help-10

[fallow rampart]

Thank you for being my listener lol

This helps me retain it better anyway

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[obtuse pebble]

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[bright vessel]

Is this correct?

[glossy basalt]

nah, average rate of change being 0 means, the calories per minute didnt change "at all"

[bright vessel]

What is the right choices?

[glossy basalt]

nah

[bright vessel]

Nah?

[glossy basalt]

as in no

[bright vessel]

So what is the right answer

?

[glossy basalt]

you'll have to work out, i can give hints, but not answer

[bright vessel]

Sorry but your explanation is not clear

[glossy basalt]

indeed

[bright vessel]

And this assignment is due in 6m

[glossy basalt]

I didn't really explain

so, what do you know about rate of change

[bright vessel]

Change is slope?

They threw in this random question that wasn’t taught

[glossy basalt]

ohhhh, you know it's something about slope

it's about slope, but not change of slope

it's just the slope (for straight lines)

so, do you know which is the correct answer now?

[bright vessel]

Straight lines?

[glossy basalt]

hint: we need the rate of change to be 0

[bright vessel]

Ohhh

[glossy basalt]

and as i just said, it's just the slope

[bright vessel]

Makes sense thank you so much

[glossy basalt]

so..... get it?

[bright vessel]

Rate of change of zero

When x=zero

?

[glossy basalt]

it's slope =0

[bright vessel]

Got it thanks

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[obtuse pebble]

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[timid silo]

we're supposed to write as one single log

how does -2lnx become ln(x/y^z)

???

[royal basin]

sounds like they wrote the answer to g instead of e by mistake!

[timid silo]

oh yeh

thanks

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[obtuse pebble]

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[timid silo]

wait sorry

@royal basin what would be the ans for e

my ans key didnt write it

[royal basin]

do you know log laws

if you do it isnt hard to come up with yourself

in particular if you know $\log_a(x^p) = p \log_a(x)$

[timid silo]

would it be -lnx^2

[warm shale]

Ann

[royal basin]

you can massage it a bit further

[timid silo]

or lnx^-2

ln 1/x^2

?

oh k thanks

.close

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[errant lark]

Are you sure that there is no more context?

Okay.

not really.

You see if you have an inequality and if you multiply both the sides with a positive value, inequality(sign) still holds.

(x+5) can be negative.

But (x+5)^2 is always positive.

So, they tried to probably make it simple by not having to take cases.

That's why they multiplied with (x+5)^2 instead of (x+5).

!nosols

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[cyan kite]

isnt it just doing x≠-5 so it doesnt do zerodevision

[errant lark]

More like - to keep the inequality sign the same.

[cyan kite]

division*

[errant lark]

Actually, you can do it by multiplying with (x+5) as well. You'll just have to make cases.

Regardless, you will do it by yourself once your doubt is cleared.

So, do you get it?

[cyan kite]

huh

[cyan kite]

ahhh

yeah because you dint wanna change the inequality by multiplying it by a negative

like

3<4

times both by -1

-3<-4 isnt true

[errant lark]

Yeah

[cyan kite]

thats why they do (x+5)^2

[errant lark]

Did you not get it??

[cyan kite]

so they inequality holds

and doesnr flip the sign

oh ok

my fauly

[errant lark]

Cool

[cyan kite]

u can do (x+5)^2

cuz its a positive number

wdym

oh

then u need to times by (x+5)^4

even power

cabtvread without latex thing srry

yes

[obtuse pebble]

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[upper fog]

how to integrate this without u substitution

[gilded needle]

whatever you do is going to be equivalent to a substitution...

[warm shale]

Dyssrupt

[upper fog]

yeah but our teacher doesnt want us to actually substitute but be able to do it in our heads

[upper fog]

thanks

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[stoic elm]

`

can someone please check my work?

[errant lark]

In 3rd, count how many names are there.

[stoic elm]

oh

8

its 40320

[errant lark]

For 4, part 2, why do you think that repitition is prohibited in each section individually?

[stoic elm]

i couldnt really figure out what they meant

was it supposed to be like

for the whole number?

[errant lark]

Yes

Rest seems fine.

[stoic elm]

wait how do i do that one then

[errant lark]

You already know this. Just keep this in mind - Whichever digit you use, you can't use it again. You have seven such spots to choose a digit each for.

For first one, you have 10 choices.

Second one - 9 now. and so on.

[obtuse pebble]

@stoic elm Has your question been resolved?

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[hushed jasper]

hey, how do i prove that in x^2 - 3x + 4 > 0, x can be all real numbers?

[hushed jasper]

i used an online calculator that shows the solution step by step, but it uses "complete the square" which i haven't learned yet and hasn't been taught in the book

am i supposed to test random values? because the quadratic formula doesn't work, as 4*4 > 3^2

[timid silo]

No

So

You know graphs of quadratic equations?

Like how they form and all

[hushed jasper]

yes, x^2 is smiley face and -x^2 is angry face

[timid silo]

Yeah

[hushed jasper]

and when the quadratic formula doesn't work, it never goes below 0 i think

[timid silo]

So in this que we have a smiley face parabola right?

