help-19

[glossy drift]

The height of a batted baseball is given by h= -4.9(t-2.4)^2 + 31 where h is height ( in
meters) and t is time (in seconds)
a) What was the maximum height of the ball? When did this occur? (2 marks)
b) What was the height when the ball was hit? (1 mark)
c) When did the ball hit the ground? (1 mark)
10. a) The graph of y= x^2 is stretched vertically by a factor of 1/2, reflected vertically in the x-axis, and then translated 3 units down and 1 unit right. Write the equation of the parabola. (1 mark)
b) Write an equation for the parabola with vertex at (-5,1), opening upward, and with a vertical stretch by a factor of 4. (1 mark)

Ion get

Any of this

[odd edge]

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[slate heath]

A group contains 9 adults and 9 children. How many ways are there to arrange these people in a line if the adults and children alternate?

[slate heath]

i dont know how to solve for it

[cursive mango]

the number of ways to arrange n items is n!

so you can find the amount of ways to arrange the adults and children then multiply to get the amount of ways to arrange them all

[slate heath]

so 9*9?

[cursive mango]

9! x 9!

[slate heath]

how do i calculate that

[cursive mango]

it is 9 multiplied by every positive number behind it

your calculator should have a function for it

look a for a ! sing

sign*

[slate heath]

is there an online calculator i can use

wolfram?

[cursive mango]

you can just google 9!

[slate heath]

is that correct

[cursive mango]

assuming I didnt mess up it should be

[slate heath]

i just used my last attempt and it said it was wrong

[cursive mango]

hmmm

sorry about that

[slate heath]

no worries

one less question to worry about

thanks anyways

.close

[odd edge]

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[odd edge]

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[ornate summit]

given that j /= 1 arbitrary, but fixed

[ornate summit]

{σ ∈ Sn : σ(1) = j ∧ σ(j) = 1} and{σ ∈ Sn−2 : σ hat keinen Fixpunkt}

i should find a bijection between those sets where σ is a permutation and Sn is the set of all bijective functions from {1,...,n} -> {1,...,n}

Both set are a set of maps

I dont know how i could fine a bijective function here

[quasi sparrow]

Try writing out all the permutations for small n

Like even n=4

[ornate summit]

ok

with the condition?

σ(1) = j ∧ σ(j) = 1

well if i take n = 3

there is only 1 potential permutation with the given condition

j = 3

σ = 1,2 3
3,2,1

and for the other set

it would be empty since n = 3 S3-2 would be S1 and it would be

1
1 whhich is a fix point

@quasi sparrow still there 🙂

[quasi sparrow]

Did you have a question for me

[ornate summit]

only if my thoguhts are right

[quasi sparrow]

Did you do n=4?

[ornate summit]

jes

for the first set of permutations there are to possible permutations

n = 4

lets say j =2

σ1 = 1 2 3 4
2 1 4 3

σ2 = 1 2 3 4
2 1 3 4

for the other set

{pi ∈ Sn−2 : σ hat keinen Fixpunkt}

pi = 1 2
2 1

i call permutations of the second set pi

so i need to find a perumutation lets call it ? such that pi = σ ◦ ?

right?

[quasi sparrow]

Don't use ? As a variable

The calculations look right

Don't know what you're trying to do with pi = sigma * something

[ornate summit]

i thought i need to find a map which takes a sigma and gives back a pi

or how could i get to the solution

[quasi sparrow]

Yes you need a bijection

[odd edge]

@ornate summit Has your question been resolved?

[ornate summit]

can you give me one more hint

i would say the map i need to find is a permutation with n-2 elements

[odd edge]

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[ornate summit]

.reopen

[odd edge]

✅

[low locust]

!original

[odd edge]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[low locust]

something is off

[odd edge]

@ornate summit Has your question been resolved?

[odd edge]

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[jolly niche]

what is the rule that allows for the limit to be moved to the exponent

[signal crown]

limit of composition of functions

but anyway, nobody writes it like in your image

[jolly niche]

oh ok

thanks

.close

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[honest turtle]

Show that for all natural numbers n, is $$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}} \cdots \frac{1}{\sqrt{n}} > 2(\sqrt{n+1}-1)$$

[clever fjord]

Good

[honest turtle]

Is this going to be a same question as a equal sign?

[odd edge]

@honest turtle Has your question been resolved?

[honest turtle]

<@&286206848099549185> Induction problem.

[odd edge]

@honest turtle Has your question been resolved?

[visual sigil]

not necessarily

you have more freedom thanks to the inequality

ie, you likely will end up showing that $2(\sqrt{n+1}-1)+ \frac1{\sqrt{n+1}}\ge2(\sqrt{n+2}-1)$

wait no

[clever fjord]

bruhh

[honest turtle]

$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}} \cdots \frac{1}{\sqrt{n}} > 2(\sqrt{n+1}-1)$$
n = k
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}} \cdots \frac{1}{\sqrt{k}} > 2(\sqrt{k+1}-1)$$
n = k + 1
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}} \cdots \frac{1}{\sqrt{k}} +\frac{1}{\sqrt{k+1}} > 2(\sqrt{k+2}-1)$$
$$=>2(\sqrt{k+1}-1) +\frac{1}{\sqrt{k+1}} > 2(\sqrt{k+2}-1)$$
$$=>2(\sqrt{k+1}-1) - 2(\sqrt{k+2}-1) > \frac{1}{\sqrt{k+1}}$$

[clever fjord]

Good

[visual sigil]

thats not even true

i think you dropped a negative

but thats not quite what i meant either

[honest turtle]

First time doing inequality proofs, so I'm very lost.

