help-23

[peak crane]

Anyone?

<@&286206848099549185>

Any1?

Anyone

Help

Help

Help

[lean otter]

do you know how to use log tables?

try taking the log of (1.08)^10

use the power law

[peak crane]

Not really

Ik how to use log tables

10log1.08=0.334

Right?

[lean otter]

yes

now find 10^0.334 by reversing the log lookup process

[peak crane]

What’s that

@lean otter

Ohhhh

I see

Thanks

.close

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[echo sequoia]

The red arrow points to my chosen answer, the white arrow is what is supposed to be the answer according to the assignment. Why? And how do I come to this conclusion with the given information?

[safe radish]

@echo sequoia Has your question been resolved?

[echo sequoia]

<@&286206848099549185>

[safe radish]

@echo sequoia Has your question been resolved?

[safe radish]

@echo sequoia Has your question been resolved?

[safe radish]

@echo sequoia Has your question been resolved?

[plucky elk]

https://proofwiki.org/wiki/Moment_Generating_Function_of_Linear_Transformation_of_Random_Variable

[echo sequoia]

Not particularly helpful given the question, but thank you for the response.

[safe radish]

@echo sequoia Has your question been resolved?

[spiral bane]

I think it helps

It’s saying the exponent on e should be 12 because that is beta

But if you simplify that expression you get 4

So it should be 16-4 to get 12

[echo sequoia]

It says that I am looking for $e^{\beta t} M_X(\alpha t)$

[flat frigate]

BlewiiQ

[echo sequoia]

But I do not know what $M_X(\alpha t)$ is. That is the question.

[flat frigate]

BlewiiQ

[spiral bane]

Oh I don’t either

But it’s enough to solve the multiple choice question

[safe radish]

@echo sequoia Has your question been resolved?

[zinc berry]

I don't know why you aren't just directly computing this using the definition.

[round mason]

What is the Pythagorean theorem

[zinc berry]

By the law of the unconscious statistician, the expected value is $M_{Y}(t) = \mathbb{E}(e^{tY}) = \sum_{k=1}^{\infinity} e^{t(4k+12)} (1-0.32)^{k-1}(0.32)$. The rest is just simplification and gives you the 5th answer as expected.

[echo sequoia]

I figured out the problem. There are two moment generating functions. Determined by the two probability density functions. Each yields a different answer, but it is customary to assume X is the total number of trials rather than the total number of failures. My answer is therefore correct. Thanks to the math department head.

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[lean otter]

Would the answer be (B)

?

[grim plover]

lemme see

Nope C I think

[muted lagoon]

shouldnt it be (a)?

[grim plover]

Its asked how many voten in favor, no?

[muted lagoon]

cuz number of people who voted in favor of the proposal is 3x

so 3x - x = 15,000

[grim plover]

OH SHIT

[muted lagoon]

x = 7500

[grim plover]

I READ THE QUESTION WRONG

AM STUPID

[muted lagoon]

yeah

[grim plover]

Sorry

[muted lagoon]

nice

[grim plover]

Yeah it's A

[muted lagoon]

its fine

[lean otter]

Do you plug and play each one or how do you get that expression so fast?

so 3x - x = 15,000

nvm I see know thanks guys the both of you.

@grim plover@muted lagoon

[muted lagoon]

🙂

[grim plover]

🙂

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[oblique idol]

If φ : G → G′
is a homomorphism then |G : ker φ| = |φ(G)|.

[oblique idol]

So this is a true or false question

I've come to the conclusion it's true, but only because I've tried a bunch of examples and they all come out true

[peak estuary]

do you know the iso theorem?

[oblique idol]

yes, I imagine you're referring to the first one?

ooooooooooooooo

G/ ker φ ∼= im φ. I just want to clarify and ask if my thought process is correct

just by the definition of the image

φ(G) = im φ

and since they are isomorphic, its true

[peak estuary]

yes

[oblique idol]

thank you

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[boreal cypress]

why is x=-7 undefined

[boreal cypress]

nvm

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[brave wolf]

it's 0/0, but the limit would be -14/1 so -14

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[lean otter]

what do i need to study topology?

[lean otter]

hello

[worthy hemlock]

Math \s

[lean otter]

?

[peak estuary]

a book

paper

time

[lean otter]

that dont answer my question

[peak estuary]

ask a bad question, get a bad answer

[worthy hemlock]

Google what you need to study for topology

[safe radish]

@lean otter Has your question been resolved?

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[raven grotto]

i know its explaining how to do it but im so confused could i get some help please?

[quasi bison]

what are you confused about?

[raven grotto]

how to work it out znd like what im suposed to do i just dont know

[misty bay]

do you understand the words in the question

[raven grotto]

sort of yeah

[misty bay]

wdym sort of

either you know the words or you don't

[raven grotto]

no i dont

[misty bay]

which words don't you know

[raven grotto]

i know the wording me just dont know what it meaning

[misty bay]

so you're telling me that you know every word in that question but you don't understand what the sentences are saying?

[raven grotto]

i know the individual words but do not understand what it means together

[misty bay]

define iterative

yeah okay I see how it is

[raven grotto]

can you help?