[timid silo]

if you haven't learned it yet, it's about time you do

it's a super useful thing

[timid silo]

Yes whenever b² - 4ac < 0 , it means the root arent real. That means graoh never cuts the x-axis

Thats what we want rn, for the smiley face to above the x-axis {positive region} for all x

Now just use the quadratic formula, show that the roots are unreal
And as coeeficient of x² is positive,
THE quadratic equation is always positive

[hushed jasper]

would it be enough as an answer to write "4*4>3^2, therefore no point below zero, every real value of x is > 0

[timid silo]

no

not enough at all

Uhhh no

[candid sigil]

Show that (b^2-4ac<0) AND (a>0) is true

[timid silo]

Showing that alone would be just telling people 16 is greater than 12

You have to show

• Determinant of quadratic equation is less than 0, therefore, unreal roots and also co-efficient of x² is greater than zero

[hushed jasper]

i haven't learned about unreal numbers yet 😅 but now i understand the problem, thank you guys

[candid sigil]

no

[timid silo]

just tell him to use completing the square

[timid silo]

We here are to provide solutions based on ones existing knowledge whenever possible. Telling him to use that method is the teachers work, not mine

[timid silo]

All it ever is, the graph would never cut the x-axis and it being an upward smiley, will tell us the quadratic is always positive for all x

[hushed jasper]

i feel like using advanced methods to solve problems in the book would be cheating

[timid silo]

See a graph , it would help

[hushed jasper]

but thank you!

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[obtuse pebble]

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[lavish drum]

what's the derivative of root4(t)

[fossil crag]

Write root4(t) as t^(a) where try to find the value of a

Then it's just a generalization of the derivative of t^n

[lavish drum]

t(1/4)?

t^(1/4)

[fossil crag]

Yes

Yeah

[lavish drum]

oh ye ok

is this right

,rotate

[teal turret]

What exactly did u do

U can simplify that

[timid silo]

She just

Used the exponent?

[teal turret]

Why’s the d/dx on the right

[timid silo]

It should be on the left i agree

She didnt differentiate or anything

[lavish drum]

I was just making sure the algebra was right but does it matter which side

[timid silo]

Just wrote it

Now just you use the uhh formula for differntiating f(x)/g(x) i guess

[teal turret]

Nah

Don’t do that

[timid silo]

Oh

[teal turret]

Easier way

[lavish drum]

could I just simplify inside

[teal turret]

@lavish drum first simplify that fraction

[timid silo]

Yeah, im new

[lavish drum]

yeah ok

[timid silo]

Oh thats what i said? x² + 1 / x

OH rihht simplify first

x + (1/x)

[teal turret]

Let them do it

[lavish drum]

yeah that's what I got

[teal turret]

Good now just go for it

Differentiate

[lavish drum]

1-(1/x^2)

are these right? I don't rly rmbr Taylor form

,rotate

[lavish drum]

is it just leaving y by itself or do I have to simplify too

[alpine raven]

taylor form ?

I guess they mean :
f'(1) (x-1) + f(1)

[lavish drum]

okay, so ig I did it right?

[obtuse pebble]

@lavish drum Has your question been resolved?

[obtuse pebble]

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[timid silo]

i don't understand how to do part e

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

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[obtuse pebble]

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[woeful gyro]

can someone help me

[obtuse pebble]

@woeful gyro Has your question been resolved?

[valid cradle]

First, let us make a geometrical representation of the situation.

Then, since we know that angle YTM is equal to angle MTS in measurement, we can make those measures equal to eachother, and solve for x.

[woeful gyro]

how do i do that with fractions

[valid cradle]

I suggest multiplying both the left and right sides by 2, takes away fractions and makes the problem easier

[obtuse pebble]

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[valid cradle]

.reopen

ah shoot

[woeful gyro]

.reopen

[obtuse pebble]

✅

[woeful gyro]

so 11/2x times 5/2?

[valid cradle]

What is your work as of this moment?

[woeful gyro]

wym

[valid cradle]

Like what is the equation you have set up?

It should look like $4x-\frac{5}{2}=\frac{11}{2}x-5$ (please work...)

[warm shale]

whynoms

[valid cradle]

cool it did work

Due to those measurements being equal, we can set up an equation to solve for x with those measurements on both sides.

[woeful gyro]

i got 5/3

[valid cradle]

That is correct.

So with this information, can you find the measure of YTS?

[woeful gyro]

and i solve for yts and its 25/3?

[woeful gyro]

i think so this another question im stuck on

[valid cradle]

Correct.

What do you know about similarity?

[woeful gyro]

what does the symbol mean

[valid cradle]

The ~ is similarity and the = with a ~ is congruence.

[woeful gyro]

ohh so the first answe and thrid answer of similar

[valid cradle]

Correct.

[woeful gyro]

so i would pick both as both true statemeants

[valid cradle]

Yep!

[woeful gyro]

am i correct

@valid cradle

[valid cradle]

that appears to be right!

[woeful gyro]

also these

[valid cradle]

Not quite there

[woeful gyro]

which answer

[valid cradle]

The first and third are incorrect.

[woeful gyro]

wouldit be rm since its congrunt

[valid cradle]

Correct.

[woeful gyro]

and for thr third one is it rym?

does it look congruent

[valid cradle]

Yep.

[woeful gyro]

last question is it 3 i counted all the dots that matched and got 3?

[valid cradle]

Not quite

[woeful gyro]

or actually it might be 1/3

[valid cradle]

That's it.

[woeful gyro]

i counted m to m prime

[valid cradle]

For the whole ratio.

So what is x?

[woeful gyro]

so the value of x i would put 1/3?

@valid cradle

[valid cradle]

No.

It is asking for the x/3

Atleast, that's the ratio between MM'/MM''

[woeful gyro]

so would would i do to get x

i got 1/3 as the ratio

[valid cradle]

Well, you have the ratio

Since it's x/3, and the ratio is 1/3, x is...?

