help-38

[shrewd summit]

like

[bright quarry]

brother please do what i suggest

[shrewd summit]

ok

[bright quarry]

@shrewd summit what is (some shit)

did you find it yet

[shrewd summit]

999a + 99b + 9c?

[bright quarry]

great job sir and can you see why this is useful

[shrewd summit]

oh then since a + b + c + d is alr div by 3

[bright quarry]

what’s special about that

[shrewd summit]

and then the something part is also div by 3

[bright quarry]

mhm

because it’s divisible by 9

[shrewd summit]

ah ty i get it

[bright quarry]

(this trick also works for divisibility by 9)

sum of digits is divisible by 9 <—> number is divisible by 9

[shrewd summit]

oh icic

tysm

[bright quarry]

can you do the reverse part now?

assume the number is divisible by 3

then show the sum of its digits is divisible by 3

it’s like the same thing but you still have to write it all out

your justifications are a bit different though of course

[shrewd summit]

yeah i see it

ty again

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[bright quarry]

you’re welcome

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[rigid orbit]

Given 4 real number abcd with 1>a>b>0>c>d>-1

Is bd greater than ac?

[burnt mulch]

Take $a=0.9, b=0.8, c=-0.2, d=-0.5$

[solid kiln]

Civil Service Pigeon

[burnt mulch]

bd = -0.4

ac = -0.18

so bd < ac in this case

[rigid orbit]

Thanks

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[final nova]

lost on how to solve this, ik i can use limit laws but other than that im stuck

[tender sluice]

!status

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[final nova]

1

[tender sluice]

simplify the denominator

[final nova]

sin^2 theta?

[tender sluice]

sign?

[final nova]

or is it -sin^2theta

[covert loom]

$\cos^2(\theta) - 1 = -\sin^2(\theta)$

[solid kiln]

alex <3

[final nova]

can i divide and just get sin(theta) to find the limit?

[tender sluice]

dont forget the absolute value

[covert loom]

it's approaching from negative side

so it becomes -$\frac{-\sin(\theta)}{\sin^2(\theta)}$

[final nova]

doesnt that mean |sin x| = -sin x since its from the negative side?

[covert loom]

yes

[final nova]

so 1/sin(theta)?

[solid kiln]

alex <3

[covert loom]

yes

[final nova]

idk why the limit for that is -inf though

im bad with trig limits

[covert loom]

because

1/0 is infinity

and since it's 0-

it's more of

1/-0.00000000000000000000001

which is - 1/0 when approaching

-infinity

[final nova]

ohh i see

so if it was 0+ itd be positive inf?

[covert loom]

yes

[final nova]

gotcha thank you!!

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[faint igloo]

hey i need a help w this question

[tender sluice]

[faint igloo]

this is the one

[hidden dew]

equate both equations

functions*

[faint igloo]

okay then

okayyy i got it

and this one

the answer for part a is 6 but idk how

[hidden dew]

i suppose n(whatever) means the number of elemnts in the diagram of this weird letter yea ?

[faint igloo]

yes

[hidden dew]

ok there are 20 elements 3 of them are not in A or B so
17 in A or B
number of elements in A is 10 and in B is 13
10+13=23
23-17 (to get the elements that are in both A and B)
=6

[faint igloo]

ohhh okayy i got it

thank yoo so much!

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[languid grove]

yo

[languid grove]

can someone help me please

How did I get this wrong

I genuinely don't get it

[split chasm]

you forgot about the leading coefficients/constant factor

(x-5)(x+3) expands out to x^2 - 2x - 15
instead of what you originally have
same issue with the denominator

[languid grove]

wait so do I have to do 2(x-5)(x+3)
and 3(x-5)(x-1)

[split chasm]

yes

[languid grove]

alr appreciate it, does this apply to all situations when i factor, assuming the leading coefficent is greater than 1

[split chasm]

yes

(more generally, not 1)

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[coarse meadow]

1/(2x5) + 1/(5x8) + 1/(8x11) + .... 100 terms

[verbal stream]

Try telescoping

[coarse meadow]

wdym?

[verbal stream]

basically, you obviously don't want to add up all 100 terms since that'd take forever

instead, try to rewrite each term so you can cancel lots of stuff out

[coarse meadow]

how would we start with that though?

