode

[umbral ravine]

whats the question?

[cedar kiln]

@final elbow Has your question been resolved?

[final elbow]

I need a reviewer for tomorrow

[upper garnet]

what's a reviewer

[final elbow]

grouped data of mean, median and mode

[cedar kiln]

@final elbow Has your question been resolved?

[final elbow]

no.

[cedar kiln]

@final elbow Has your question been resolved?

[final elbow]

<@&286206848099549185>

grouped data of mean, median and mode

[final elbow]

<@&286206848099549185>

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@final elbow Has your question been resolved?

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[crimson sedge]

I need all area

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[rich flint]

How many ways can i distribute 15 balls in 4 lockers if uno must stay empty

[rich flint]

can anyone explain or send a video of the model please?

[cedar kiln]

@rich flint Has your question been resolved?

[spice sparrow]

So one locker must stay empty?

[rich flint]

yes

[spice sparrow]

are the lockers numbered or not

are they unique?

[rich flint]

yes

[spice sparrow]

Then its 4 times the problem but with 3 lockers

Look at the problem like this

You have 4 lockers but 1 must stay empty

So you have 4 equal situation

1 with locker 1 empty 1 with locker 2 empty etc

So you calculate the ways in which you can distribute 15 balls over 3 lockers

and do that times 4

[cedar kiln]

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[cedar kiln]

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[obtuse basin]

How can I go about proving via induction that 4n + 5 < 5^(n+1)?

[obtuse basin]

The original proof was 1 + 4n < 5^n, but I already did the initial inductive step. Not sure where to go from here.

[dire geode]

might want to use $5^{n+1} = 5^n \cdot 5$

[wraith dagger]

riemann

[obtuse basin]

Hmmm

4n + 5 < 5^n * 5

It makes sense in my head

[crystal raptor]

Get the form of your inductive hypothesis on the left

[obtuse basin]

What do you mean?

Oh, turn it back into 4(n+1)?

[crystal raptor]

4n+1 right?

That's less than 5^n from the inductive hypothesis

[obtuse basin]

4n+1 was my original bit

Then I did n + 1 which became 4(n+1) + 1

Which I converted into 4n + 5

Was I not supposed to do that?

[crystal raptor]

Well you need to use at some point that 4n+1 < 5^n

[obtuse basin]

Yea

Right now I got 4(n+1) + 1 < 5(5^n)

[crystal raptor]

So write 4n+5 as 4n+1+4

Now you can apply the inductive hypothesis

[obtuse basin]

Wait I started with 4n + 1 < 5^n

And then I set n = n + 1

So then you get 4(n + 1) + 1 < 5^(n+1)

Which becomes 4(n + 1) + 1 < 5(5^n)

[crystal raptor]

Yes but you need to use this to prove the n+1 step

That's the whole point of induction

[obtuse basin]

I'm a bit confused

I'm trying to figure out how to prove this

I got my initial inductive step

But I'm not sure where to go from there

[crystal raptor]

When you do induction, you assume it holds for n and then use that assumption to show it holds for n+1

[obtuse basin]

Yea

But to start I suppose that n+1 is true, and then actually go on to prove it

[crystal raptor]

You are assuming 4n+1 < 5^n so you need to use that to show that 4(n+1)+1<5^(n+1)

[obtuse basin]

Yea

Like here's an example of an inductive proof I did a bit ago

[crystal raptor]

Why would you assume n+1 holds? That's what you're trying to prove

[crystal raptor]

Yeah look how you used the case for n when changing 7^n to 6k+1

It's the same thing here

[obtuse basin]

And I got that step before

So now to turn it into n+1 form, I turned 4n + 1 < 5^n into 4(n+1) + 1 < 5^(n+1)

And now I need to simplify that down to prove it

But I'm stuck on how to actually simplify it down

[crystal raptor]

Write the left hand side as 4n+1+4

And then use your hypothesis!

[obtuse basin]

That's the part I'm struggling with lol

[crystal raptor]

Don't write out the full n+1 inequality, we haven't proven it yet, just start with 4(n+1)+1 and write it as 4n+1+4 and use your inductive hypothesis

What's the only thing we know about 4n+1?