[woeful gyro]

im confused 😭

do i have to find the ratio of 1:3

[woeful gyro]

it would be 1 ?

x is = to 1

@valid cradle

[valid cradle]

Yeah

[woeful gyro]

can yu double check this to see if im correct

[valid cradle]

Correct.

Okay that's everything that I'll give you

This is more akin to a math help for a singular problem, not a checking service 😅

[woeful gyro]

thank you you have been very helpful

[valid cradle]

no problem!

[obtuse pebble]

@woeful gyro Has your question been resolved?

[obtuse pebble]

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[ivory bolt]

how many solutions does this have between -9pi to 9pi?

[zenith spade]

Alright, so my first thought is to get rid of that pesky absolute value

[ivory bolt]

same here, but i dont think that would help actually

i believe it would be a case where we would need to analyse using the absolute value rules

as to when its positive and then negative

tried getting rid of it, got nothing

[zenith spade]

Well, think about the range of sin(x)

[ivory bolt]

did that aswell

i think its a good idea to analyse between, lets say 0 and pi

and then -pi to 0

there should be a pattern there where it would repeat to -9pi

and to 9pi

[zenith spade]

Yeah I agree, I was thinking to let y=sin(x), find some y that’s plausible from 0 to 2pi (or from -pi to pi) and then generalize to -9pi to 9pi

Since sin(x) is periodic

[ivory bolt]

eys

okay

i got some sort of patter on the right hand side

cant begin to think for the log though

<@&286206848099549185>

[timid silo]

There seem to be many images and questions. Would you mind providing me a specific image and question?

[ivory bolt]

.

[timid silo]

Are we only considering real numbers?

[ivory bolt]

solved it

just analysing the functions

[timid silo]

Okay

[obtuse pebble]

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[tight cave]

Having some trouble setting up the cross sections

Also don't know if I am taking the proper cross sections or not I think they are triangles but I am not fully sure

[ivory bolt]

no actually, 10

is it 10?

[tranquil sonnet]

yes

[ivory bolt]

there you go

base * height / 3

[tight cave]

yes the total volume is 10, but I am confused on how you arrive?

[ivory bolt]

here

[tight cave]

so would the base of the triangle be root 34 and the height root 41 like I made or would they be different?

[ivory bolt]

no bro

the base is 5 * 4 / 2

its just a right triangle

the height is just 3

[tight cave]

I see now. I was trying to take the cross sections horizontally instead of vertically. Thanks

[ivory bolt]

you are welcome

remember to not overcomplicate certain things 😉 life is simple my guy

[tight cave]

yeah this is for calc 2 so I thought I would have to set up an integral to solve it but I guess not

[ivory bolt]

i mean, you could

but why go through the trouble when there are simpler ways of doing things?

[tight cave]

true

[obtuse pebble]

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[obtuse pebble]

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[light condor]

how to solve the blank one?

[keen quarry]

wherever there is an x in f(x), replace it with (x-5)

then simplify

[light condor]

OHHHH

okay

that makes sense ty

.close

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[spice chasm]

If $\sin(2x)=\sin(x)$, does $x=0$?

[warm shale]

Jash

[placid badge]

depends on the domain

if its unbounded domain, then x = pi is a valid solution as well

[spice chasm]

how do you find every solution

[placid badge]

use double angle identity on the LHS

2sin(x)cos(x) = sin(2x)

[spice chasm]

oh ok

.close

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[hushed gust]

Hi everyone! Just started sets and would just like to confirm whether or not this is the correct thinking. Thanks!

[rich silo]

There are many more!

[hushed gust]

ill take another look

[rich silo]

To make it easier, consider your set something like {A,B,C,D}

Helps you find all of them

[hushed gust]

ok. im a bit confused about the {1,2,1,2}, there shouldn't be any duplicates in sets right?

[rich silo]

{1,2} and {1,2,1,2} are different elements here

[hushed gust]

is {1,2} not equivalent to it?

[rich silo]

mmmm.. no

[hushed gust]

Ok so I would just treat them as different things and match them together to create distinct subsets right?

[rich silo]

it's like the first set contains a black pen and a red pen

and the second set contains 2 black pens and 2 red pens

They are not the same!

[rich silo]

yess

Like {A,B,C,D}

[hushed gust]

{∅}, {{1}}, {{1, 2}}, {{1, 2}, {1, 2}}, {{1, 2}, {1}}, {{1, 2}, {1, 2, 1, 2}}

6 distinct subsets?

[rich silo]

Please, let's work with A,B,C,D 😛

And then just replace each letter with the element!

[hushed gust]

{A}, {B}, {C}, {D}, {B, C}, {B, D}

[rich silo]

Correct, but there are many more!

From a set of 4 elements, you can have

Subsets containing 1 element from the original set

Subsets containing 2 elements from the original set

Subsets containing 3 elements from the original set

Subset containing 4 elements from the original set

Did you understand me?

Let's start with the subsets containing 1 element

you already listed them

A, B, C, D

Now, the subsets containing 2 elements

There are not just BC and BD!

[hushed gust]

{A}, {B}, {C}, {D}, {A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C,D}, {A,B,C}, {A,C,D}, {A,B,D}, {B,C,D}, {A,B,C,D}

i've got this so far

is this what it means?

I'm not sure I can find anymore

[rich silo]

oh my goooood

Perfect

💯

[hushed gust]

😯 really

[hushed gust]

thank u so much. i will remember this

[rich silo]

Yeah, it's logical, 15

[rich silo]

You are welcome

[hushed gust]

Is there any way to know how many there should be?

[rich silo]

Yes

But not 2^n - 1

Do you know what a combination is?

$$C^{n}_{r}$$

[warm shale]

Madelyn

[hushed gust]

from a quick google search apparently it's a formula to check the possible arrangements possible. What are n and r though?