[trim musk]

Observe that the difference of the terms on the bottom is constant

[verbal stream]

yes ^

that often hints that you can cancel out denominators

For example

1/(1*2) + 1/(2*3) + ... + 1/(9*10)

you can rewrite it as
1/1 - 1/2 + 1/2 - 1/3 + 1/3 ... - 1/9 + 1/9 - 1/10

and all the middle shit cancels out

to 1 - 1/10

You can do something similar here

[coarse meadow]

how would we start it though?

cause 2x5 seems more complex than just a 1x2

[verbal stream]

what is 1/2 - 1/5

let's try that

[coarse meadow]

3/10

[verbal stream]

right

but that's 3 times as much as you want

what about 1/5 - 1/8

[coarse meadow]

3/8

[verbal stream]

3/40

But yeah notice how it's three times more than you want it again

[coarse meadow]

o oops

[verbal stream]

You can generalize:
$\frac{1}{n} - \frac{1}{n+3} = \frac{3}{n(n+3)}$

[solid kiln]

Ari

[coarse meadow]

why do we not want the 3 tho?

[verbal stream]

So what we can do is take
$\frac12 - \frac15 + \frac15 - \frac18 + \dots + \frac{1}{299} - \frac{1}{302}$
and then divide the result by 3

[solid kiln]

Ari

[verbal stream]

because the problem asks for 1/10 + 1/40 + ...
not 3/10 + 3/40 + ...

[coarse meadow]

how did we get to 1/2-1/5 etc then?

[verbal stream]

That's just something we tried

[verbal stream]

I'm basically trying to apply this strategy to this problem

and adjusting as needed

[safe loom]

^

[coarse meadow]

im kinda lost rn how do u get the subtractions etc though

[verbal stream]

Your goal is to make things cancel out right

I'm noticing that you can rewrite 1/(2*5) as (1/2 - 1/5) / 3

And if you write all the fractions in this form

almost everything cancels out

Why did I come up with rewriting it as subtraction? It's just a general approach for these long sums (we call them telescoping sums)

[trim musk]

This is also useful (I don’t know latex sorry)

[coarse meadow]

wait does this apply to any fraction thats similar to like the questions one

[verbal stream]

when you have a constant pattern in your denominators

like 2, 5, 8, 11, etc. like we did here

and you have "matches"

then yes

but not always

For example 1 + 1/2 + 1/3 + 1/4 + ... is infinite so this won't work for this sum

[coarse meadow]

wait but js wondering how did you know that they would cancel out if you did it like that?

[verbal stream]

because I've done this problem before lol

or something similar

[coarse meadow]

oh lol

[coarse meadow]

wait i searched up an answer for it and they did 1/3 x [(3n+2)-(3n-1)]/[(3n+1)(3n+2)] after 1/[(3n+1)(3n+2)] can u kinda explain that step a bit

[verbal stream]

Notice how (3n+2)-(3n-1) is just 3

[coarse meadow]

yeah but why not just write 3 instead of writing all of that

theres gotta be like a reason right

[verbal stream]

idk man maybe they were just showing it

more "algebraic" ig

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[dark oar]

There is just so much going on here...I sort of get why they picked the u, but then they started to just do things

[dark oar]

how did they get 1/(2(2x+1)^2 and why they multiplying du by it...

[red mountain]

this not use u yet, this uses the fact that 1/(2x+1) = (2x+1)/(2x+1)^2

then this becomes du

and with this they rewrite (2x+1)^2 into 1+u^2

[dark oar]

ooooooooo

okokokok

im understanding

ty for being awake

close!

close

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[dense snow]

I wanna ask is my answer correct because the computation is quite tedious and annoying to do

[dense snow]

Should not be (n-1(n-2)) it should be n-1(n-1)because there are n-1 (-1)

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[dense snow]

<@&286206848099549185>

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[dense snow]

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[jolly saffron]

Is this question ill-posed? Since orthogonal matrices of different orders can't be added anyways

[jolly saffron]

But for the sake of not leaving this blank, I guess it would be for matrices of the same order?