[obtuse basin]

It's less than or equal to 5^n

So I got

4(n+1) + 1 < 5(5^n)
4n + 5 < 5(5^n)

And I don't know where to go from here

[crystal raptor]

Stop writing the thing you're trying to show

Just work with the left hand side

$4n+1+4 \leq 5^n + 4$ right? Using our inductive hypothesis

[wraith dagger]

iCaird

[obtuse basin]

How did you get + 4 on the right side?

[crystal raptor]

Because we have a +4 on the left side?

[obtuse basin]

That +4 came from distribution

I didn't just add it in there

[crystal raptor]

Nor did I?

I know where it came from, that's why I put it there

[obtuse basin]

I started with showing 4n + 1 < 5^n

And then did n+1 stuff

4(n+1) + 1 < 5^(n+1)

Simplifying the left side gives 4n + 4 + 1 < 5^(n+1)

[crystal raptor]

!!!!

[obtuse basin]

This is how I'm supposed to do proofs lol

This is literally how I was taught.

n+1 right after the assumption for n, and then prove why that n+1 is true

[crystal raptor]

Yes but you didn't immediately write 7^(n+1) = 6m+1 did you

[obtuse basin]

Oh I see what you're saying

4n + 1 < 5^n
4n + 4 + 1

Now what?

I think I see something

[crystal raptor]

.

[obtuse basin]

Just gotta switch around that 4 and the 1

Okay I see what you mean

So then you get 4n + 1 + 4, and 4n + 1 < 5^n, so thus 4n + 1 + 4 < 5^n + 4

[crystal raptor]

You see how you get yourself into a pickle doing what you did lmao

Uh huh

Now just convince yourself that adding 4 is less than timesing by 5

[obtuse basin]

I convinced myself

Not sure how to write it out though as a proof

[crystal raptor]

Uh ig think about the worse case scenario, 1+4=1*5

(For positive numbers bigger than or equal to 1)

We change the 1 to anything bigger and timesing by 5 is deffo more than adding 4

If you like you could show it by induction

[obtuse basin]

heh.

[cedar kiln]

@obtuse basin Has your question been resolved?

[obtuse basin]

Nope

Still working on it

So I'm not sure if a sub-induction proof is the way

So I'm still stuck on 4n + 1 + 4 < 5^n + 4

[crystal raptor]

Okay here's a better way

4 < 4*5^n

For n >= 1

No need to prove that, very clearly true, worst case is 4 < 20 and the right hand side only gets bigger for bigger n

[obtuse basin]

How did you get 4 < 4*5^n?

[crystal raptor]

Just think about it

a < ac for positive c

And ig positive a

[obtuse basin]

I don't get how you converted it though

[crystal raptor]

Oh I see why you're confused

I'm doing this to the 4 that's all the way on the right

I'm saying 5^n + 4 < 5^n +4*5^n

Carrying on our chain of inequality

[obtuse basin]

I'm still left off on 4n + 1 + 4 < 5^n + 4

But on the left side you got rid of the 4n + 1 and replaced it with 5^n

[crystal raptor]

^

[obtuse basin]

I don't really get what your basis for just throwing in a < 5^n + 4(5^n) is though

[crystal raptor]

You'll see exactly why

Because that's 5*(5^n)

Which is 5^(n+1)

[obtuse basin]

But it just kind of appeared out of nowhere

I don't get how you derive 4n + 1 + 4 < 5^n + 4 < 5^n + 4(5^n) from 4n + 1 + 4 < 5^n + 4

It just kind of appears

[crystal raptor]

You work backwards, you know where you need to end up

[obtuse basin]

Why do you multiply by the 4 by 5^n?

Isn't the end goal 5(5^n)?

Not 5^n + 4(5^n)?

[crystal raptor]

Simplify this

[obtuse basin]

So I threw it into a calculator online that's supposed to explain it step by step

It magically turns 4 * 5^n + 5^n into 5 * 5^n

I don't get how

It just says "Add similar elements"

[crystal raptor]

Call 5^n "sheep"

We have 4 sheep + 1 sheep

5 sheep

[obtuse basin]

Oh duh.

I'm actually stupid.

Is this right?