[misty bay]

yeah I can, but you have to put in some effort in first

this problem seems as straightforward as reading the steps and doing them in order

but for some reason, you can't do that, and I suspect that's because you don't understand the words

[raven grotto]

okay

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[astral matrix]

Can i pull out the 8 from the inatgral?

[icy lance]

no

you can factor out 2 though

[misty bay]

you can write $\int_0^4 8 , dx + \int_0^4 -2x , dx$

[flat frigate]

Saccharine

[misty bay]

if you so prefer

[astral matrix]

oh i see

thanks

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[obtuse plover]

what i did wrong?!?!

[obtuse plover]

is it not just plugging in?

o wait

[devout shale]

What value of T results in the point (1, 0)?

[obtuse plover]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

got it

thanks Mr. AustinU, appreciate it

[devout shale]

np

[obtuse plover]

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[gilded scroll]

Best way to figure out what the upper function is without graphing it? finding the are between two graphs

[gilded scroll]

in this case I have 8/x and 2x

now that ive seen the graph it makes sense how 2x is the upper function on the bounds from 2 to 4

but I wont have my computer during the exam tommorow and was wondering if there was a mathamatical way of doing things

oh

maybe if I plug in my bounds and see what y values I get

is this right and will it work

ping me if you respond please

[devout shale]

@gilded scroll

It depends on your bounds

first find where your two functions intersect

if they only intersect at the beginning/end of your bounds or elsewhere

this tells you that one function must be higher than the other throughout the entire interval

you don't have to worry about them swapping if that is the case

if they do intersect in between your bounds

then split the integral there

and on either side

evaluate the functions to see which is greater

[gilded scroll]

i do this by setting them equal to eachother right?

[devout shale]

that is how you find their intersection yes

[gilded scroll]

ok

[devout shale]

for example if you had

x^2 and 1/x

from 0 to 1

check their intersection

see that it is only at x=1

[gilded scroll]

1

[devout shale]

this is at the boundary

so you know that 1 of those must be greater than the other

which one?

try plugging in a point like x=.5

that will tell you

if instead you had

x^2 and 1/x from 0 to 2

checking their intersection tells you they meet at 1

so they must flip flop between which is higher at that point

so split your integral

and check on both sides

first check between 0 and 1

x=.5

and then check between 1 and 2

x=1.5

and then you will know which is higher for each interval

see what I mean?

[gilded scroll]

Hi sorry had to help someone, thank you sm that makes sense

.close

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[undone grotto]

Assume the function f is continuous everywhere for all real x's and that f(0)=f(2). Show that there is at least a number a in the interval [1,2] such that f(a) = f(a-1).

[undone grotto]

I did the proof by using words and I understand that it is correct and etc but in the answers they kinda used IVT. And I'd really appreciate it if someone can explain to me how I can prove this using IVT.

[finite yacht]

If f(1) = f(2) then it is trivial. Otherwise, define g: [1,2] → ℝ by g(x) = f(x) - f(x-1) and apply IVT to g

[safe radish]

@undone grotto Has your question been resolved?

[undone grotto]

but how?

the only thing Ik about f is that f(0)=f(2) so g=f(a)-f(a-1)=0

[finite yacht]

What are g(1) and g(2)?

Is g continuous?

[undone grotto]

I'd think so considering that f is continuous

[finite yacht]

And what about this?

[undone grotto]

g(1)= f(1) - f(0)

g(2) = f(2) - f(1)

[finite yacht]

Do those two numbers have different sign?

Remember that f(0) = f(2)

[undone grotto]

f(0) = f(2) = L

g(1)= f(1) - L

g(2) = L - f(1)

[undone grotto]

Yeah one will be negative and the other positive

in the same interval

and IVT states that then it will have at least once crossed 0

but what about g'(x)

[finite yacht]

Exactly

[finite yacht]

What happens with g'(x)?

[undone grotto]

what if g'(x)=0 in the interval

[junior smelt]

nowhere are you told that f (and hence g) is differentiable

[the differential] may or may not exist

[finite yacht]

Also what would be the problem with this?

[undone grotto]

so that means the there may or may not be a solution

[finite yacht]

No

[undone grotto]

nevermind

g'(x) = 0 will only give us more solutions

but in the question they only asked us to prove there is at least one so the proof holds!

[finite yacht]

Exactly

[undone grotto]

Thank you so much for the help!

[junior smelt]

[+ differentiability implies continuity so you may as well just argue as per above]

[undone grotto]

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[undone grotto]

Assume that $a_1, a_2, . . . , a_n$ are real numbers such that $a_0+\frac{a_1}2+\frac{a_2}3+. . . +\frac{a_n}{(n+1)} = 0$.
Show that the equation $a_0 + a_1 x+ . . . +a_nx^{n} = 0$ has at least one root in the interval (0, 1).

[flat frigate]

afeAlway

[undone grotto]

Can anyone help me with this?

How should I take on this?

[plucky elk]

try proving it for n=1, 2 first

[undone grotto]

I got nothing

[junior smelt]

Look at the form of the condition, and the form of what they're asking you to work with...