[verbal stream]

I think you could claim No using this reason

but maybe ask

[jolly saffron]

I'm gonna check for the same order anyways

[somber ginkgo]

it's possible that whoever wrote this question was thinking of "the set of all nxn orthogonal matrices"

[jolly saffron]

Does it even satisfy closure

[solid kiln]

dyxn

[jolly saffron]

I don't think it does lol

I mean tbf scalar multiplication is an even easier giveaway

Alright then that's that

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[jolly saffron]

Thank you everyone!

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[red mirage]

<@&286206848099549185>

[umbral hazel]

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[mystic veldt]

cauchy schwarz inequality

[red mirage]

Dang you re right

And it gives me equality

So they are proportional right?

And what are the rational and real solutions?

[mystic veldt]

yes

[mystic veldt]

ur not getting the answers just like that

[red mirage]

Well it gives me just one solution

If they re equal to their denominator

Or are multiples

And just i see what multiples are perfect squares?

@mystic veldt

[mystic veldt]

for them to be rational, ( k ) would need to somehow cancel out ( \sqrt{3}, \sqrt{5}, \sqrt{7} ), but that’s not possible unless ( k = 0 ). So the only rational solution is ( (0,0,0) )

[solid kiln]

yajat

[red mirage]

!done

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[red mirage]

.close

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[mystic veldt]

yeh

[red mirage]

Ok thanks

[mystic veldt]

wlcm

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[torpid lantern]

guys i dont understand this part

[torpid lantern]

circled in green

[jolly pebble]

Subtract both sides by 7g

[torpid lantern]

yes

that lead us to

9= 8g?

[jolly pebble]

No

Both sides

[torpid lantern]

how

[jolly pebble]

7g + 9 = 8g
7g + 9 -7g = 8g - 7g

[torpid lantern]

ohh

so its not a simultanous equation

we r solving for g

tysm

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[lone narwhal]

<@&286206848099549185>
For question 12 i can't seem to solve the last question being
*hypotenuse: 5 adjacent: 5x and opposite: 7x *
x has to have an exact answer
what i tried is
(7x)^2+(5x)^2=5^2
(12x)^2=25
12x=25$
12x=5
x=5/12 : but this doesn't give you an exact answer.

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!15min

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[mystic iris]

the second line shown is wrong

[lone narwhal]

can you explain how

[opal grove]

@red mountain fr

[mystic iris]

you cannot combine (7x)^2 and (5x)^2 into (12x)^2

[lone narwhal]

oh

[mystic iris]

if you square both sides normally you get the right answer probably

[lone narwhal]

but if you get 74x

and divide that by 5

still

[spring current]

Btw I think theres another concept youve misunderstood
"exact answer" doesnt mean it has to be an integer

it just means you dont round

[mystic iris]

can you show your steps

[spring current]

2.5 is an exact answer

even sqrt(2) is an exact answer if you leave it as that and don't round it

[lone narwhal]

i know

[spring current]

then why did you say at the top x = 5/12 but thats not an exact answer

[lone narwhal]

doesn't that go on forever though?

[spring current]

and?
sqrt(2) goes on forever
its still an exact answer

as long as you dont round its an exact answer

[lone narwhal]

oh

[spring current]

ie:
pi is an exact answer
but if you rounded it to 3.14 that would make it not be an exact answer anymore

[lone narwhal]

so thats why i had trouble

[spring current]

theres still another mistake with your working though im pretty sure

[lone narwhal]

well that makes everything easier

[spring current]

nvm

[lone narwhal]

well

[mystic iris]

?

[lone narwhal]

ye

wait

cause when you do 5x^2 plus 7x^2 you get 74x

or does it have to be sqaured?