[visual sigil]

so my general strategy would be along the lines of

basis: $1>2(\sqrt2-1)$

IH: suppose that $1+\frac1{\sqrt2}+\frac1{\sqrt3}+\ldots+\frac1{\sqrt n}>2(\sqrt{n+1}-1)$

Then, we have $1+\frac1{\sqrt2}+\frac1{\sqrt3}+\ldots+\frac1{\sqrt n}+\frac1{\sqrt{n+1}}>2(\sqrt{n+1}-1)+\frac1{\sqrt{n+1}}$

and want to show the RHS here is greater than or equal to $2(\sqrt{n+2}-2)$ because if you can satisfy that then the induction holds

[honest turtle]

I don't quite see how the row got changed from $\frac{1}{\sqrt{n}}$ to $1\sqrt{n}$.

[clever fjord]

Good

[visual sigil]

oh wait thats just me forgetting texit doesnt have my latex shortcuts

[clever fjord]

bruhh

[odd edge]

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[honest turtle]

.reopen

[odd edge]

✅

[honest turtle]

So I'll be left comparing $2(\sqrt{k+1}-1) + \frac{1}{\sqrt{k+1}} \geq 2(\sqrt{k+2}-1)$?

But why greater or equal, instead of greater?

[clever fjord]

Good

[visual sigil]

yeah if you can prove that inequality the result follows

[honest turtle]

Alright, I'll try. Thanks for the help:)

.close

[odd edge]

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[viscid isle]

i'm going crazy

[viscid isle]

I tried (random ball * second black ) + (first black * random ball) = 1/3, i get the same answer when i list all possible draws and see which contains black which are RB BR BG GB, the answer i got is 1/3. is it correct, and also the same asnwer when i do (1 black + any / total) 1C1*5C2/6C2 = 1/3

never mind

is the answer 1/3?

[hushed comet]

how do you get 19/30?

[viscid isle]

forgot GB

[hushed comet]

ah ok so you got it then 👍

[viscid isle]

All my answers lead to 1/3, but there is F on the question althought the answer on the paper is 1/3

i don't think i got it

<@&286206848099549185>

[visual sigil]

try drawing a probbability tree

[viscid isle]

i did

[visual sigil]

show?

[visual sigil]

dont worry about the colours that arent black, thatll just make it more confusing

think along the lines of a coin is either black or not black

also, note that the coins are not replaced

[viscid isle]

what do you mean

[visual sigil]

the answer to the question would be identical if there were 5 red coins and 1 black coin

because we dont care about the colour of the coins that arent black

so that can massively simplify your probability tree

[visual sigil]

ok, now write the probabilities on each branch

for the first coin and the second coin

a probability tree is kinda called a probability tree because you write in the probabilities

[visual sigil]

right

so now add up the probability for each branch containing a black coin

[viscid isle]

add up?

[visual sigil]

yeah

[viscid isle]

you mean * or +

[viscid isle]

why

i thought it's a *

since it's black "AND" not black and vice versa

am i wrong

[visual sigil]

no i meant
add the probabilities of the distinct branches containing blacks

[viscid isle]

how do i get a branch probability (confused)

just to be sure

?

[viscid isle]

don't leave me here bro

man

<@&286206848099549185>

[odd magnet]

the chance is P(the first coin is black)*P(the second coin is anything) + P(the first coin is not black)*P(the second coin is black)

[viscid isle]

I need the answer bro 😭😭😭

[odd magnet]

yes, this is correct

so what is the probability of each branch, and what if you then add them together

[viscid isle]

is this a question

[odd magnet]

I'll do the bottom branch for you: it's 1/6 * 5/5 = 5/30

now what's the chance of the top branch

[viscid isle]

i'm all ears

[odd magnet]

you've drawn the diagram right, you just need to read it now 🙂

[viscid isle]

I did that at first but without a Tree, here on the first method

I think it's my instructor fault then

[odd magnet]

all I can think of is that the question is poorly written

...no that doesn't work either

[viscid isle]

what

[odd magnet]

was gonna say maybe it needs to consider the chance of 2 black coins

but there's only 1 in the box lol

[viscid isle]

oh i thought you were talking about my answers lol

[odd magnet]

everything you've done here is correct - it's 5/30 + 5/30 = 10/30 = 1/3

[viscid isle]

tysm

.close

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[rain tundra]

How do I find the smallest possible value of S?

[rain tundra]

A infinite GP series converges iff $|r| < 1 \$
Given : $a_2 = ar = 1 \implies r = \frac{1}{a} (a \neq 0)\
\implies S = \frac{a}{1-\frac{1}{a}} \
\implies S = \frac{a^{2}}{a-1}$

[clever fjord]

Normed

[round bone]

S = 1/r + 1 + r + ...

rS = 1 + r + r² + ...

S - rS = 1/r
S(1-r) = 1/r
S = 1/(r(1-r))

[rain tundra]

Ahh nvm I see it now, it says S is positive. I missed that and was confused.

.close

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[round bone]

The product ab is maximized when a = b. Therefore, S is minimized when r = 1 - r. r = 1/2 minimizes S.

[rain tundra]

hmm this is also a nice way to solve it,thanks

[odd edge]

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[quiet berry]

heya

[mystic saffron]

hey

[quiet berry]

one second

[mystic saffron]

alr

[quiet berry]

okay

so

[mystic saffron]

mhm?