[nocturne minnow]

Same meaning as this, just different formattting

[rich silo]

There you go, he posted a very nice picture

[hushed gust]

What does "Number of items selected from the set" mean?

oh

it's the 1/2/3/4 combination right

so i would keep n = 4 and take r =1/2/3/4

and then take the sum of the 4 outputs?
so c1 + c2 + c3 +c4 = 15 in this case

[rich silo]

Yesss, exactly

nCr gives you the number of distinct ways

$$C^{4}{1}+C^{4}{2}+C^{4}{3}+C^{4}{4}$$

[warm shale]

Madelyn

[rich silo]

n always 4, because the original set has 4 elements

we are selecting, choosing from it

[hushed gust]

great!

[rich silo]

4C1 is for the subsets with 1 element in them

4C1 = 4

That's why we found 4

4C2 is for the subsets with 2 elements in them

4C2 = 6

That's why we found 6

And so on..

4C3 = 4

4C4 = 1

[hushed gust]

gotcha

[rich silo]

Lovely, isn't it 😋

[rich silo]

Good job, you're a fast learner!

[hushed gust]

id imagine this would be amazing for large calculations

[rich silo]

Exactly

A set of 4 elements, even if you do not use any math, you can find all the possibilities

But imagine if the set had, idk, 10 elements for example

How many distinct subsets are there?

would take forever to think it

But with nCr, you get your answer quick :P

[hushed gust]

yup

could i ask you math geniuses about some other questions also

[rich silo]

im no genius, im bad at math 😋

but yeah, you can open a new channel with a new exercise/question anytime

[hushed gust]

I've got an explanation for this but I feel like this method isn't optimised at all and not even sure if it's right

[nocturne minnow]

I'm not even a math genius, I did engineering and I just google everything lmao

[hushed gust]

We want to prove that every element in the intersection of sets A and B, i.e., (A ∩ B), is also an element of set C.

Let's take an arbitrary element "x" from (A ∩ B), which means "x" is both divisible by 2 and divisible by 9. Mathematically, we can write this as:

x ∈ A ∩ B

This implies that x is a multiple of both 2 and 9. In other words, x can be written as:

x = 2k1 = 9k2

where k1 and k2 are integers.

Now we want to show that x is also divisible by 6, which would mean x ∈ C. To do this, let's consider the prime factorization of 9:

9 = 3 * 3

Since x is a multiple of 9, it must also have a factor of 3. And since x is a multiple of 2 (because it's in set A), it must also have a factor of 2. Therefore, x can be written as:

x = 2 * 3 * (some other factors)

Notice that x has both a factor of 2 and a factor of 3, which means it has a factor of 6. This proves that x is divisible by 6, and thus:

x ∈ C

Since "x" was an arbitrary element from (A ∩ B), we have shown that any element in (A ∩ B) is also an element of set C. Therefore, we have proven that (A ∩ B) ⊆ C.

Set A consists of integers that are divisible by 2:
A = { ..., -4, -2, 0, 2, 4, ... }

Set B consists of integers that are divisible by 9:
B = { ..., -18, -9, 0, 9, 18, ... }

Set C consists of integers that are divisible by 6:
C = { ..., -12, -6, 0, 6, 12, ... }

The intersection of sets A and B is:
A ∩ B = { ..., 0, 18, 36, ... }

Looking at this intuitively. I can see that the statement is true but I'm not sure how to prove it with a decent method..

[rich silo]

Okay look, A is the set of all multiples of 2, right?

B is the set of all multiples of 9

C is the set of all multiples of 6

A∩B must then be the set of all multiples of 2 & 9 at the same time

in other words, the set of all multiples of 18

[hushed gust]

ye and 6 is a factor of 18

[rich silo]

Exactly, so all the multiples of 18 are eventually multiples of 6

[hushed gust]

[rich silo]

you can show it like this

$$\text{Let x } \in A \cap B , \to x=18K$$

[warm shale]

Madelyn

[rich silo]

$$x=18K=6 \times 3 \times K = 6 \times (3K)$$

[warm shale]

Madelyn

[rich silo]

You agree that 3K is just another integer

3*integer is just another integer

Say, K'

[hushed gust]

yup

[rich silo]

$$x=6K'$$

[hushed gust]

sum/product of integers = integer

[warm shale]

Madelyn

[rich silo]

Like this, you've shown that all multiples of 18

Are multiples of 6

[compact shadow]

The question didn’t say “non-empty”

[rich silo]

Oh yeah, right

The empty set is a subset of every set

I always forget to include it 😝

Thank you!

[hushed gust]

what does that mean? 4C0 would be needed too?

[rich silo]

Yeah, an extra set which contains nothing, the empty set is also a subset

That makes them 16

[compact shadow]

Set A having n elements, subsets B of A one-to one correspond maps from A to {0,1}. Mapping the elements in B to 1, elements not in B to 0. So how many such maps? Each element x has two choices of its image, n elements so 2^n many

Is an alternative way

[rich silo]

interestiiing

i did not know about the 2^n

Awesome!

[rich silo]

{A}, {B}, {C}, {D}, {A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C,D}, {A,B,C}, {A,C,D}, {A,B,D}, {B,C,D}, {A,B,C,D}, ∅

[hushed gust]

The {A} in question is the empty set here. After including the empty set I only get 15 combinations

im confused haha

[rich silo]

What do you mean by {A} is the empty set here

in your exercise, the first set is {0}, not the empty set!