[spring current]

"74x^2

[lone narwhal]

i want to know

[spring current]

yes thats the mistake i was talking about
you definetely cant forget the square

[lone narwhal]

how does it stay like that

[spring current]

look you have (5x)^2 + (7x)^2
first you distribute teh square to get
25x^2 + 49x^2

now imagine that x^2 is an apple or something

[mystic iris]

if you add two like terms you keep the square

[spring current]

25 apples plus 49 apples is 74 apples

[lone narwhal]

ye

[spring current]

so its gonna be 74x^2

the x^2 stays as an x^2 just like the apple stays as an apple

[lone narwhal]

ohhhhh

so i just can't forget about squares

[spring current]

definitely not no
idk where you even got that idea from ToT

[lone narwhal]

okay

well

thx

im gonna close now

[spring current]

Yw bye

[lone narwhal]

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[fiery ember]

Recent test question and I have some idea about the answer

2 coins with one side 4 and one side k are flipped, added together, and result in P(X=2k) = 1/4, P(X=k+4) = 1/2, P(X=8) = 1/4, its a basic tree diagram if u want to visualize it

[fiery ember]

They provide that Var(X) is 0.5 and to find k

[slate crescent]

!occupied

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[fiery ember]

no calculator btw

Im confident that the correct formula would be to use Var(X) = E(X^2) - (E(X))^2 and sub values and end up with k being 3 or 5

My friends got -3 by using Var(X) = np(1-p) but isnt that for a binomial dist and not work in this situation

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[glad grotto]

yo

[glad grotto]

Can someome explain these 2 examples to me again? I forgot how to do it 💀

For the first one is just adding all the numbers up in a, b, & c?

Cuz if so it'd be 14, 27, & 29

<@&286206848099549185>

[tardy ledge]

i think your right

B' js means everything thats not B

AnB js means only the stuff in a and b's intersection

[glad grotto]

Ah ok

[tardy ledge]

AuBnC js means all the stuff in a and b that are intersecting with C

[glad grotto]

What about the 2nd example?

Like the answers

U means or

n means and?

Right

[tardy ledge]

n means intersect

thats how i was taught

[glad grotto]

alr

[tardy ledge]

U means union

so like add ig

[glad grotto]

Alr

Ty

[tardy ledge]

np

[glad grotto]

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[quick garnet]

Turn the infinite periodic decimal 0.58 to a normal fraction: answer is 7/12.

It says to use the formula for the sum of infinite regressing geometric progression (S=a1/1-q). I think a1 = 1/2

[empty orchid]

More like $0.58+0.00\bar{3}$

[solid kiln]

;(

[quick garnet]

Yeah

[empty orchid]

1/2 isn't necessarily your first term then

[quick garnet]

There's also a (3) Infront of 0.58

[empty orchid]

The repeating part of this is just 0.003+0.0003+0.00003... which should be what in terms of a sum?

[empty orchid]

That is accounted for by the 0.003(bar)

[quick garnet]

Ah I didn't know

[empty orchid]

^

[quick garnet]

So the q is 1/10? And a1?

[empty orchid]

Look at the first term of the sequence

[quick garnet]

0.58

Got it

Thx

[quick garnet]

0.58/(1-0.1) but it's not it

[empty orchid]

It is not

[empty orchid]

I was talking about this sequence

[quick garnet]

Then how do I solve it

[empty orchid]

0.58+sequence

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[static sigil]

Why is this part B wrong?

I've been doing integral from 0 to 22.58 (-.1q^2 +51)q dq

[zinc ginkgo]

,w int 0 to 22.58 (-0.1x^2+51)xdx

[solid kiln]

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[zinc ginkgo]

Is your formula for total possible revenue wrong

[static sigil]

That's possible...

I'm using

integral from 0 to horizonal intersection point D(q)*q dq =revenue

D(q) is demand

Does that sound right?

[zinc ginkgo]

No

Total possible revenue sounds like the range of revenue possible

So it should be an interval of the form [0, N]

[static sigil]

What's N?

[zinc ginkgo]

The max revenue

[static sigil]

Oh, okay. Why would that be the case?

[zinc ginkgo]

Just how I interpret the question

[static sigil]

I guess if what I was doing was wrong, maybe that could be it

Does everything else about what I'm doing seem right?

[static sigil]

It looks like N = 0 though unless I'm confused

[zinc ginkgo]

if your revenue in A) is greater than 0, how can the maximum possible revnue be 0

[static sigil]

Sorry, I meant that the max revenue is found at q=0

I don't get it though, wouldn't the maximum revenue be in relation to D(q) not q?

[zinc ginkgo]

yes probably. find the max revenue

[static sigil]

1151.58

you want me to set that as the upper bound?