[quiet berry]

$y=\pm\sqrt{4a}\sqrt{x}$ find when y = x

[clever fjord]

Combustion

[quiet berry]

because it's telling you that the ordinate is equal to the abscissa

[mystic saffron]

yeah

[mystic saffron]

then what ?

what about the angle asked ?

[odd edge]

@mystic saffron Has your question been resolved?

[gritty temple]

do you know about the parametric form of a point on parabola (at^2,2at) ?

[mystic saffron]

yup

[gritty temple]

so the point youve found above with the x=y relation you can find the t of that

[mystic saffron]

4a,4a

[gritty temple]

so now you need to find the other point on the parabola

[mystic saffron]

would be the point of normal

[gritty temple]

for that you have the relation t*=-t-2/t

[mystic saffron]

yeah

then ?

[pearl axle]

just math till the math is mathing

[mystic saffron]

what does the second statement means ?

[pearl axle]

https://tenor.com/view/brain-gif-338783462287106063

[gritty temple]

you have two points of the intersection on the parabola

[mystic saffron]

yeah

[gritty temple]

uhh you can probably find the angle now

[mystic saffron]

ah i see

we have to join the two intersections to the focus

and then the angle between those ?

yeah we got intersection points

and we know the focus as well

yup

done !

thanks @gritty temple

[gritty temple]

👍

[mystic saffron]

.close

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[bright iris]

dont' know how to approach

[bright iris]

if i use direct substituion i get 1/0

but infinity aint' the asnwer

[tropic copper]

${a} = a - [a]$

[clever fjord]

Bettim

[mortal trench]

backslash

[bright iris]

got it

[tropic copper]

also i dont really see why direct sub would give you 1/0

[bright iris]

for lhl

soo

[tropic copper]

${-1} = 0$

[clever fjord]

Bettim

[bright iris]

yeah but it's approaching 1 not exactly 1 yes?

so the fractional part itself is close to 1

[tropic copper]

approaching -1 yes

we dont see lhl rhl for direct sub do we?

[bright iris]

i don't get it

but wait

[bright iris]

i'm trying to do this

hold on brb

[tropic copper]

it changes for negative numbers

you have to use absolute definitions

$[-0.1] = -1$

[clever fjord]

Bettim

[bright iris]

ok so

can we write -{-x} as {x}?

i'm confused at that part

[tropic copper]

no

$-{-x} = - ( (-x) - [-x]) = [-x] - (-x)$

this makes sense

[bright iris]

what

here

and you're substituting {x} in the {-x} though?

man i feel so fucking dmub wtf

[tropic copper]

hold on

wait that is wrong

[bright iris]

idk man i think it's correct since the solution is similar to that

[clever fjord]

Bettim

[bright iris]

oh i noticed something

${ 1^{+} } = 0$

sigh where's the fractional part

[clever fjord]

princedhq

[tropic copper]

well yes

[bright iris]

anyway basically {1^+} is zero

i can solve this question logically but i want to understnad this step

[tropic copper]

but the more accurate representation is ${1^+} = {1+h} = \lim_{h \to 0} h = 0$

[clever fjord]

Bettim

[tropic copper]

so yes it is 0

and the lhl would be 1

right?

[bright iris]

no what

lhl tends to infinyt

infinty

[bright iris]

can you explain me how they got to first step to the second one

[tropic copper]

when x tends to -1^-

it s like -1.00000000000000000001

so fractional part is 0.999999999999999999999

which is basically 1

so {x} = 1

[bright iris]

yes i know but how did they rearrange stuff in the denominator pls tell

[tropic copper]

wait fuck it im confused now

lets start from scratch

[bright iris]

ok

[tropic copper]

$\lim_{x \to -1} \frac{1}{\sqrt{|x| - {-x}}}$

right?

[clever fjord]

Bettim

[bright iris]

yes

[tropic copper]

now as by definition

${a} = a - [a]$

right

[clever fjord]

Bettim

[bright iris]

yep

[tropic copper]

so

${-x} = (-x) - [-x]$

[clever fjord]

Bettim

[tropic copper]

right?

[bright iris]

yep

[tropic copper]

now let the shit under root be alpha

so

$\alpha = |x| - [(-x) - [-x]]$

[clever fjord]

Bettim

[tropic copper]

righT?

[bright iris]

yep

[tropic copper]

now lhl is

-1^-

which is veryyyy less than -1

say

-1.00000001

[bright iris]

yep true

[tropic copper]

so

|x| = 1

[bright iris]

yep

[tropic copper]

oka?

(-x) = 1

ok?

[bright iris]

hmm okay yep

[tropic copper]

and [-x] = 0

wtf

1 over root 2

no shit

im going insane hold o

[bright iris]

this is the sol btw

yeah ikr i hate gif's and fractional parts as well

[tropic copper]

okay so let us ignore every thing and the solution is pretty clear

so we now have to find just two things

{-x} when x tends to -1^- is x+2

{-x} when x tends to -1^+ is x+1

alright?