[undone grotto]

I mean like I am not getting anywhere

[plucky elk]

show what you tried

[undone grotto]

$x^2 + \frac{a_1}{a_2}x + \frac{a_0}{a_2} = 0$ and $a_0 + \frac{a_1}{2} + \frac{a_2}{3} = 0$

[flat frigate]

afeAlway

[undone grotto]

this is how far I got (not even sure if this was what you told me to do tbh)

[junior smelt]

Have this here? @undone grotto

[undone grotto]

what is it?

[junior smelt]

Notice how you want
[
a_0+\frac{a_1}2+\frac{a_2}3+ \ldots +\frac{a_n}{(n+1)} = 0
]
and
[
a_0 + a_1 x + \ldots +a_n x^{n} = 0
]

[flat frigate]

@junior smelt

[undone grotto]

yeah

but like how does that help me? Can you maybe give me a hint?

[plucky elk]

0 = 0

[junior smelt]

The proof is actually trivial and left to the reader

[undone grotto]

wait I think I'm getting somewhere

[junior smelt]

[also it's that the first line here is given to you, and you want the bottom line to have the solution on the open interval (0, 1) btw]

[undone grotto]

$a_1x - \frac{a_1}2 + . . . + a_nx^n - \frac{a_n}{n+1} = 0$

[flat frigate]

afeAlway

[plucky elk]

promising

[undone grotto]

was this what you told me to do?

[undone grotto]

wdym? I am I on the right path or totally lost (I bet on totally lost)

[plucky elk]

if n=1, then the x solution is obvious

find n=2, 3 and try to find a pattern

[safe radish]

@undone grotto Has your question been resolved?

[safe radish]

@undone grotto Has your question been resolved?

[safe radish]

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[lean otter]

Did I solve this correctly? And how would I find the other measure of angle x?

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

<@&286206848099549185>

[safe radish]

@lean otter Has your question been resolved?

[safe radish]

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[lament oyster]

so what ive done so far

[flat frigate]

chenyboi

[lament oyster]

$a+ar+ar^2=273$

[flat frigate]

chenyboi

[lament oyster]

so

$a(r^2+r+1)=273$

[flat frigate]

chenyboi

[lament oyster]

r^2 + r + 1 has a range of [1,inf) so it basically just works with any whole number

therefore, because 273 has prime factorization of 3,7,13

there are 8 factors

and each one can become a

so i got 8

but thats not an answer choice

[drowsy moss]

because each one can't be all of them

for example, there's no way to make r^2+r+1 = 1

[lament oyster]

oh yeaaaa cause r cant be 0

okay but all the others can though

[drowsy moss]

can they?

[lament oyster]

what doesnt work then

im asuming u mean 273 doesnt work for r^2+r+1 but why

[drowsy moss]

work it out

[lament oyster]

.close

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[icy lance]

A-B=C then 1-B/A=C/A sure so long as A isnt 0

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[mint veldt]

Okay so I was about to use the product rule for this, but I realize I need to do something about sin theta /cos theta

I just need one hint here on how to continue from here

[drowsy moss]

simplify your fraction first before you take the derivative

[mint veldt]

Yes

That's where I'm at

The question required me to rewrite f(x) in terms of sin(x) and cos(x)

Now, I don't know how to simplify it more

[drowsy moss]

$\frac{a}{b/c} = a\left(\frac cb\right)$

[flat frigate]

Zybikron

[mint veldt]

ah

That's a new one

[drowsy moss]

in other words, dividing by 1/cos(x) is the same as multiplying by cos(x)

[mint veldt]

wait

I'm not sure why I would rewrite cosine

That wasn't my problem

it's the numerator no?

[drowsy moss]

sure. You can take the derivative however you want to

[mint veldt]

okay

how do we handle sin theta/ cosine theta

[drowsy moss]

quotient rule

[mint veldt]

wait

so I have to use the quotient rule twice then?

[drowsy moss]

if you want to take the derivative this way, then yes

[mint veldt]

rip

Is there another way

[drowsy moss]

yes

[mint veldt]

wouldn't that just give me (sin theta / cos theta - 1 ) * cos theta

[drowsy moss]

yes

[mint veldt]

Then I still have the problem of sin theta / cos theta ...

[drowsy moss]

keep going with this

[mint veldt]

I could multiply cos theta into the 1st term\

[drowsy moss]

you'd distribute the cos(theta) to both terms, yeah

[mint veldt]

oh

so I'd end up with sin theta - cos theta

[drowsy moss]

yes

a considerably easier derivative

[mint veldt]

totally

wow

thank you so much

for sticking through that with me

[drowsy moss]

np

[mint veldt]

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[wind shoal]

Define a function to be convex if the line segment between any two points on the graph of the function lies above the graph between the two points. show that if you connect any two points on the function, the resulting function is still convex

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@wind shoal Has your question been resolved?

[safe radish]

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[torpid marsh]

I'm unsure on how I find the x value

[torpid marsh]

<@&286206848099549185>

[junior smelt]

Well, those angles they labelled are on a straight line, aren't they?

[torpid marsh]

They are but I don't know how to get it from the equations

[junior smelt]

What do you mean "get it from the equations"? which ones?

[torpid marsh]

I need to find the value of x on this part of the line in degrees

[full pilot]

you know those 2 add up 180

[worthy hemlock]

Concept of supplementary angles

[full pilot]

^ ^ ^

[torpid marsh]

Yes

[full pilot]

yea

so you can solve

[worthy hemlock]

You can create an equation based off that info

[torpid marsh]

How do I do that?