It kind of seems like I pull the 4*5^n out of my butt though

[crystal raptor]

That's right

I mean if you wanted to come up with it yourself you would work backwards, you know you needed to end up less than 5^(n+1) so you break that up and compare it a bit

You be like well 5^(n+1) = 5*(5^n)

Hmm I have only 1 5^n in what I've done so far

So I'll write it as 5^n + 4*5^n

[obtuse basin]

🧠

[crystal raptor]

And then you'd compare and see that the only logical step is using 4 < 4*5^n

[obtuse basin]

I see

ty

Now I have to solve this venn diagram thing

A survey of a group of 150 tourists was taken in St. Louis. 
The survey showed the following:

70 of the tourists plan to visit the Art Museum
80 plan to visit Gateway Arch
60 plan to visit the zoo
40 plan to visit the Art Museum and the zoo
45 plan to visit the Art Museum and the Gateway Arch
30 plan to visit the Gateway Arch and the zoo
25 plan to visit all three places.

Total = 150
F(A) = 70
F(B) = 80
F(C) = 60
F(A and C) = 40
F(A and B) = 45
F(B and C) = 30
F(A and B and C) = 25

I need to find the amounts for only A, only B, and only C

Not sure how to do that though with the info I've been given

Oh wait, but I know that N(A or B or C) = N(A) + N(B) + N(C) - N(A and B) - N(A and C) - N(B and C) + N(A and B and C)

so then N(A or B or C) = 70 + 80 + 60 - 45 - 40 - 30 + 25 = 120

And none of the places is the complement of any

So then it's 150 - 120 = 30

Now I need to only solve for A

Okay I figured it out

On the diagram, A and B is 40, but I want ONLY A and B

Which should be A and B - A and B and C

And from that, I can do the same for A and C

And then the total for A should be 70, or 70 - (20 + 25 + 15) which is 10 and hopefully correct

Thanks

.close

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[tacit bolt]

Hey for any french math people

[tacit bolt]

donc j'ai tout fait mais je ne vois pas quoi faire pour le 5)

[cedar kiln]

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[lean barn]

<@&286206848099549185>

[lean barn]

Pls help

I am kinda stuck

How can I find the greatest change of rate

Of this question

I try graphing but I couldn't because I mess up

[frosty oxide]

y = mx+b where m is the rate of change

[lean barn]

Ok thank you let me try real quick

I know that the change of rate of A is 6 and B is 2 and C is 3

I don't know if I am doing It right though

But C is (3,3) because up 3 to the right 3

[frosty oxide]

You should choose an interval that makes it easy for you to see the slope/rate of change

It seems like [2, 3] would be a good choice

[lean barn]

The thing is I need to find which of the 4 answers is the greatest change of rate

[frosty oxide]

When you go from 2 to 3, by how much do you go up?

[lean barn]

1

[frosty oxide]

No, I meant on the x axis, from 2 to 3 (referring to x)

When you go from 2 to 3 on the X axis, how much do you go up on the Y axis?

[lean barn]

You go 3 up

[frosty oxide]

Looks 4 to me

[lean barn]

Oh ok

Thank you

For your help

[frosty oxide]

Np

[lean barn]

.close

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[west cliff]

Hello, i was wondering how i would do these questions

[west cliff]

so for 2b) i got -2p 3√2pq

and for 3 i got -13v 3√2v^2 + 5w 3√3w

but im pretty sure they are wrong

and for 2a) 5m^2 √7n

[cedar kiln]

@west cliff Has your question been resolved?

[cedar kiln]

@west cliff Has your question been resolved?

[loud anchor]

i wonder if you've tried simplifying it with wolfram alpha.

which helps ppl to self-check their work

[cedar kiln]

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[thin roost]

Hello

[thin roost]

How do I graph a fucntion??

let's say $f(x)=x+1$

[wraith dagger]

! बाशुदेव

[thin roost]

How do I graph it

[obsidian coral]

Make a table of values and plot those

[thin roost]

Ok

Close

.close

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[lament sedge]

I have this graph(https://www.desmos.com/calculator/y0spruyhh3) were I take in a power of the points (t,w) where t=a-2,a-1,a,a+1,a+2. w is the same thing but a is replaced with b. So you get a square with a center of (a,b) that you can drag around. This square represents complex points and I am trying to make a other bunch of points be some power of the original points. The way I do this is by first making 2 functions. One is this. f(z1,z2)=sqrt(z1^2+z2^)^s*cos(s**tan^-1(z2/z1)) and the other is the same thing but with sin and is g(x). s is the power your trying to take. so then I plotted (f(t),f(w)) but for some reason it did not work with s=1/2 on the left half of the plane. So I thought it was a problem with the inverse tangent so I made a peicewise function in both g(x) and f(x) where if z1 is>=0 then take the normal inverse tangent and it is less then add pi. But the still did not work and I am a very confused. I tried many methods but they all made weird graphs on some part of the plane.