[bright iris]

how thooo

[tropic copper]

from the solution

[bright iris]

yes that's what i want to understnad 😭

[tropic copper]

yes lets try to do that

okayso

${-x} = (-x) - [-x]$

[clever fjord]

Bettim

[bright iris]

also btw just send me your solution my replies will be late as i'm solving others questions rn too

[tropic copper]

and -x = -(x-h)

so

-x = h-x

let {-x} be alpha

alpha = (h-x) - [h-x]

we can ignore the first h

$\alpha = (-x) - [h-x]$

[clever fjord]

Bettim

[tropic copper]

so x tends to 1^-

nah nvm

[h-x]

where x tends to -1

so its [h+1]

since h is almost 0

[h+1] = 1

so

$\alpha = -x - 1$

[clever fjord]

Bettim

[tropic copper]

now

$\lim_{x \to -1} \frac{1}{\sqrt{|x| - {-x}}}$

[clever fjord]

Bettim

[tropic copper]

so it is just

$\lim_{x \to -1} \frac{1}{\sqrt{|x| - \alpha}}$

[clever fjord]

Bettim

[tropic copper]

so $\frac{1}{\sqrt{1-(1-1)}}$

which is 1

[clever fjord]

Bettim

[bright iris]

is this lhl?

[odd edge]

@bright iris Has your question been resolved?

[odd edge]

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[ocean pendant]

can someone check this. im not sure ive done it correctly

[odd edge]

@ocean pendant Has your question been resolved?

[ocean pendant]

@tidal glacier

[ocean pendant]

<@&286206848099549185>

[odd edge]

@ocean pendant Has your question been resolved?

[ocean pendant]

<@&286206848099549185>

[crimson topaz]

send me the question

[ocean pendant]

the question is the first 5 lines ish

[odd edge]

@ocean pendant Has your question been resolved?

[odd edge]

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[odd edge]

@dusky pebble Has your question been resolved?

[odd edge]

@dusky pebble Has your question been resolved?

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[mystic saffron]

anyone know how to do this

[mystic saffron]

So parralell to P means perpendicular to the normal of P.

I will denote the line we are trying to find by $L_1$

[clever fjord]

Derivative

[mystic saffron]

so $L_1: (x,y,z) = (1,1,1) + t_2v$ where v is a vector in $R^3$

[clever fjord]

Derivative

[mystic saffron]

now we know that $n\cdot v = 0$ and $n = (3,2,2)$

[clever fjord]

Derivative

[mystic saffron]

also, $v = (v_1, v_2, v_3)$

[clever fjord]

Derivative

[mystic saffron]

this is where i am stuck

i thought that v could be anything, as long as $n \cdot v = 0$ but that did not work because it must meet 2 other conditions

[clever fjord]

Derivative

[mystic saffron]

hmmm

thinking i am onto something

this problem reminds me of a problem i did yesterday

I will come back if i get stuck in this solution i will try to fabricate

ok i am stuck

maybe i gave up too early but i know that i am on to something

Here is my solution so far:

Let $L_1$ be the line we are trying to find and \
$L_1: (x,y,z) = (x_0, y_0, z_0) + t_2(v_1, v_2, v_3)$. In parametric form we have:
$\begin{cases}
x = x_0 + t_2v_1 \
y = y_0 + t_2v_2 \
z = x_0 + t_2v_3 \
\end{cases}$
Let $(x_0, y_0, z_0)$ be the point of intersection between $L$ and $L_1$. Thus $t_2 = 0$ at this point. \
The parametric equation of line L is:
$\begin{cases}
x = 2 + t \
y = 2-t \
z = 3-3t \
\end{cases}$
Thus the point of intersection is, in parametric form: \
$(2+t, 2-t, 3-3t)$. \
Now, we know that for L to parralel with P, n must be perpendicular to v. In other words:
$3v_1 -2v_2 + 2v_3 = 0$. \
The parametric form of the $L_1$ is now:
$\begin{cases}
x = 2+t-t_2v_1 \
y = 2-t + t_2v_2 \
z = 3-3t+t_2v_3 \
\end{cases}$
Since we want it to pass through $A(1,1,1)$, then:
$\begin{cases}
1 = 2+t-t_2v_1 \
1= 2-t + t_2v_2 \
1 = 3-3t+t_2v_3 \
\end{cases}$

[clever fjord]

Derivative

[mystic saffron]

this is what I have done so far. I am so close

i just need to know how to find v1, v2,v3

I have a 5 variable equation, but max I can reduce it to is 4.

<@&286206848099549185>

here is problem for more context

refer to my work above

it's so close I know it's close

there is like 1 more key step and I can't find it

[odd edge]

@mystic saffron Has your question been resolved?

[odd edge]

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[pastel orbit]

!status

[odd edge]

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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[pastel orbit]

2 (for part a; havent yet gotten to part b)

so since f is cont. we know theres a number c in [a,b] st. f(c) <= f(x) for any x in [a,b] because of EVT

[pastel orbit]

here is Problem 3a that the question quotes

[pastel orbit]

we know that c =/= a because if it were, then we'd have to conclude that f'(c+) = f'(a+) >= 0 by 3a, which is a contradiction to the hypothesis that f'(a) < 0. Similarly, c =/= b (again, contradiction using 3a).

now I am a bit lost

I want to say that since c is in (a,b), i.e. c =/= a,b, we know that f'(c+) = lim_c+ [f(x)-f(c)]/[x-a] >= 0 by 3a. Similarly, f'(c-) <= 0 by 3a. Thus, f'(c) = 0, and we are done.

[pastel orbit]

but I am not sure this is valid

any suggestions?