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@static sigil Has your question been resolved?

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[marsh forum]

I believe I have to use the $\varepsilon -\delta$ defn here

[solid kiln]

What a wonderful world it is !

[willow mortar]

Have you considered using the inner product hint ?

[quiet zodiac]

@marsh forum lmk if u need the step by step solution

[trim jolt]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[marsh forum]

Thanks, but i was intending to post my soln

[trim lichen]

<@&268886789983436800> this is the second time

i already warned this guy in another channel about using AI like this

[trim lichen]

do you not have any basic results at your disposal

well. ok nevermind i guess it would take too long to state absolutely everything

[marsh forum]

This is my first epsilonic proof in R^n

so I'd rather do the entire thing

[willow mortar]

You don’t need to go epsilonic here

You can go sequential instead

[marsh forum]

We haven't done that yet 😭

so $\norm{x-a} < \delta$

[solid kiln]

What a wonderful world it is !

[trim lichen]

is that it

[marsh forum]

For now

Trying to makes sense of $\abs{x-a} < \varepsilon$

[solid kiln]

What a wonderful world it is !

[marsh forum]

because that isn't the out put

ah

I have to show that $\norm{x-a} < \delta \implies \abs {\norm{x} -\norm{a}} < \varepsilon$ ?

right

[trim lichen]

i think you might not want to say "i have to show" twice in one sentence unless you are a redundancy fan of redundancy

[solid kiln]

What a wonderful world it is !

[trim lichen]

you have to show that there EXISTS A DELTA that makes this implication work.

also it might be a bit more convenient to split this into cases based on whether a=0 or not

essentially to first show the function is cts at 0

which is a tad easier

and then something different can be done for a ≠ 0

[marsh forum]

okie

[trim lichen]

the solution i came up with eventually involves cauchy-schwarz at one point during the ineq chain

[marsh forum]

we start with $a=0$

[solid kiln]

What a wonderful world it is !

[marsh forum]

in that case $\delta = \varepsilon$ works

[solid kiln]

What a wonderful world it is !

[marsh forum]

Now e look at the case where $a \neq 0$

[solid kiln]

What a wonderful world it is !

[marsh forum]

now what

The sheer amount of norms is confusing me

[trim lichen]

$\absv{\nrm{\bd{x}} - \nrm{\bd{a}}} = \frac{\absv{\nrm{\bd{x}}^2 - \nrm{\bd{a}}^2}}{\nrm{\bd{x}} + \nrm{\bd{a}}} \leq \frac{1}{\nrm{\bd{a}}} \absv{\nrm{\bd{x}}^2 - \nrm{\bd{a}}^2}$

[marsh forum]

hmm, yea

[solid kiln]

ann.in.a.teacup

[marsh forum]

ohh

but $\norm{x}^2 - \norm{a}^2 = (x-a) \cdot (x-a)$

[solid kiln]

What a wonderful world it is !

[trim lichen]

incorrect

[slate knoll]

wrong

[trim lichen]

$(\bd{x}-\bd{a})\cdot(\bd{x} - \bd{a}) = \nrm{\bd{x}}^2 + \nrm{\bd{a}}^2 - 2 \bd{x} \cdot \bd{a}$.

[solid kiln]

ann.in.a.teacup

[marsh forum]

ah right

I was thinking about (x-a) \cdot (x+a)

[slate knoll]

you are considering the triangle inquality?

it helps with continuity

[trim lichen]

yes you were

[vestal wharf]

Do not use AI to help people. If they wanted to use AI, they could just use it themselves.

[trim lichen]

may i reiterate this is the 2nd time ive seen that guy in particular post AI slop

[slate knoll]

true

[marsh forum]

yeah

[vestal wharf]

Actually since this you have been warned repeatedly by mods to stop spamming help channels with AI, I'm muting you again.

[trim lichen]

im not a mod tho

or did other mods already bonk him

[marsh forum]

$\norm{x}^2 - \norm{a}^2 =( x +a) \cdot (x-a) = (x-a+2a) \cdot (x -a)$

Now cauchy schwartz ?