[cedar kiln]

@lament sedge Has your question been resolved?

[lament sedge]

<@&286206848099549185> If I need to clarify anything please tell me

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[eager kayak]

I have a question about solving through this trig sub because i don’t exactly know which part of the denominator should even be substituted after find the u and du and from there where i should plug it in because normally it would be the entire square root but the 4 before the x^2 makes it more complicated and different

[cedar kiln]

@eager kayak Has your question been resolved?

[eager kayak]

can i get clarification on how i should do the trig sub?

i only need to know beginning part but the 4x^2 makes it completely different from how i did it in class

where it’s always been just x^2

<@&286206848099549185>

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[crimson sedge]

<@&286206848099549185>

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[past lance]

question: When we deal with questions where binomial distribution probability formula is required, the order does matter right? So why are we using combinations instead of permutations?

[modern compass]

Because nCk counts the number of ways to have k successes in n trials

[past lance]

what would nPk count?

[modern compass]

Not really sure, but it would count the same outcome multiple times, giving you an over count

Basically, this is a box/object problem.
If you have 5 trials and want 3 successes, it’s the same concept as “you have 5 boxes and 3 balls, how many ways can you place the balls so there is at most 1 ball in a box”

If you put the balls in boxes 1,2,4 it’s the same as putting them in boxes 4,2,1 and 1,4,2

So you count the outcome with a combination

And you can really see this at the extremes
If we have 5 trials and 5 successes, there’s really only 1 way to do that, SSSSS
But if we used permutations it’s be 5P5 = 5!

[past lance]

i see what you mean for the ball case but in the case of binomial distribution its success/failure so say we want out of 3 events 1 to be a success we can have
success fail fail
fail success fail
fail fail success
these are the same thing but the order of when success occurs still matters

[modern compass]

Right, you’re choosing 1 box to put the success in

The order you choose the boxes doesn’t matter.

[past lance]

ohhh

ok i see what you mean

if out of 3 events 2 should be a success we would visualize this as 3 boxes and choose 1 box to put the successes in
so success A in box 1 and success B in box 2 is same as success A in box 2 and success B in box 1

but we can put it in box 3 too which is different

[modern compass]

Yep

[past lance]

alright this makes more sense now!

thanks!

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[muted dust]

Hi I need help here

[delicate ember]

,rotate

[muted dust]

Qn 3

[delicate ember]

what have you tried?

[muted dust]

I made this but idk what to do

[fervent mason]

32 divided by 5 has a quotient of 6 and a remainder of 2. This can be written as
32 = 6 × 5 + 2.

64 divided by 7 has a quotient of 9 and a remainder of 1. This can be written as
64 = 9 × 7 + 1

In general, “a divided by b has a quotient of q and a remainder of r” can be written as
a = q·b + r

That is the clue @muted dust

[muted dust]

Thanks i got it

.close

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[visual bloom]

hi can some one helb me?

[wispy talon]

Did you do part a?

[visual bloom]

i did but am not sure if it's correct

i said that a is vertically opposite

[wispy talon]

Vertically opposite will require lines to be parallel

But they aren't

Try to solve it in the reverse order, to find ACB, you need BDC and CDB

Now you need to find CBD

[wispy talon]

Try to do this

[visual bloom]

BDC and CDB= 90? is that right?

[wispy talon]

Yeah

BDC and CDB us the same thing

[visual bloom]

so am right

[wispy talon]

Yes

[cedar kiln]

@visual bloom Has your question been resolved?

[visual bloom]

so how dose that help me with finding a?

[cedar kiln]

@visual bloom Has your question been resolved?

[cedar kiln]

@visual bloom Has your question been resolved?