<@&286206848099549185>

[sweet gazelle]

Hey I'm looking at everything

Might be able to help

Well I mean for a) wouldn't the idea just be that if it attains a minimum at a then after a it has to start increasing, meaning that f'(a+)>=0

And the same idea for b

[pastel orbit]

well yes, but that wouldn't cut it for rigor i dont think

[sweet gazelle]

Yes definitely you're right

Like you have to write up a formal proof

What class is this for?

Calculus I? Or II?

I'd think it's Calc I

[pastel orbit]

eh, not sure what it would be tbh

technically, it's calculus 1, but its more like an introduction to analysis than a 'standard' calculus course

[sweet gazelle]

I see

Ok back to te question

[sweet gazelle]

With these ideas try writing a rigorous proof and show me

Or do you need help doing that ?

[pastel orbit]

honestly, I really want to still use this idea because it seemed promising

but I don't know if it works

[sweet gazelle]

What I don't get about that idea is where did c come from?

I don't see in any of the two photos a variable called c, only a and b

[pastel orbit]

here

[sweet gazelle]

Ah ok

Yeah I got the idea you're saying in that initial idea

It's good

I'm not sure how to help you tbh

[pastel orbit]

no worries, I think I've convinced myself the idea has to be right

so I'll go with it

thanks for your help anyways

I appreciate it

[sweet gazelle]

No problem

[pastel orbit]

.close

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[timber tangle]

If you want to prove that an integer divides (an expression)

[timber tangle]

if you show that integer(k) = (an expression that can be factored out completely without remainder)

is that enough to prove it?

[toxic rose]

what do you mean by this?

[timber tangle]

I want to prove this statement

[toxic rose]

ah

should've just posted the problem

[odd edge]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[timber tangle]

its because i dont want the answer

[toxic rose]

remember this for future occasions

[timber tangle]

just hints and stuff

[toxic rose]

we are not supposed to that, we r supposed to help you (if we r free)

anyways

coming back to the problem

[toxic rose]

well think about it yourself

if n is not a multiple of 3, and neither is n+1 nor n-1, is the expression divisible by 6?

[timber tangle]

since it can be factored out completely then I think it shows 6 divides n^3-n..

[toxic rose]

no, it does not

until and unless atleast one of those factors is divisible by 6

until then the whole expression is not divisible by 6

u have to prove that out of n, n-1, and n+1, atleast one is divisible by 6

or alternatively, one is divisible by 3, and one is divisible by 2

[timber tangle]

ahh ok

[toxic rose]

and the above is true because

until and unless n, n-1, or n+1 = 6j for some integer j

[timber tangle]

wait wait I'll try

[toxic rose]

kk

[toxic rose]

nuh uh, u cant take n as a multiple of 6 else it would've been as simple as saying well just let n^3 - n be a multiple of 6 and we would've been done

[toxic rose]

try this @timber tangle

this is the intended method

[timber tangle]

what I was thinking was that 6|n -> 6| n^3-n

no? you can't do that?

if 6 divides n then 6 divides n^3-n

and 6 have to divide n because...ohhh for ALL n in natural num

[toxic rose]

yes

heres a hint

take 3 consecutive numbers

and try to prove atleast one of them is divisible by 3

using induction, or proof by contradiction

I would recommend induction

@timber tangle u there?

[timber tangle]

yes, im working it out

[toxic rose]

aight

u can also try prove the more stronger argument that amongst n consecutive numbers, atleast one of them is divisible by n

[timber tangle]

induction is not going to work?

[toxic rose]

write the whole thing out

[toxic rose]

do you mean to ask how to do the induction step?

[timber tangle]

no, I didnt try the induction step yet, but just looking at it, if 3k=n it doesnt make sense how 3k=n+1

because n=/= n+1

[toxic rose]

ur doing it wrong

Heres how you do it

Base Case: One of the numbers in (1;2;3) is divisible by 3

Induction Hypothesis: Suppose one of the numbers in (n;n+1;n-1) is divisible by 3

Induction Step: Consider (n+1, n+2, n)

Case 1: n was divisible by 3 i.e. n = 3k. Then we have that one of (n+1, n+2, n), namely n, is divisible by 3.

Case 2: n+1 or n+2 was divisible by 3 and n was not divisible by 3. If n wasnt divisible by 3, then n must of the form 3j+1 or 3j+2. Which means n+1 or n+2 was divisible by 3 as one of n+1 or n+2 must be of the form 3(j+1) .Then again we have the above.

By principle of mathematical induction, amongst (n;n+1;n-1) one must be divisible by 3.

[timber tangle]

if n is in both cases, can't you just say that since they are in both cases it is proved that n,n+1, or n+2 is divisible by 3?

is that what you are saying for case 2?

[toxic rose]

not really

let me specify it more

@timber tangle is it clear now?

[timber tangle]

no

oh wait

[toxic rose]

i am waiting

[timber tangle]

you did the first case to show that 3|n but for case 2, you assumed 3|n+1 and 3|n+2 because you want to check all 3 cases of consecutive numbers.

[toxic rose]

yes

tbh the more 'elegant' proof is based on modular arithmetic

but i dont think u know that

[timber tangle]

how is induction required tho? can’t you just say n, n+1, n+2 and split them into two cases where 3 divides n and 3 does not divide n

modular arithmetic?

yes.. i think i know it?

you gotta give an example

[toxic rose]

well what about n+1,n+2,n+3?