The empty set is denoted { } or ∅

Add it to the listed 15 subsets

{A}, {B}, {C}, {D}, {A,B}, {A,C}, {A,D}, {B,C}, {B,D}, {C,D}, {A,B,C}, {A,C,D}, {A,B,D}, {B,C,D}, {A,B,C,D}, {}

[hushed gust]

oh the ∅ in {∅, {1, 2}, {1}, {1, 2, 1, 2}} is not the empty set

[rich silo]

No, that's not a ∅, that's a zero 0

it is one of the elements of the set

[rich silo]

⬆️

A=0

[hushed gust]

I think I made a critical mistake in copying the question

[rich silo]

oh nooo, omg 😛

[hushed gust]

would it be then:

ABC combinations + {}

and so 8 combinations including empty set?

[compact shadow]

What he told you is the same
(1+x)^n=ΣC(n,k)x^k
Plug in x=2, you also have
2^n=ΣC(n,k)

if three elements, n=3, then 2^3

But here n=4 though, still

φ is a valid element of the set

{φ} and φ={} are completely different thing

[rich silo]

Well said

Thank you!

So, 16 still in here

Those 15 we found + ∅

[obtuse pebble]

@hushed gust Has your question been resolved?

[hushed gust]

i see

[rich silo]

Did you get it

[hushed gust]

the 0 in the set is an element right

the question asks for subset, does that mean we do not include {A} by itself?

[compact shadow]

We do

[rich silo]

We dooo

{A} is {φ}

[compact shadow]

A though, not {A}

A is an element of {subsets of A}

A , φ, and others

[hushed gust]

understood

thank you guys. i appreciate it 😊

ime back to grind lol, assignment is due and i've barely done anything, lesson learnt for next time

oh my goodness

(a) should just be 15 ways right

excluding the empty set

[compact shadow]

Checking

[hushed gust]

no my bad

i didnt read the question properly

[compact shadow]

Is one position only taken by one person

I know one people can take many positions, but is any position taken by exact one person

A position is taken by
<=1 person
or exact 1 person
or any number of people?

[hushed gust]

im not sure, this is all the info given

[compact shadow]

I calculate first assuming a position can only be taken by <=1 person
(Make sense right, we never heard of having two head masters for example)

I got different answer

(5)(5)(5)(6)-(4)(4)(4)(5)

Four positions, D for dean, A,B,C for other three
5 persons, l for lord, h for Harry,x,y,z for other three

Position A can be taken by 4 people hxyz or no one, 5 cases similar for B,C.
D can be taken by all 5 people or no one, 6 cases, this gives us (5)(5)(5)(6)

And if h got no position

Each position we have one less case

So minus (4)(4)(4)(5)

[hushed gust]

sry ima just take a few min and try to absorb this haha

[compact shadow]

Sure

[hushed gust]

mah I,Q not high enuf

[compact shadow]

Me neither. Good thing that math doesn’t require that , early stage

[hushed gust]

For (B)

would we use a similar idea to combination

like 4C2 + 4C3 + 4C4 / 4C0 + 4C1 + 4C2 + 4C3 + 4C4

[compact shadow]

No

h is offered two positions among ABC: C(3,2)(4)(5), h is offered D and one position among ABC: C(3,1)(4)(4)

Divide by number of all cases (5)(5)(5)(6) in the end

(I used notation C(n,m) for your nCm)

[hushed gust]

I'm a bit confused here. Is this literally 5 * 5 * 5 * 6 - 4 * 4 * 4 * 5?

[compact shadow]

5 times 5 times 5 times 6

There is no multiplication symbol on my keyboard so I typed that way

Yeah

And * often disappeared in my text so I don’t like using it

[hushed gust]

C(3,2)(4)(5) * C(3,1)(4)(4) / (5)(5)(5)(6) ?

[compact shadow]

No plus

[hushed gust]

shieeee

right

cuz it's or right

like in logic

[compact shadow]

(C(3,2) * 4 * 5+C(3,1) * 4 * 4)/5 * 5 * 5 * 6

Yeah

Other two are easy I leave hints here when you do them
c): clearly only || 4 || cases where a candidate takes all positions
d): clearly position ||D|| is offered to ||l||, so ||3|| remaining positions for ||4|| people, so ||(4)(3)(2)||

[hushed gust]

C = 80%?

[compact shadow]

No 4 cases

Divided by (5)(5)(5)(6)

[hushed gust]

I got 0.144

[compact shadow]

I didn’t calculate it

Correct

[hushed gust]

for d would we also need to take in to acc the total (/total case)

i thought i got it but nvm

[compact shadow]

Yeah

[hushed gust]

c(4,1)

[compact shadow]

No need, it wasn’t asking for probability

[hushed gust]

i think

[compact shadow]

No

(4)(3)(2)

3 remaining positions for 4 remaining people

First position 4 choices, second position 3 choices left…

[hushed gust]

Just 24 ways in which l can get a position

since he only applies for D rig

[compact shadow]

Yeah

[hushed gust]

man this is so hard for me

[compact shadow]

I see

[hushed gust]

engineering students hhaha

measly cs student here

[compact shadow]

I see. They require something different

Other than math subjects I like, I sometimes see engineering students focus more on things like Fourier transform, or statistics , some learn little algebra

[hushed gust]

For listing distinct prefixes, suffices, and substrings. Would the string itself belong to all 3 categories?