[loud anchor]

this can give a proof of the famous pythogoras theorem

[cedar kiln]

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[dire geode]

2 * undefined is undefined

[crystal raptor]

because you can't start applying rules like that to values that arent even defined

[dire geode]

only for x >0

[pulsar skiff]

depends on how you define ln(x)

[fair hare]

,w ln(-|x|)

[wraith dagger]

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[pulsar skiff]

technically ln(x) = 1 means (k ∈ Z) x = 2kiπ

[fair hare]

,w ln (|x|)

[wraith dagger]

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[sick ruin]

ln(-x) = 1/2 * ln(x^2)

By that reasoning

Hmm

[fair hare]

,w is ln(-x) = 1/2 * ln(x^2)

[pulsar skiff]

and ln(-1) = (k ∈ Z) (2k + 1)iπ

[wraith dagger]

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[pulsar skiff]

so ln(x) = a means x = e^a + 2kiπ and ln(-x) = ln(x) + iπ

[sick ruin]

Christ

[pulsar skiff]

so if you have 2 ln(-x) you get 2 ln(x) + 2iπ

[sick ruin]

Maybe that’s a little advanced

I mean

Complex numbers?

[fair hare]

cause

[lament elbow]

lmao

[pulsar skiff]

the right side reduces to 2 ln(x) since the period is 2π

[fair hare]

,w log(0)

[wraith dagger]

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[fair hare]

that's why

[sick ruin]

Actually 1/2 ln(x^2) is the same function as ln|x|

Interesting

[pulsar skiff]

so in conclusion if (k ∈ R | k < 0) ln(k) is defined, then 2 ln(k) = ln(k^2)

[dire geode]

isn't that just cuz $\sqrt{x^2} = |x|$

[wraith dagger]

riemann

[sick ruin]

Yeah

[pulsar skiff]

for x ∈ R yes

[fair hare]

yeah

[dire geode]

,w solve log(x^2) = i

[wraith dagger]

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[sick ruin]

Cool thing though cause it extends the domain to negatives while maintaining the properties we want

whereas absolute value is cringe and unbased

[pulsar skiff]

(x ∈ R) e^(ix) = cos(x) + i sin(x)

[pulsar skiff]

so x = +/- [cos(1/2) + i sin(1/2)]

[cedar kiln]

@crimson sedge Has your question been resolved?

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[balmy schooner]

Hi

[balmy schooner]

make 3√x^14 into the form of X^a * 3√x^b
with A being as big as possible whole number, and B as small as possible whole number

I need help for solving this question and an explanation.

The answer I got is x^2 . 3 √x^6

I’m not sure if it is correct

[dire geode]

is this your starting expression?

$3\sqrt{x^{14}}$

[wraith dagger]

riemann

[balmy schooner]

Yes

[dire geode]

do you know that $\sqrt{y}$ means $y^{1/2}$ ?

[wraith dagger]

riemann

[dire geode]

Also, do you know that

$(a^{1/n})^m = (a^m)^{1/n} = a^{m/n}$

[wraith dagger]

riemann

[balmy schooner]

No.. I learnt A^m/n but not the other ones

And root y = y^1/2 that I learnt

[dire geode]

the last one i sent is similar to this:

[dire geode]

ignoring the 3 for the moment, can you use the square root rule i sent to re-write this without the square root?

i.e. somewhere there should be a 1/2

[balmy schooner]

So is it x^1/14 and x^1/b ?

Hmmm…. Somewhere should be 1/2?

I’m still confused

[dire geode]

think of it like $y=x^{14}$

[wraith dagger]

riemann

[dire geode]

plug $y = x^{14}$ here

[wraith dagger]

riemann

[balmy schooner]

y = x^14

[dire geode]

now write this but with x

[dire geode]

using this

Do the substitution

[balmy schooner]

y^1/2 = x^14/2 ——> y^1/2 = x^7

[dire geode]

perfect

[balmy schooner]

So if root 3

It is y^1/3

[dire geode]

Are you asking if this is true
$\sqrt[3]{y} = y^{1/3}$

[wraith dagger]

riemann

[balmy schooner]

Yes

[dire geode]

yes it is

https://www.mathsisfun.com/numbers/nth-root.html

don't make these mistakes

this is similar to what you're doing

[balmy schooner]

make 3√x^14 into the form of X^a * 3√x^b ——-> x^14/3 into x^a * x^b/3 ———> x^4 * x^2/3 ?