[timber tangle]

oh okok

[toxic rose]

u need to assume the 2 cases to prove that

this is how induction works

[timber tangle]

yes because it’s usual order (i think that’s what it’s called) of n in natural nums

hence why you gotta show for two cases of consecutive nums

[toxic rose]

well ordering principle you mean?

[timber tangle]

i think so?

[toxic rose]

well

if u mean that

[timber tangle]

ok

thank you so much for helping!

[toxic rose]

https://math.stackexchange.com/questions/1139579/why-is-mathematical-induction-a-valid-proof-technique

[timber tangle]

i’m gonna sleep now

cause it’s pretty late

but

I am curious about the modular arithmetic thing you were talking abt

[toxic rose]

this explains why induction is true, through WOP i.e. how induction works

[timber tangle]

would it be possible to pm me that method?

[toxic rose]

You can open a channel sometime later, if I'll be there I will tell what I was thinking (nvm this)

[toxic rose]

eh yeah sure

[timber tangle]

thanks

[toxic rose]

(u can do .close now)

[timber tangle]

yup

.close

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[echo vapor]

Can someone check my answer please?

[visual totem]

It’s correct

[west merlin]

yea looks good

[echo vapor]

nice thanks

.close

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[viscid jacinth]

3 points given A(acost,asint), B(bsint,-bcost),C(1,0) find the locus of the center of mass of the triangle connecting those points

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@viscid jacinth Has your question been resolved?

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[merry pulsar]

Which of the following statement is true for an m-by-n matrix A and n-by-k matrix B. Pick one of the choices

1. min( m, n ) <-rank( A ) <= max( m, n )
2. rank (AB) = min( rank(A),rank( B))
3. If m = n-k,rank( A ) + rank( B ) <= rank (A + B)
4. If A is an m-by-n matrix of rank m, then rank( AB) = rank( B)

[merry pulsar]

I think it's option number 2 just from eliminating the others

[warped glacier]

yep option 2 is correct

https://math.stackexchange.com/questions/978/how-to-prove-and-interpret-operatornamerankab-leq-operatornamemin-ope

[merry pulsar]

ah ok thank you!

.close

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[warped glacier]

wait I didn't read it properly SORRY

it looked so familiar

@merry pulsar sorry it's not 2

[merry pulsar]

.reopen

[odd edge]

✅

[merry pulsar]

what could it be 👀

oh wait i;m so sorry i typed the question wrong

Which of the following statement is true for an m-by-n matrix A and n-by-k matrix B. Pick one of the choices

1. min( m, n ) <=rank( A ) <= max( m, n )
2. rank (AB) <= min( rank(A),rank( B))
3. If m = n-k,rank( A ) + rank( B ) <= rank (A + B)
4. If A is an m-by-n matrix of rank m, then rank( AB) = rank( B)

option 2 is <=

@warped glacier

[merry pulsar]

Originally I had written it as = but that is wrong

So I think 2 is still correct?

[warped glacier]

yeah with <= that's now correct

[merry pulsar]

ah ok sorry about that

thank you!

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[errant heart]

hello how can i derive this thing

[odd magnet]

hint: split the fraction into 3 parts, simplify, and then differentiate each piece

[odd edge]

@errant heart Has your question been resolved?

[fervent hound]

Or use quotient rule

[errant heart]

How?

Can you show me the 1st step

[fervent hound]

Wait u're deriving with respect to x right?

[odd edge]

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[mystic saffron]

it's related trigonometry.

all parts.

i really hate these type of questions.

[zenith tartan]

start by making a digram please

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

thats hard

[zenith tartan]

anything that you have tried making?

[odd edge]

@mystic saffron Has your question been resolved?

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[steel wave]

@mystic saffron are you able to draw an image for the given conditions?

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

lil bit

but i think its incorrect

@steel wave

<@&286206848099549185>

yes sir

[odd edge]

@mystic saffron Has your question been resolved?

[steel wave]

Construct a horizontal line Passing through P

Let it cut RS at say T

Then use trigonometry(sine, cosine, etc) to get the length of PR

Then length of PT using length of PR

And finally length of PS using that of PT

@mystic saffron can you do this?

[mystic saffron]

tysm i undertstand now

.close

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[tepid hornet]

The profile of an embankment is described by the graph of the function f with f(x) = √x.

a) Calculate the slope angle of the slope in B_1 (1,1) and B₂(9,3).

b) A ramp with a gradient angle of 14° is built on the embankment, which ends at point B(1,1) on the embankment. Explain why this ramp does not end on the embankment without kinks. Where does this ramp start and how long will it be?
c) A ramp with a 14° incline should be built onto the embankment without kinks. Where does the ramp start on the embankment and how long will it be?

I did a

Its 26.57 and 9.46 degrees

[odd edge]

@tepid hornet Has your question been resolved?

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@tepid hornet Has your question been resolved?

[tepid hornet]

.close

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[odd edge]

Please don't occupy multiple help channels.

[mystic saffron]

"(b) Investigate the convergence of the series:"

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[teal adder]

I’m stuck in this

[zenith tartan]

dawg this is physics

but im physicist

[teal adder]

Ok pookie

Ur smart figure it out

[zenith tartan]

so what have you tried so far?

ok wait

[teal adder]

Idk

[zenith tartan]

youve written tthat over there

let me see

[teal adder]

Sum of forces or something but idrk what to do next

[zenith tartan]

dawg have you tried making a free body diagram?

[teal adder]

Dawg🤯

[zenith tartan]

yea dawg?

have you?