[worthy hemlock]

Concept of supplementary angles

You know that those two angles sum up to be 180

What does sum mean?

[full pilot]

First angle: 9x + 5
Second angle: 6x + 20
First angle + Second angle = 180

[torpid marsh]

I know but what is the angle in °

[full pilot]

solve for it

algebra?

[worthy hemlock]

Use what panda gave you to solve for x

Because that's what the question asks for

Find the value of x

[torpid marsh]

Alright but do either of you guys know how I might be able to do that on my ti89 to speed up the process?

[full pilot]

???

[torpid marsh]

I will solve it on my own first but if there's a way it'd be useful

[worthy hemlock]

It's not a hard process

[full pilot]

Do it on your own

about 2 steps

[worthy hemlock]

First angle: 9x + 5
Second angle: 6x + 20
First angle + Second angle = 180
Can you plug in the info and create the equation?

[torpid marsh]

Okay so I have 9x-5+6x+20 written down then do I combine like terms?

[worthy hemlock]

That's not an equation

[torpid marsh]

How do I get the angle from one of them?

[worthy hemlock]

.

[full pilot]

solve for x

set your equation equal to what it should sum to

[worthy hemlock]

Did you create the equation based on that info?

[worthy hemlock]

And you are not solving for an angle, you are finding the value for x

[torpid marsh]

How do I solve for x? I'm sorry if I'm completely missing it

[full pilot]

First angle + Second angle = 180

plug in those angles

combine like terms

solve for x

[torpid marsh]

So I do need to write 9x-5+6x+20?

And combine like terms

[full pilot]

yup

[worthy hemlock]

As I stated, that is not an equation

An equation is equal to something

[full pilot]

and set it equal to 180 to make it an equation!

[worthy hemlock]

There's not an equal something there

[torpid marsh]

9x-5+6x+20=180?

[worthy hemlock]

Yes

[full pilot]

💪

[worthy hemlock]

That's better

Now combine like terms, solve for x

[torpid marsh]

15x+15?

[full pilot]

equals?

[worthy hemlock]

Stop making the equals part disappear

[torpid marsh]

15x+15=180

[worthy hemlock]

What should you do next to solve for x?

[torpid marsh]

No idea

[worthy hemlock]

If you have x - 60 = 90, what do you do to find x?

[torpid marsh]

150-60=90

See how much is left?

I don't know what it's called

[worthy hemlock]

That was an easy example. What if it was x + 1234123434 = 1232456904, what now?

What should you do to isolate x?

[torpid marsh]

1232456904-1234123434

I did it in reverse my bad

Oh wait

[worthy hemlock]

So basically subtracting 1234123434 on both sides, correct?

[torpid marsh]

Yeah

[full pilot]

so can you do that with your equation you had before?

[torpid marsh]

So 180-15?

[full pilot]

you gotta subtract it from both sides of the equation

[torpid marsh]

15x-15, 180-15

Or do I not touch the 15x

[full pilot]

keep the equals

[torpid marsh]

15-15, 180-15

[full pilot]

your equation is 15x + 15 = 180

subtract 15 from both sides

[torpid marsh]

15x=165?

Is that correct?

[worthy hemlock]

No

[full pilot]

subtract 15 from both sides

[torpid marsh]

I meant that

My bad

[worthy hemlock]

What should you do now to isolate x?

[torpid marsh]

X=165°

[worthy hemlock]

No

Whatever you do to one side, you need to do to the other

[torpid marsh]

X1=165° X2=15°

Is that right?

[worthy hemlock]

[torpid marsh]

I don't know either

I only have one X value

[worthy hemlock]

You are at this point, correct?
15x=165

[torpid marsh]

Yes

[worthy hemlock]

What should you do to isolate x?

[torpid marsh]

I don't know

[worthy hemlock]

If you have 3x = 159, what should you do?

[torpid marsh]

-3x from both sides but since 159 has no x it isn't affected

[worthy hemlock]

No

https://youtu.be/7DPWeBszNSM?t=788

[torpid marsh]

Can you just tell me what to do?

How do I isolate X?

[worthy hemlock]

No, you can watch that video

You seem like you are having too much trouble solving linear equations

Hence the video

[torpid marsh]

Because you help me to a certain spot and leave me to drown, I don't know how to isolate X.

Can you tell me how to isolate X?

[worthy hemlock]

That is why I presented the video

[torpid marsh]

I will watch the video but why can't you help me with that?

[worthy hemlock]

Because I'm not going to give you what to do

You are capable of doing it yourself if you are pushed in the right direction

Hence why I presented the video

It's not a hard process

[torpid marsh]

It is if I have no idea, hence why I'm here

[worthy hemlock]

And that is why I presented the video

So you can see what you should try to do

[torpid marsh]

Why are you unable to? I just need help with this.

[worthy hemlock]

Because I'm not going to give you what to do
You are capable of doing it yourself if you are pushed in the right direction
Hence why I presented the video
It's not a hard process

[torpid marsh]

X=11?