[compact shadow]

I don’t know what these are, googling

Found it

I think so

Since the whole is in the beginning also in the end

Yeah wiki confirmed it

[hushed gust]

mad lad

u should test ur i.q haha

a lot of the concepts i see on wikipedia uses language i dont understand

personally

[compact shadow]

There is no actual concepts unfamiliar to me, just called by different names by different people

For example, I saw one person call “permutation” as “shuffle” in one of these channel. I am perfectly familiar with permutations

So I look at how he define his “shuffle” then I can answer his question, just that.

High i.q people solve competition problems, I can’t

[hushed gust]

haha ok XD

i guess im low i.q

[compact shadow]

I don’t think so. Besides it’s not needed in math, at least in early stage

Easiest subject, irrelevant with what people feel or imagine. Irrelevant with the real world. Everything white or black no grey area. Pure logic step by step, no need for memorization, observation, experiment, money… no extra information, every information is contained from definition, axioms in the beginning…

[hushed gust]

fair

true, the logic is great, it's not like writing an essay where u can get marked up or down based on how the marker feels about it

what profession in engineering are/did you taking/take?

[compact shadow]

I major in math

I mean I saw engineering students focus on different things compared to what I am interested in. I know nothing about Fourier transform , statistics, PDE…

[hushed gust]

over here in new zealand we have like electrical, software, chemical, civil engineering etc

are u in the states?

[compact shadow]

No, living in Japan

[hushed gust]

shieeee

i heard it's a good place for visiting but not living

it's pretty harsh there apparently

[compact shadow]

US is probably best in terms of math. Europe second

[hushed gust]

exchange student?

or are u japanese

[compact shadow]

I will become a Japanese in the future

Right now I am a refugee in my own eyes even though I am not one technically

[hushed gust]

thoughts on japan?

[compact shadow]

Good, maybe a little sad that math in Japan in general are declining

More and more resources gathered in few top universities things like that

Best universities in local , for example mine, doesn’t even have algebraic geometry lessons, (even though my professor know a lot about it, gap between elder and younger is also an issue. They are far better)

[compact shadow]

I don’t know why, it seems like the center of mathematics always follow the center of economics. When Paris was the economic center, France was the first in terms of math. It is and always will be one of the tops of course.

Anyway that will be some total new discussion in discussion channels

[hushed gust]

i guess u could put US on ur bucketlist

[compact shadow]

Too hard, almost impossible to achieve. Anyway, I am satisfied with Japan very much already. I don’t have the need. I am not good at math, I also don’t major in main subjects in math. Japan is more than sufficient for me to find everything I need.

[hushed gust]

can u speak japanese?

nvm that's a dumb question ofc u can

[compact shadow]

I really need something more I guess I will consider applying for Kyoto , Tokyo Universities again in the future when I finish my current PhD and get money.

Yeah I can

[hushed gust]

alright bro, ima head off to finish my physics lab report. good talking to you and i really appreciate ur help today

https://tenor.com/view/love-hug-gif-25661526

[compact shadow]

No problem

Any time

[hushed gust]

can i add u?

[compact shadow]

Sure

I always welcome people. I reply when I check my DM

[hushed gust]

.close

[obtuse pebble]

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[sleek sinew]

Not sure if it is the first or last interval

And what they mean by "f increases the most"

I think the last interval, the total change in y is the most

[polar fossil]

yeah total change in y

[sleek sinew]

The first one has width 1

The last interval has width 2

So it would be D) because f increased by a total of 12?

[sleek sinew]

?

Would it be the last interval then?

[polar fossil]

yeah

[sleek sinew]

Thank you

[obtuse pebble]

@sleek sinew Has your question been resolved?

[obtuse pebble]

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[fossil spoke]

i got y=(x-4)/9

but answer says

[fossil spoke]

x'=3x+5y
y'=x+2y

x=(x'-5y)/3
y=(y'-x)/2

i plugin these expressions into y=2-3x

(y-x)/2=2-3((x-5y)/3)

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

y-x=4-2x+10y

9y=x-4

y=(x-4)/9

im confused

<@&286206848099549185>

[obtuse pebble]

@fossil spoke Has your question been resolved?

[obtuse pebble]

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[floral canopy]

Hello, I have a question in Linear Algebra 1.
If I have 2 matrices A,B and I know that if i≠k then Ci(A)=Ci(B) but for i=k Ck(B)≠Ck(A)
Then how can I find det(A+B)?

[royal basin]

what is Ci(A)?

[floral canopy]

The i'th column of A

[royal basin]

right

so A and B match in every column except the k'th.

[floral canopy]

yea

[royal basin]

so then (A+B)/2 will have the same columns (except the k'th) as both A and B, and its k'th column will be 1/2 C_k(A) + 1/2 C_k(B)

and you can use the multilinearity of the determinant to say det[(A+B)/2] = 1/2 (det(A) + det(B))

[floral canopy]

wait how can you say that?

wouldn't it be det[(A+B)/2] = 1/2 (det(A+B))?

[royal basin]

that's wrong, and also not what i'm referencing.

[floral canopy]

oh right

[royal basin]

if you fix all columns but one in a matrix, and consider the determinant as a function of the last column, then said function is linear.

[compact shadow]

This is correct

[floral canopy]

no its not

how?

Im so confused

[compact shadow]

All other column is the same as A , or B

[royal basin]

no it isn't.

you know determinants don't scale like that

[floral canopy]

isnt it like 0.5^n or smth

[royal basin]

don't you?

yes precisely, det( (A+B)/2) = 1/2^n * det(A+B).