[dire geode]

Observe that your problem is essentially to write 14 = a * 3 + b where 0 <= b < 3

14 = 7 * 2 + 0 from the previous problem

[dire geode]

oh yea 12 / 3 = 4 this is right

[balmy schooner]

So I am right?

Thank you so much for your time it really helped me.

.close

[cedar kiln]

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[kindred sundial]

Can someone help me understand how cos^2x-sin^2x ends up equaling cos2x

[dire geode]

#help-11 message

[kindred sundial]

Wow how did I not realize that

Thx

.clote

.close

[cedar kiln]

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[cedar kiln]

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[rich flint]

Hi, if have 9 different size books and i have to order them in a bookshelf having two of them together (the biggest one next to the following one( in size)) how many ways can I order them

[rich flint]

i did 8!2! but idk if it's ok

[fair geyser]

yes

that should work very well

[wraith dagger]

SirRiceBurger

[rich flint]

a friend said it's 7!2! but I think 8 makes more sence

[fair geyser]

8 makes more sense

[wraith dagger]

SirRiceBurger

[rich flint]

?

[dire geode]

stop spamming vico's help channel

[wraith dagger]

SirRiceBurger

[rich flint]

thanks for helping

.close

[cedar kiln]

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[crimson sedge]

there’s a bag with 14 marbles.
5 red
7 green
2 blue
what is the possibility of picking three marbles of the same color

[lament elbow]

well you have 2 ways to do that

P(all 3 red) + P(all 3 green)

[crimson sedge]

what about blue

do i just ignore it?

[lament elbow]

there are only 2

[crimson sedge]

correct

[lament elbow]

so you cant get 3 blue

how would you find the probability of all 3 red

[crimson sedge]

1/14 x 1/14 x 1/14 ?

[lament elbow]

nope

first of all, is this with or without replacement?

[crimson sedge]

without

wait

sorry with

it’s with replacement

it stays 14

sample size remains the same

[lament elbow]

are you sure?

does it say in the problem that you put the marble back in the bag after each pick?

if so, then we need to consider blue

[crimson sedge]

can we do it both ways

i’m not entirely sure

[lament elbow]

ok lets do without replacement first

[crimson sedge]

sure

[lament elbow]

there are 3 picks, all of which are independent of one another

for the first pick, what are the chances of getting a red ball?

[crimson sedge]

1/14

[lament elbow]

but there are 5 red balls?

so the chances are 5/14 to get a red one

[crimson sedge]

oh wait

ahh yes

that makes sense

sorry

yeah

ok so the chance of getting a red marble is 5/14

[lament elbow]

ok

that for the 1st pick

if we dont replace the marble in the bag, for the 2nd pick, how many total balls are there, and how many are red

[crimson sedge]

yea

13 total

4 r red

so now we have 4/13

[lament elbow]

yup

and the 3rd pick?

[crimson sedge]

3/12

[lament elbow]

great

so to get the chances of getting all 3 red (assuming we pick only 3 balls) would be those 3 probabilities multiplied together

[crimson sedge]

5/14 x 4/13 x 3/12

[lament elbow]

i.e. $\frac{5\cdot4\cdot3}{14\cdot13\cdot12}$

yeah

whatever that is 🙂

[crimson sedge]

haha

yes for it

[wraith dagger]

JamesH

[crimson sedge]

got it

makes sense

and the same principle applies for green as well

if it’s a non replacement then blue isn’t constituted

considered*

[lament elbow]

exactly

[crimson sedge]

and if we do replacement it’s just 5/14 ^ 3

[bold hinge]

Yeah just don't worry about the blue balls

[lament elbow]

yes

[crimson sedge]

and if we do do replacement then the blue would be 2/15 ^ 3

[lament elbow]

right

[crimson sedge]

ok

so say with replacement

it asks what is the possibility of picking 3 of the same color

would i list out each one

[lament elbow]

you would add P(3 red) + P(3 green) + P(3 blue)

that would give you P(all 3 same color)

[crimson sedge]

ok got you

so

(5/14)^3 + (7/14)^3 + (2/14)^3

would be

all three same color

[lament elbow]

right

[crimson sedge]

awesome

thank you so much

your a lifesaver

[lament elbow]

np

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[faint dirge]

@lethal sedge

[lethal sedge]

hi

@faint dirge

thanks for helping

[faint dirge]

hmm this shouldn't be that hard of a double integral

it should be perfectly straightforwards, no?