[worthy hemlock]

And what did you do to get that? From 15x = 165

[torpid marsh]

Divided both sides by 15

[worthy hemlock]

Good

And you found x, like it asked for

[torpid marsh]

I would like to apologize for earlier. I do appreciate and understand your teaching method. Thank you for your help

Now that I understand what I need to do, I see that the way you taught me to do it was effective and worked

[worthy hemlock]

No worries, there is a methodical reason on why I taught like that. Just spitting out what you needed to do isn't going to help, because there is a high chance you could be still confused as well as chance of you not learning what to do. There are hundreds of videos that teach everything you need to do. And it presents it so much better than text/words

[torpid marsh]

I agree with you on that, thank you for your patience as well

[worthy hemlock]

Well, if you are done with the channel, .close to close the channel

[torpid marsh]

I have one more question

I'm at the isolating X part and I'm dividing both sides by 11 and I'm getting 11 over 153

[worthy hemlock]

Can you post the original question and the steps you have so far?

[torpid marsh]

Here's the problem and I've done 6x-1+5x+28=180 to 11x+27=180 to 11x=153

[worthy hemlock]

The reason why that's wrong, is because you applied the wrong parallel line concept

[torpid marsh]

Oh wait

You're right

Both are the same value

Same angle

[worthy hemlock]

And what does that mean?

Equation wise

[torpid marsh]

I only need to do one of them

[worthy hemlock]

Not exactly

If they are the same angle measure, that means they are equal, correct?

[torpid marsh]

Yeah

[worthy hemlock]

Meaning you can set those expressions equal to each other

[torpid marsh]

So 5x+28=6x-1

[worthy hemlock]

Yes

[torpid marsh]

Do I combine like terms for this?

[worthy hemlock]

Sorta, you need to get all the terms with x on one side, and everything else on the other

[torpid marsh]

So 11x+27?

Or 6x+5x+28-1?

[worthy hemlock]

one side
other side
Means one side of the equal sign

And the other side of the equal sign

[torpid marsh]

6x+5x=28-1
11x=27

[worthy hemlock]

So with 5x+28=6x-1
You want all the terms with x on one side of the equal sign and everything else on the other

[worthy hemlock]

Not quite

Let's start with the 6x, what should you do to remove it from the right side?

[torpid marsh]

Divide?

Wait nvm

That's how you get rid of it

I'm pretty tired atm cause I was up til 1:30am last night doing math

I'm not thinking clearly

[worthy hemlock]

How about this? Here's a video with the concept you are applying, and it should help out https://www.youtube.com/watch?v=avt1w4Fuk-0&ab_channel=MathisSimple!

[torpid marsh]

-1x+28-1?

[worthy hemlock]

Let's reset, here's the original problem again
5x+28=6x-1

What's the first step?

[torpid marsh]

-6x from both sides?

[worthy hemlock]

Good

And what do you get when you do that?

[torpid marsh]

-1x+28=-1

[worthy hemlock]

Good

What's the next step?

[torpid marsh]

-28 from both sides?

[worthy hemlock]

Yes

[torpid marsh]

-1x=-29

[worthy hemlock]

Is x isolate yet?

[torpid marsh]

Not yet

[worthy hemlock]

So what should you do next?

[torpid marsh]

Divide both sides by -1

[worthy hemlock]

Yes exactly

[torpid marsh]

X=29

[worthy hemlock]

Yep that's it

[torpid marsh]

Thank you for helping me out with this

[worthy hemlock]

I hope you understand why I'm not spoonfeeding and giving you the step by step answer

This gives you a chance to learn what to do

[torpid marsh]

Yeah I understand

It's definitely more effective than a step by step

[worthy hemlock]

And how to approach it. I'm sure after that, you have a much more better understanding on what to do now

[torpid marsh]

Yea thank you for this

This should be it for me

[worthy hemlock]

Well if that's it, feel free to .close

[torpid marsh]

Enjoy your day or night

.close

[safe radish]

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[worthy hemlock]

Same to you

[safe radish]

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[keen birch]

help

[keen birch]

how does this make sense

[flat frigate]

MathIsAlwaysRight

[brave wolf]

where F(x) is the antiderivative of f(x)

[keen birch]

i understand antiderivatives and what not

but how do i get an equation out of this

[brave wolf]

$\int_{b}^{a}f\left(x\right)dx$ try to express this

[flat frigate]

MathIsAlwaysRight

[brave wolf]

Sorry, I forgot dx in the last equation

[brave wolf]

So this will be F(a)-F(b) by definition

Now you need to find relation between F(b)-F(a) and F(a)-F(b)

[keen birch]

ok got it

[safe radish]

@keen birch Has your question been resolved?

[safe radish]

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[nova berry]

How would I go about calculating the strength of a throw and determining whether the throw could have caused damage to a person hitting there leg If the throwing person is the man with the furthest throw ever and a 200g ball was used and it had around 3 seconds to accelerate

[safe radish]

@nova berry Has your question been resolved?

[zinc token]

show your work

[nova berry]

I only know that 145km/h is like what the speed is at It's peak

[buoyant shadow]

how can a man accelerate a ball for 3 seconds, this is absurd

[nova berry]

Not me who made the task 😅

[safe radish]

@nova berry Has your question been resolved?