[compact shadow]

Oh nvm, sorry

Being dumb again

I thought he wrote det(A)+det(B) RHS

[floral canopy]

But isn't this just assuming the original statement to be correct? I still dont understand how we got from det[(A+B)/2] --> 1/2 (det(A) + det(B))

[compact shadow]

A=(c(1),…,a,…,c(n))

The k-th column is a
B=(c(1),…,b,…,c(n))
The k-th column is b
(A+B)/2=(c(1),…,(a+b)/2,…,c(n))

[royal basin]

if you fix all columns but one in a matrix, and consider the determinant as a function of the last column, then said function is linear.

[floral canopy]

But is there a proof or something to this? Like how do you know said function is linear?

[royal basin]

yes there is a proof

boils down to laplace expansion

if you laplace-expand along the k'th column, you get the entries of the k'th column scaled by some (n-1)×(n-1) determinants which don't involve any other entries from the k'th column

with some minus signs sprinkled in which don't detract from my point

i'll be honest i cbf to write it all out in its full gory detail.

[floral canopy]

Ill look it up

thanks!

[obtuse pebble]

@floral canopy Has your question been resolved?

[floral canopy]

I think I figured the proof myself

thank you!

.close

[obtuse pebble]

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[obtuse pebble]

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[obtuse pebble]

Please don't occupy multiple help channels.

[rough nexus]

For (b),

[alpine raven]

and in english, what does it mean ?

[rough nexus]

There's uh, English translation under Malay

[obtuse pebble]

@rough nexus Has your question been resolved?

[obtuse pebble]

@rough nexus Has your question been resolved?

[obtuse pebble]

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[delicate onyx]

Hi! I have a small problem and unsure how to proceed. Here it is below:
\begin{equation}
\int(\mathbf{Y}\times\mathbf{X}^q)d\mathbf{X}
\end{equation}
The main difficulty is that X and Y can be any combination of matrices or vectors such that the expression (cross product) is valid. What is the right way to solve it?

[warm shale]

ekh

[delicate onyx]

*Y is independent of X

[obtuse pebble]

@delicate onyx Has your question been resolved?

[delicate onyx]

Here is my attempt

[obtuse pebble]

@delicate onyx Has your question been resolved?

[delicate onyx]

Nope

[obtuse pebble]

@delicate onyx Has your question been resolved?

[delicate onyx]

<@&286206848099549185>

[obtuse pebble]

@delicate onyx Has your question been resolved?

[obtuse pebble]

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[obtuse hatch]

I have to write an absolute value function as a piecewise function, I know how to get to the answer and what the answer is, but I dont know how to flip the inequality so that it is right.

[obtuse hatch]

damn, I wasnt thinking

ignore that one

but the first question

[tacit scarab]

,w graph (x+4)(x-1)

[warm shale]

Results provided by WolframAlpha

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

Requested by _wherewolf_

[tacit scarab]

You should get $x\le-4 \ \text{and} \ x\ge 1$

[burnt vale]

u fortgot ,,2"

[warm shale]

WhereWolf

[tacit scarab]

It's the same anyways

[obtuse hatch]

so what would be the right way to set up the inequalities to solve them, or do I have to graph?

going from (x+4)(x-1)

[warm shale]

Dyssrupt

[obtuse hatch]

idk what a.b means, but at least that a.b is greater than or equal to 0

[tacit scarab]

it should mean a times b

[obtuse hatch]

ahk

[frank monolith]

a and b are (x+4) and (x-1) for your problem

if that helps

[obtuse hatch]

for it to be equal to 0 at least one of them has to be 0

and for it to be positive they have to be same polarity

[frank monolith]

yeah, same polarity

so either both of them positive or both negative

you had one condition correct

[obtuse hatch]

so I like resonate my way into changing the inequality after doing both of them as >=0?

[frank monolith]

you forgot negative one

[frank monolith]

wdym

[obtuse hatch]

I can explain but my thought process is prob wrong

if its not too much to ask I just want the way to solve it after I get (x+4)(x-1)>=0

I usually learn best by looking at solving methods

or if even my method up until that point is the best one

[frank monolith]

just dont nosol me wolf

[tacit scarab]

Lol ok

[obtuse hatch]

oh, you cant give me the exact solution ig?

[warm shale]

Dyssrupt

Dyssrupt

[frank monolith]

i dont know how to do line breaks sorry

[obtuse hatch]

its fine

I'm following along so far

[frank monolith]

do the rest yourself lmao

if you get stuck, then ask for help

[obtuse hatch]

do I use the "most negative" number for the inequality?

[frank monolith]

cant send the solution, sorry

[frank monolith]

take the intersection of 'and' conditions

and union of the 'or'

[obtuse hatch]

was wondering if you could send the solution to like (x+7)(x-3) and get around it that way

since I am mostly after teh method and not the actual answer itself lol

[warm shale]

Dyssrupt

[obtuse hatch]

OOOH

[frank monolith]

now do for when both are negative

[obtuse hatch]

thank you

so i just solve both this one

and this one

and then take what I get in common

[frank monolith]

no

not common in this case

[obtuse hatch]

I mean as in for both

[frank monolith]

its an 'either' situation ig

[obtuse hatch]

oh right

ig I'll just experiment instead of writing my thoughts

I got what I need now tho ty

do I need to do something to mark it as done?

[frank monolith]

type .close

[obtuse hatch]

.close

[obtuse pebble]

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[dawn sonnet]

Hi, for this question I subsituted x^2 + 1 as t and got integral dt/t

[dawn sonnet]

but im confuse how i would solve after that

[robust sleet]

basically inyegral of 1/x

[burnt vale]

do you know the lnx derivative?