[teal adder]

dawg

Sum of forces is .11 N downwards

[zenith tartan]

dawg let me rotate that

,rotate

[teal adder]

So would it be .11N to cancel it out

Or do I need angles for this lmao

[zenith tartan]

dawg do you know the formula to find friction?

[teal adder]

dawg

yes

did I find it wrong?

Ff/Fn= mu

[zenith tartan]

and what is the normal force here dawg?

[teal adder]

.2 newtons

Dawg

[zenith tartan]

no dawg

incorrect

its F dawg

[teal adder]

Dawg whatttt🤯

[zenith tartan]

yea dawggg

[teal adder]

Wdym

I did normal force

Sum of forces

.11 Newtons

[zenith tartan]

dawg calm down

listen

[teal adder]

Fgravity is .2 newtons

Dawg did u even look at my work dawg

[zenith tartan]

no dawg

[prisma cairn]

Meow

[teal adder]

Hush up dawg @prisma cairn I’m tryna learn

[zenith tartan]

dont try to hush kushu dawg

he my frnd

[teal adder]

sorry dawg

[zenith tartan]

it's ok dawg

[teal adder]

Can we get back to the problem though dawg

[zenith tartan]

sure dawg

okay so the object is at rest

[teal adder]

How u just tell me the answer instead of making me figure it out dawg

nevermind dawg continue dawg

[zenith tartan]

no dawg, aint telling any answers dawg, that's against the rule dawg

[pulsar elbow]

lol what is going on

[teal adder]

Hush dawg

[zenith tartan]

physics smay

lmao💀

[teal adder]

All these kids hro

[zenith tartan]

she's a graduate tho

[teal adder]

Private conversations between me and dawg smh

okay

u prolly think I’m dumb too

It’s ok tho

I just didn’t pay attention

[zenith tartan]

okay so we know that it is at rest

[teal adder]

yes

[zenith tartan]

that means we can equate its horizontal and vertical forces

right dawg?

[young void]

J

[teal adder]

Yes dawg

Austin shut up pleade

[zenith tartan]

who tf is this dude now

okay

[teal adder]

omg continue with the explanation

[zenith tartan]

so what are all the forces in the horizontal direction?

[teal adder]

Just F

[zenith tartan]

@prisma cairn pls dont😭

let me get that helpful role

[prisma cairn]

Hint: the normal force is normal to the surface

[zenith tartan]

yea that might help

[teal adder]

Wdym normal to the surgece

Yall are really bad at explaining

I’m telling u give me the next step and I’ll understand it

I’ve learned like 12 of these on my own

[zenith tartan]

damn

[teal adder]

I got a 1540 SAT dawg

[zenith tartan]

ur smart dawg

what is SAT dawg?

[teal adder]

Dawg

What

Are u not from the United States

It’s a standardized test that I scored better than 99.6% of people on

[prisma cairn]

Bold of you to assume I'm tryna explain

[teal adder]

omg

[zenith tartan]

damn

[teal adder]

Dawg

I don’t have time for this dawg I’m tryna go to Harvard dawg

Ur not making any sense dawg

[zenith tartan]

okay so no there is one F and one normal force in the horizontal direction

[teal adder]

What

What are u yappin about

You are NOT getting the helpful role lmao

[zenith tartan]

alr then do it on ur own

[teal adder]

NOOOO

[desert marlin]

!redir

[odd edge]

This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[teal adder]

I WAS KIDDING DAWG

dawg please

Dawg i was just messing with you

@zenith tartan

[zenith tartan]

bruh austin im helping only calm down please

[prisma cairn]

At least 99.6% of the USA can't find the normal force

[zenith tartan]

okay so

[zenith tartan]

did you get this

[teal adder]

Yeah you’re pushing the eraser and it pushes back

[zenith tartan]

calm down kushagra boy

yes

[teal adder]

That’s what you’re trying to find though

Is this not correct?

[prisma cairn]

I don't remember telling you my name but alr

[teal adder]

There’s a frictional force stopping the block from sliding and gravity pulling down

So 0.09 Newtons of force upward and .2 Newtons downward, for a total vertical force of .11 Newtons

[zenith tartan]

i mean ur name is kushu, let's be honest no one has name kushu here its just a small name for kushagra, isnt

[teal adder]

But idk how to use that to find how much I need to push the block to prevent movement

[zenith tartan]

how did you get 0.09 i wanna know

[teal adder]

The .45 frictional coefficient times the force of gravity times the mass

[zenith tartan]

tf

ok listen now

[teal adder]

That might be wrong though

[zenith tartan]

yea it is

[prisma cairn]

Hmm true

[zenith tartan]

okay, so fresh starting, is is in equilibrium so we can say that $\sum \mathrm{F_y}=0$\
$\mathrm{F-N}=0$\
$\mathrm{F=N}$

N is normal force here dawg

[teal adder]

Is this an angles problem

[zenith tartan]

no

[clever fjord]

यजतलमाओ

[zenith tartan]

did u get this? ^

[teal adder]

Yes

Yeah I understand that

[zenith tartan]

okay also ill be denoting friction force as F_f

[teal adder]

Ok

[zenith tartan]

okay so $\rm{F_f=\mu\cdot N}$\
$\rm{F_f=\mu\cdot F}$ we are replacing N by F because we've already proved it above

[clever fjord]

यजतलमाओ

[teal adder]

I follow

[zenith tartan]

okay now we have friction force

[teal adder]

What is it

[zenith tartan]

and now coming on the vertical forces now

this mu times F

[teal adder]

mu time F

Ok

[zenith tartan]

lets just not put value of mu here

we'll put that in end

ok?