[safe radish]

@nova berry Has your question been resolved?

[safe radish]

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[coarse kelp]

is u sub always allowed in factoring?

[icy lance]

its just a way to make a disguised quadratic more obvious

like if you had e^(2x)+2e^x+1 you could let u=e^x then u^2+2u+1 and you get (u+1)^2=(e^x+1)^2

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[hollow elbow]

how to approach this

hell]

o

[lean otter]

😃 i have not done 3d geometry yet as its in next grade but does this help?

https://www.shaalaa.com/question-bank-solutions/find-perpendicular-distance-point-1-0-0-line-x-1-2-y-1-3-z-10-8-also-find-coordinates-foot-perpendicular-equation-perpendicular-equation-of-a-line-in-space_46603

[safe radish]

@hollow elbow Has your question been resolved?

[hollow elbow]

ok mate

[safe radish]

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[tranquil pebble]

Here, Q = ysinA-xcosA

[tranquil pebble]

can anyone please help

[silver cliff]

If you equate Q = 0, you get tanA = x/y

Use that to find sinA and cosA

Then find L.H.S

And show that its equal to R.H.S

[tranquil pebble]

I see

wait lemme try

wow thank you

.close

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[static yacht]

Discrete maths

[static yacht]

Have I made the NFA incorreclty? I don't think so. However the epsilon-closure step has my eyes spinning now and I can't create the DFA myself

Please help 😄

.delete

.close

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[low mist]

I don't get this proof

[low mist]

i get that triangles AQB and PQA are similar

but i don't quite get how that allows me to conclude that |PQ|^2=|AP||BP|

[delicate glacier]

Write the ratios of the sides for the similar triangles

[low mist]

ahh, i get it

it was really that simple lol

thanks :)

.close

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[weary bear]

idk what to do from here

[gritty glacier]

Substitute again 3x+1 in the original fucntion

[weary bear]

3x+1 = 2x+1?

[gritty glacier]

6x+3=k(2x+1) i think

[weary bear]

where did that come from

[gritty glacier]

f(3x+1)=k(2x+1)

f(x)=2x+1

Consider 3x+1 to be x

[weary bear]

okay

x = 3x+1?

[gritty glacier]

So 2(3x+1)+1

K=3 ig

U have solutions?

[weary bear]

sorry I'm slightly confused, how is x = 3x+1?

[gritty glacier]

Consider it to y then

See the first function

[weary bear]

okay y=3x+1

[gritty glacier]

First function

F(x)=2x+1

[weary bear]

yes, at the very top? f(x) = 2x+1?

yes

[gritty glacier]

Now what would f(y) be?

[weary bear]

could it not be anything?

[gritty glacier]

No bro

We have y here

[weary bear]

okay so 2y+1?

[gritty glacier]

So whereever we have x we substitute y

Yes

Now write y as the expanded term

The one we considered it as

[weary bear]

y = 3x+1

[gritty glacier]

2(3x+1)+1

[weary bear]

oh right, I think I'm getting it

[gritty glacier]

Yea sure try it

[weary bear]

yeah I'm pretty sure I got it, thank you🙏

.close

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[topaz smelt]

Can someone please help with this

[thin bridge]

what have you tried?

[topaz smelt]

i have no clue how to start

[thin bridge]

would you be able to identify the integer solutions to
$$|p| + |q| = 1$$

[topaz smelt]

sorry for the late reply

[flat frigate]

ℝamonov

[gloomy sail]

Can’t you just square both sides

To get rid of the modulas function

Or is that illegal algebra. It’s been while

[topaz smelt]

but still how wud u find all possible values of y

and then add them

i think

modulus is like

sqrt a^2 or somth

[wintry condor]

Since an absolute value is always positive or equal to 0, the possible values for the 3 abs would be 0,0,1 or 0,1,0 or 1,0,0

[topaz smelt]

so x wud have to be 3 or 4

[wintry condor]

Well you can also notice that |2x-y| and |2x-y-1| has a difference of 1

So 1,0,0 wouldn't be possible

[topaz smelt]

hmm

im so confused

[wintry condor]

What part do you not understand

[topaz smelt]

like

what i wud do next

from there

to figure out the possible values of y

[wintry condor]

After that it's just system of equations
Do you understand that the possible pairs for (|x-3|,|2x-y|,|2x-y-1|) are (0,1,0) or (0,0,1)

[topaz smelt]

oh ok

i se

so after solving

y = -5 or -6

so b

-11

[wintry condor]

Hm?

You can define the value of x first, since |x-3|=0

[topaz smelt]

ya after solving

like whole thingy

i got -11

as answer

ty

.close

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[supple shore]

A regular pentagon was cut out of lined paper.The pentagon is rotated around its center in the opposite direction clockwise, from 21° to 21°. The drawing opposite shows the situation after the first movement. What will we see when the pentagon overlaps the hole again for the first time?

[supple shore]

.close

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[zenith pilot]

HI

[zenith pilot]

Hold ip I forgot whete the WUESTION was

[safe radish]

@zenith pilot Has your question been resolved?