[hexed gull]

you don't need to, you can use inverse chain rule for it

since the derivative of x²+1 is 2x

[sage geode]

Inverse chain rule is basically the same as substitution

Either way they need to realize differentiating which function yields 1/t

[hexed gull]

ys except you don't need to do it explicitly here :)

[sage geode]

Right

[hexed gull]

@dawn sonnet are you still present

[obtuse pebble]

@dawn sonnet Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[pallid owl]

hello

[timid silo]

hello

[pallid owl]

should i post the question here

[timid silo]

yes

[pallid owl]

A person is remodeling the front entrance to his home and needs to cut an arch for the top of an entranceway. The arch needs to be 30 ft wide and 5 ft high. To draw the arch, he will use a stretched string with chalk attached at an end as a compass.
a) Using a coordinate system, locate the center of the circle.
b) What radius should the person use to draw the arch?

[timid silo]

Where are you stuck?

[pallid owl]

finding the center of the circle

[timid silo]

Do you know the equation for a circle?

[pallid owl]

yes i do

[timid silo]

okay

so

I want you to

draw a picture of the entranceway

[pallid owl]

i did look

it’s a little messy

[timid silo]

That

circle, is so

[slender nest]

it, works

[frank monolith]

symmetric

[pallid owl]

😭

[slender nest]

https://tenor.com/view/perfect-circle-class-gif-12604813

do this

[pallid owl]

it’s such a stupid question like it’s gonna be the end of me

[pallid owl]

yes

[timid silo]

What is pythagorean theorem?

[pallid owl]

15^2+ x^2= x+5

[timid silo]

no

[pallid owl]

a^2+b^2=c^2

[timid silo]

yes

[pallid owl]

yes i plugged that in

[timid silo]

$$c^2=a^2+b^2$$

[pallid owl]

yes

yup

[warm shale]

Free Geoffrey

[timid silo]

why are you saying yep 😭 all of those were wrrong

[pallid owl]

anyways

[timid silo]

Plug in values

what is x

[pallid owl]

15

[timid silo]

wut

[pallid owl]

oh

we don’t know x

that’s the midpoint that we’re tryna figure out

[timid silo]

Okay

solve for x

you have

pythagorean theorem

[pallid owl]

x=20

[timid silo]

Yes

where is center

[pallid owl]

(20,0) ?

[timid silo]

wut

where is (0,0)

draw picture if you don't understand

[pallid owl]

it’s in the origin

ugh

we’ll i got (0,-20)

[obtuse pebble]

@pallid owl Has your question been resolved?

[obtuse pebble]

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[delicate onyx]

/reopen

[delicate onyx]

.reopen

[delicate onyx]

Anyone?

🙏

[tardy epoch]

[delicate onyx]

Hi! I have a small problem and unsure how to proceed. Here it is below:
\begin{equation}
\int(\mathbf{Y}\times\mathbf{X}^q)d\mathbf{X}
\end{equation}
The main difficulty is that X and Y can be any combination of matrices or vectors such that the expression (cross product) is valid. What is the right way to solve it?

[warm shale]

ekh

[delicate onyx]

Here is my attempt

Sorry for handwriting

[obtuse pebble]

@delicate onyx Has your question been resolved?

[obtuse pebble]

@delicate onyx Has your question been resolved?

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[trim canyon]

help?

[unkempt flicker]

"for every 8 duck decoys MM carves he must give up 4 fishing lures" is the big key here

if [# decoys goes up by 8] then [# lures goes down by 4]

can you perform that pair of actions on the first line in order to fill in the second line of the table?

[trim canyon]

so the next line is 16 lures and 8 decoys yes?

[unkempt flicker]

yes

[trim canyon]

ok so i got

20,0
16,8
12,16
8,24
4,32
0,40

i believe its correct?

[unkempt flicker]

looks good to me

[trim canyon]

tysm <3

[unkempt flicker]

👍

[obtuse pebble]

@trim canyon Has your question been resolved?

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[wind creek]

Hi everyone, I'm having a hard time trying to figure out this one. If someone could help me

I don't even know where to start. I've tried using polar coordinates but it lead me to nowhere

[fierce lagoon]

Write out your u(x,y) and v(x,y), find the jacobian

Mb not u(x,y) and v(x,y)

Write the region as a set of equations which it is bounded by

And then find your u = and v = equations

And then find the jacobian

[wind creek]

Should I separate the region in 4 small regions?

and how do I know what the correct u and v are?

.close

[obtuse pebble]

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[obtuse pebble]

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[dreamy blade]

what is 2^^2?

[frosty river]

Too vague question

2^^2 can be 2²=4 in some programing languages

[obtuse pebble]

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[dreamy blade]

I figured out it is tetration

[obtuse pebble]

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[warm atlas]

isn't it: 3x^2+3y^2*y'-3y'=0 the implicit differantiation?

[grizzled shore]

What happened to the product rule on -3xy

[warm atlas]

Thankssssss!!

.close

[obtuse pebble]

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[obtuse pebble]

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[lethal ice]

sin^4 (x) dx
(its actually part of a bigger problem but this is the part im having trouble with)
im pretty sure i have to use the half angle formula thing but Im not sure how to go about it

[shut lagoon]

Just write sin^4 (x) = (sin^2 (x))^2 and apply half angle to the inside. Then develop and apply half angle again.

Might be an easier way depending on what the rest of the integral is.

[lethal ice]

the rest was mostly just extra numbers like the inside of the sin was 6x and it's a definite integral

o to pi/6

[naive sigil]

https://www.khanacademy.org/math/integral-calculus/ic-integration/ic-integration-with-trig-identities/v/trig-identies-for-trig-integral

[lethal ice]

bros the resource provider

ty