[teal adder]

Okay

[zenith tartan]

Okay so, now for vertical forces, $\sum \rm{F_x=0}$\
$\rm{F_f-mg}=0$\
$\rm{F_f=mg}$

[clever fjord]

यजतलमाओ

[zenith tartan]

$\rm{\mu \cdot F=mg}$

[clever fjord]

यजतलमाओ

[zenith tartan]

im pretty sure now you can that value of F

[teal adder]

Wait so F_friction = mu times F, where F is mass times gravity?

Correct?

[zenith tartan]

no

why do he thing so?

[teal adder]

What dawg

[zenith tartan]

???

[teal adder]

Ohhhh

Okay

So F= mg/mu

[zenith tartan]

right

[teal adder]

So .02 times 10

Divided by .45

.44 newtons

That’s the answer

Can we solve all of these real quick

[zenith tartan]

yes

[teal adder]

Thanks dawg

[zenith tartan]

no problem dawg

[teal adder]

Can we solve the rest now though

There’s 6 more

Do u have time for me dawg

[zenith tartan]

holy moly dawg r u fkin serious😭

[teal adder]

No

I’ll figure it out

[zenith tartan]

yea

[teal adder]

But yeah I have no idea what any of your work means

[zenith tartan]

i can give you a hint though

[teal adder]

Took 30 minutes and I still don’t understand

Nah it’s fine

[zenith tartan]

bruh

[teal adder]

We just didn’t do one like this in clwss

Where you’re applying force to something against a surface

The angled ones I’ll figure out though

Fsin theta or something

Where can i give u a 5 star review

[cold urchin]

good work my apprentice keep it up

[zenith tartan]

yea you just equate thier horizontal and vertical components

[zenith tartan]

yessir👍

[teal adder]

Why are u equating the components though

[zenith tartan]

because it is at equilibrium

[teal adder]

Ok

That doesn’t make sense

[zenith tartan]

bruh

ok

look

[teal adder]

It’s not moving yeah

But why would you then equate the horizontal and vertical forces

[zenith tartan]

suppose ur pushing a carton in the right side

[teal adder]

Yes

[zenith tartan]

so someone have to push that carton with the same amount of force from to left side right?, in order to keep it at rest

[teal adder]

Yes

OHHHHH

[zenith tartan]

suppose in this case only u pushing it with A newtons and the person pushing it with B newtons

so you say

that A=B right?

[teal adder]

so it’s going down at A newtons

And you’re pushing it from the side

So the B newtons, the side force has to be equal?

[teal adder]

Yeah this makes sense

[zenith tartan]

not has to be equal

not everytime dawg

[teal adder]

So the downward force is equal to the frictional force?

[zenith tartan]

only when questions gives you that the object is at rest

[teal adder]

????

[zenith tartan]

right!!

[teal adder]

Since it’s at rest

[prisma cairn]

Or constant velocity

[teal adder]

Ok

[zenith tartan]

yes

[zenith tartan]

right, mb

[teal adder]

So when you did the sum of the vertical forces equals 0

Zero because they cancel out, it’s at rest

[zenith tartan]

yes yes yes

[teal adder]

What are F - N = 0

F is the frictional force

[zenith tartan]

dawg u gettin it

[teal adder]

N is the downward force

And they’re equal

[zenith tartan]

bruh

[teal adder]

Wait no

F is normal force

N is force due to gravity

[prisma cairn]

Bruh

[teal adder]

Up down gravity whatever

[zenith tartan]

kushu wanna help you, he's saying me

kushu

go ahead

[prisma cairn]

Ok so

You equate the horizontal forces differently

And vertical ones differently

You don't connect them

Okay?

[teal adder]

What are F and N

In the part in the middle where is says equilibrium

What are F and N

Like the names

Because ik they’re the upward and downward y forces

[prisma cairn]

N is the normal force that the board is applying on the eraser

[south plume]

F - force
N - normal force to the force F

[teal adder]

But it’s vertical?

[prisma cairn]

Horizontal

[teal adder]

Aren’t we saying the sum of the y forces equals 0

[prisma cairn]

Yes

Vertical and horizontal both

[teal adder]

So then why are you talking about the horizontal force in the vertical part of the calculations

I’m trying to understand what the F and N are

[prisma cairn]

Because friction (which is vertical) requires normal force( which is horizontal)

[teal adder]

Ok

[prisma cairn]

F is the force twitch which you push the eraser

N is the force with which the eraser gets pushhed back towards you by the board, so that the eraser is at rest

So F must equal N

You get this?

[teal adder]

Yes

But why is the the sum of vertical forces

Y values

Wouldn’t that be x values

Horizontal forces

[prisma cairn]

What?

[teal adder]

You see where it says sigma Fy

And sigma Fx

[prisma cairn]

I haven't come to Fy yet

[teal adder]

To it me it seems like the y and x should we switched

bro

[prisma cairn]

I just explained the Fx

[teal adder]

you never said that

that’s why I was asking 30 questions

[prisma cairn]

horizontal means Fx

[teal adder]

I know that

You weren’t explaining the Fx though

You were explaining this

Which the guy notated as Fy

Which I don’t think is correct

He notated Fx as dealing with gravity

I think the dude from before just notated it wrong

It makes sense to me if I flip the y and x