[zenith pilot]

HOLD UP

GOT IT YAYY

Unit: compound interest

Please tag me or @ me when u can help

[unique fossil]

Do you have answer just to make sure i am correct

[zenith pilot]

Yes I do, let me show my working out I will tag u

I know its wrong because money wouldn’t go to 73 cents @unique fossil

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[zenith pilot]

.reopen

[safe radish]

✅

[zenith pilot]

My answer is 2622.73191358 rounding would be; 2622.73

[safe radish]

@zenith pilot Has your question been resolved?

[zenith pilot]

<@&286206848099549185>

This is my question to state again

This is my working out

Please tag / @ me if u can help

[young nexus]

can you explain a bit more why you think it is wrong? @zenith pilot

[zenith pilot]

Since its taking about compound interest its obviously money and the cents go to 73 and money cant pass 60 hence I believe my answer is wrong

[thin bridge]

and money cant pass 60
where's that coming from

[young nexus]

sorry, i dont know what you mean. i would say the answer is correct.

[zenith pilot]

Like after 60 it becomes a dollar

[young nexus]

you could do the calculation with a rounding each year, which would give you nearly the same result - cents would be 74 instead of 73

[thin bridge]

Like after 60 it becomes a dollar
what?

[zenith pilot]

60 cents = dollar

[thin bridge]

are you saying that if a shop prices something at 60c
that you'll need to pay one whole dollar for it at the checkout?

60c isn't a dollar

[zenith pilot]

So what would be correct 73 or 74

[zenith pilot]

OMG

OMG

OMG

I WAS THINKING ABOUT TIME

WTF IM SO SORRY

[thin bridge]

The Latin root word “cent” which means “one hundred”

[zenith pilot]

YES YES

OMG I WAS THINKING OF TIME THIS IS EMBARRASSING

[young nexus]

74 is a sort of nitpicking.

[zenith pilot]

Oh

So would u go for 74 or 73

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[deft mesa]

can someone help me ?

[deft mesa]

this is about Inequality of Quadratic Mean (QM), Arithmetic Mean (AM), Geometry Mean (GM) and Harmonic Mean (HM)

[pearl sundial]

Do u want the steps or only solution?

[deft mesa]

the steps pls

[safe radish]

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.

[deft mesa]

@young nexus thanks y for your last help

[young nexus]

solve one equation in real numbers with five unknowns?

[deft mesa]

yes

[young nexus]

ok, then im out since i do not know how this should make sense.

[deft mesa]

thank u bro for your trying

i have the answer key ---> x=y=u=v=1/4 and z = 3, but i dont know how to get it

[deft mesa]

but its about using this

[young nexus]

just one last question: "solve the following equations ..." means for me that there are more then one equation

[deft mesa]

maybe by using QM or AM or GM or HM we can add more equations ?

[pearl sundial]

First, it's not possible that x, y, z, u, v are integers

If x, y, u, v are 1 or 0, then $2\sqrt{z-2} = z$, z doesn't have a real solution

[flat frigate]

阿皮pea

[pearl sundial]

Create a graph of $2\sqrt{z-2}$ and $z$, $2\sqrt{z-2} < z$ always

[flat frigate]

阿皮pea

[pearl sundial]

Therefore x, y, u, v some of them are greater than 0 and less than 1

[deft mesa]

does it have the solution ?

[pearl sundial]

It has but i don't find it and i cant tell you

I can only lead you to solve it

[deft mesa]

so it doesnt have ?

[pearl sundial]

Lol

[deft mesa]

i just realized it

im really sry

[pearl sundial]

I'm sorry I didn't see this message

[deft mesa]

no, its my fault

sry cuz i didnt realize it

thanks y

[pearl sundial]

No need to sorry

[deft mesa]

hehehe thanks

.close

[safe radish]

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[safe radish]

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[safe radish]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[silver axle]

1

[dense prism]

Can you try calculating the number of favourable cases?

Which has 1 man and 1 woman, how many ways are there to achieve such a team

[silver axle]

um 4?

[dense prism]

Not quite

Let's think about it

We have 6 men available and we need to pick 1 from them, and we also have 4 women available and we need to pick 1 from them. How many combinations are possible?

[silver axle]

i dont get it

[dense prism]

Let's say we have M1,M2,m3,m4,m5,m6 as 6 men.
And w1,w2,w3,w4 as 4 women

How many combinations of a man and a woman can we make here ?

M1,M2 etc are their names

[silver axle]

uh m1 f1, m1 f2, m1 f3

theres a lot

how would i caclulate it

[dense prism]

Well can you see that for each man you choose you have 4 further woman options to choose from?

Let's say i pick M1, and then I have 4 women to choose as the second person right?

[silver axle]

oh so 24

[dense prism]

Exactly

Now that's our number of favourable conditions

Now recall probability of an event = number of favourable cases/total cases

You just calculated favourable cases

Try figuring out total cases and divide them to get your answer

[silver axle]

how would i get total cases

is it ok if i think about it and then come back and ask if i dont know

[dense prism]

Yeah sure

[silver axle]

is the answer just 10 choose 2

like the total

events

[dense prism]

Yes

Which is equal to?

[silver axle]

uh idk i forgot the formula

[dense prism]

Well the formula is nCr=$n!/[(n-r)!r!]$

You can apply that, and after you get the answer I'll tell you how to solve even if you in case forget the formula

[flat frigate]

Radiation 𝕏

[silver axle]

so 10!/8!*2!