[lethal sedge]

i believe so

[faint dirge]

what would you need my help with, then?

[lethal sedge]

i just having trouble since its fubini

we didnt get much instruction on type 1 vs type 2 etc

I can do the double integral evaluation but need help on steps 1 & 2

[faint dirge]

a quick look just says it's the theorem that lets us integrate dxdy or dydx

ok so

[lethal sedge]

oh ok got it

[faint dirge]

that's all there is to it

it's type II, pretty much a single integral

[lethal sedge]

how do I draw it

sketch*

[faint dirge]

literally just graph the four curves

y=0, y=sqrt(3)/2, x=f1(y) and x=f2(y) (I'm lazy so 😓 )

the shaded area between them is your domain

[lethal sedge]

how do I graph x = f1(y)

im sorry

what is that

[faint dirge]

I just used f1(y) as a stand in for your x bounds

I'm just lazy

y=0, y=sqrt(3)/2, x=(-y+6)/3 and x=6arccosy

[lethal sedge]

oh ok so I graph that then write type 2

and then what is the integral setup pls

[faint dirge]

yup

[lethal sedge]

could you use mathway.com and screenshot it

im sorry i got no instruction whatsoever

[faint dirge]

most of the domain is pictured here

and I use desmos because desmos is the best

but it's the area between the parallel blue and red lines (the y=0 and y=sqrt(3)/2 ones)

[lethal sedge]

oh ok got it

[faint dirge]

and beneath the green curve and above the black curve

[lethal sedge]

and do you know the integral setup

im rlly sorry

[faint dirge]

it's no problem

[lethal sedge]

im extremely grateful for your help

[faint dirge]

dw about it

gl with your future work

the integral itself involves the integral of arccos I think

[lethal sedge]

oh ok

could you use mathway.com

it would be easier for me to see

[faint dirge]

so you might be better off doing it in three parts along the x axis, from the black blue to black red intersection, black red to green blue, then green blue to red green intersection

[lethal sedge]

the setup

[faint dirge]

wdym?

[lethal sedge]

the integral setup

like the actual setup

this lets you actually setup an integral

@faint dirge

[faint dirge]

we have latex for that lol

[\int_0^{\frac{\sqrt{3}}{2}}(6\cos^{-1}{y}-\frac{6-y}{3})(7)dy]

[wraith dagger]

Scythe

[lethal sedge]

isnt it a double?

[faint dirge]

well yes

but

[\int_0^{\frac{\sqrt{3}}{2}}\int_{\frac{6-y}{3}}^{6\cos^{-1}{y}}(7)dxdy]

is the same thing

[wraith dagger]

Scythe

[lethal sedge]

also over here isnt it technically not bounded by y=0

i mean the region D

[faint dirge]

it's not

[lethal sedge]

oh ok

[faint dirge]

the region D has a leftward bound at the purple line

[lethal sedge]

so its not bounded by the four lines/curves?

[faint dirge]

and continues until the black line meets the blue

[lethal sedge]

oh ok

[faint dirge]

the distance between the black line and the purple line at any given y is

[lethal sedge]

so this

[faint dirge]

no

[lethal sedge]

or this whole thing

[faint dirge]

that

the whole area between the curves

[lethal sedge]

so its a huge area

[faint dirge]

yes

well, "huge"

[lethal sedge]

ye ig

relatively

[faint dirge]

you integrate from y=0 to y=sqrt(3)/2

and at each y, the distance between the purple and black lines is

[6\cos^{-1}{y}-\frac{6-y}{3}]

[wraith dagger]

Scythe

[faint dirge]

hence why I put it in one integral

even if it's the same thing as the double

I just skipped an easy step

[lethal sedge]

ohhhh ok

thanks

this is not included right

[faint dirge]

nope

[lethal sedge]

everything to the left of the purple

got it

[faint dirge]

the purple line is your left bound

[lethal sedge]

got it

[faint dirge]

or bottom bound, if you look at it as the y axis being the independent one

[lethal sedge]

oh ok

[faint dirge]

to the right

the purple line is the far left. Everything to its right before the big black curve is in the domain

[lethal sedge]

got it