[dense prism]

Yep

[silver axle]

what would that be

[dense prism]

$10!=1098765432*1$

[flat frigate]

Radiation 𝕏

[silver axle]

so 10!/2! is 10! without the 2*1

and that divided by 8! is just 10*9 which is 90

[dense prism]

90/2 actually

[silver axle]

how

[dense prism]

The 876.... part gets cancelled from the 10! due to the 8!

Then we are left with 10*9 in the numerator

And still 2 in the denom

[silver axle]

ohh

yeah

so 24/15 is the answeR?

[dense prism]

24/45

90/2 is 45

So your answer is 24/45

[silver axle]

oh yeah mb

[dense prism]

Or 8/15

[silver axle]

tysm

[dense prism]

Np

[silver axle]

.close

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[safe radish]

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[remote spade]

Can anyone please confirm my bounds?

[remote spade]

I felt fairly certain I was correct with my bounds but the integral evaluates to 0 which I'm pretty sure is not correct

the problem is the term 4 - x^2 because that thing just makes every term go to 0

[safe radish]

@remote spade Has your question been resolved?

[safe radish]

@remote spade Has your question been resolved?

[safe radish]

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[spring jasper]

Why does $\int_{-\infty}^{\infty}x,dx \ne \lim_{a\rightarrow\infty}\int_{-a}^{a}x,dx$

[flat frigate]

Farmer John

[spring jasper]

Also, why does $\int_{-\infty}^{\infty}x,dx \ne 0$?

[flat frigate]

Farmer John

[spring jasper]

Because the graph is symmetrical on both sides of the y axis right

so the areas should cancel out?

[hasty wagon]

i think it's a similar concept to
infty - infty is indetermined

[split ether]

Consider $\lim_{a\to\infty}\int_{-a}^{a^2}xdx$ too

[flat frigate]

A Lonely Bean

[hasty wagon]

i think wiki explains well here
https://en.m.wikipedia.org/wiki/Indeterminate_form

[split ether]

It seems to be the same thing, but it approaches infinity

And not 0

[spring jasper]

yes but in this case the right bound grows faster than the left bound

[split ether]

Nonetheless they both approach the original bounds

[spring jasper]

true

[spring jasper]

wait so whats this equal to

is it 0?

[split ether]

So there is not really a good formulation for the integral

[split ether]

It diverges to infinity

But if you take a and -a as bounds then it's 0

[spring jasper]

oh

[split ether]

That is the contradiction I am talking about

[spring jasper]

wait thats weir

so how would i know when it should be 0 and when it should diverge

do the upper and lower bounds have to be the same

[split ether]

Thinking about the graph could help

E.g. here x obviously diverges to infinities as you look at the right and at the left of its graph

[spring jasper]

yes but

why is it equal to 0 when its from -a to a

[split ether]

Ah

[spring jasper]

and divergent when its from -a to a^2

if the graphs are the same

[split ether]

Because x is an odd function

And (-a, a) is a symmetric interval

[spring jasper]

yes but the area under the graph is the same right?

[split ether]

If you are considering different values of a, then no, they are different

Again, this is all centered about the fact that you can "approach infinity from different paths"

[spring jasper]

yeah but a isnt finite in this case

[split ether]

This also comes up when dealing with limits of two-dimensional functions

Some of them are undefined because going across, let's say, y = x and y = x^2 gives different values

[split ether]

That's the thing, we don't talk directly about a = inf

[spring jasper]

ohhh

so we just ignore a = inf

and think about a as if it was finite?

[split ether]

We first look at what happens when a is finite

And then apply the limit

This is why we write a -> inf instead of a = inf

[split ether]

Yes

[spring jasper]

i see alright that makes more sense

thanks for the help

.close

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[boreal cypress]

top right, isnt it 0 not 2?

[stray socket]

No

[glass carbon]

x + 1

when x = 1

[quasi bison]

why'd it be 0

[boreal cypress]

1-1 = 0

0/0 = 0

[stray socket]

0/0 is not 0

Lol the instant reaction

Oh Doug you goofed up man

You woke the entire server

[flat frigate]

Modus

[stray socket]

How can 0/0 = 0

[boreal cypress]

so you cant just be plug happy

and instantly go in there

[stray socket]

Not in this case

[boreal cypress]

damn it

def goofed up

thanks

[faint seal]

you can but if you get an indeterminant form then you have to do more work

[boreal cypress]

got it

thanks all 9 of you math wizards

for helping me solve this very hard problem

.close

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[mint veldt]

I'm not sure why this isn't the correct set up for the chain rule

[mint veldt]

Note, I rewrote index 3 root x as x^2/3

[light tartan]

x^2/3 would be cube root of x^2

the cube root of x would be x^1/3

[mint veldt]

oh

oooppsie

tyu

.close

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[mint veldt]

what am I doing wrong here?

[obtuse plover]

it doesnt look like the 3 is an index

[faint seal]

maybe the function is $e^{3 \cdot \sqrt{x}}$, not $e^{\sqrt[3]{x}}$

[flat frigate]

kheerii