[trim lichen]

uhhh

for starters there's a stray period in there

but also, you have not made any reference at all to the dot product of any vectors

so this is like

unusable

[solid kiln]

What a wonderful world it is !

[trim lichen]

mmm

ok thats better but there's still no invocation of cauchy-schwarz yet & im not sure that rewriting x+a as x-a+2a does us any good yet.

[marsh forum]

So what do I do

so I can re-write this as

$(x-a) \cdot (x-a) + (2a) \cdot (x-a)$

[solid kiln]

What a wonderful world it is !

[marsh forum]

taking it's norm and using the triangel inequality

we find that this gives is $\delta^2 + 2\norm{a} \delta$

[solid kiln]

What a wonderful world it is !

[trim lichen]

stop

overcomp

$\frac{1}{\nrm{\bd{a}}} \absv{(\bd{x}-\bd{a})\cdot(\bd{x} + \bd{a})} \overset{\text{C.S.}}{\leq} \frac{1}{\nrm{\bd{a}}} \cdot \nrm{\bd{x}-\bd{a}} \cdot \nrm{\bd{x} + \bd{a}}$

[solid kiln]

ann.in.a.teacup

[trim lichen]

now comes the time to artificially impose a cap on delta. if you require $\delta$ to be at most $\nrm{\bd{a}}$, then it becomes possible to write down the following bound: $\nrm{x + a} \leq \nrm{\bd{x} - \bd{a}} + 2 \nrm{\bd{a}} \leq 3 \nrm{\bd{a}}$

[solid kiln]

ann.in.a.teacup

[marsh forum]

I se

*see

Thanks

.cose

.close

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[quick garnet]

For what values of a and b does P(x)=ax¹⁰⁰-bx⁹⁹-1 divide by (x-1)²? Using Horners method preferably

[trim lichen]

are you sure you want to spend hours and hours writing out the evaluation of a 100th degree polynomial with Horner's method???

[quick garnet]

That's what the guide in the book says. I think it should become some sort of system idek

There's a, b, c, d answers also

[exotic pine]

there should be a pattern in horners if you want

[quick garnet]

a=101, b=-100
a=-99 b=-100
a=100 b=99
a=b=-100

[exotic pine]

ok thats really helpful

hint: (x-k) is a factor of f(x) <=> f(k)=0

[placid radish]

well it shouldnt be hard with horners rule, do you know how to apply it in this case?

[jaunty scarab]

1 is a root => the sum of the coefficients of the polynomial add to 0

but you also have that 1 is a root of multiplicity 2, so you can divide the polynomial by x-1, and then apply the rule again to the resulting polynomial

[quick garnet]

Can someone do it with the explanation? English isn't my first language

[placid radish]

do you know how to apply horners method?

[quick garnet]

Yeah but

Only in normal polynomials

[placid radish]

try and apply it to x-1 and see what you get

[jaunty scarab]

why is this one not normal?

[quick garnet]

Like this?

[placid radish]

can you try doing it (i think there is too little space between the -b and -1, there are 98 terms inbetween)

[quick garnet]

😧

Like I've only used Horners method to find the roots of a polynomial how do I do it here

[placid radish]

but before we begin, and you do a simple example so I can see how you are applying horners method?

[quick garnet]

a*(x-1) -b , then (a*(x-1) -b)(x-1)-1

And that should be equal to 0

[placid radish]

OH, I thought it was a different method...

[quick garnet]

What does that look like?

[placid radish]

ig if this is how you do horners method this isn't really applicable for this question

[marble wharf]

horners method can refer to two things, either evaluating P(1) or doing the polynomial division P(x) divided by x-1. however, as it turns out, its much easier to use a different condition here for when P(x) is divisible by (x-1)^2

[quick garnet]

Um and what's that?

I'm getting lost idk

[marble wharf]

derivatives

[placid radish]

I think they might not have learned that

[marble wharf]

the last person who asked this question had learned about them

[quick garnet]

I have but how does that help

[marble wharf]

1 is a double root iff 1 is a root of P and of P'

[placid radish]

mbmb lol

[exotic pine]

(x-1) is a factor of p(x), so p(1)=0

[quick garnet]

I get a=b+1 but theres 2 answers that match that

A) a=100 b=99
B) a=-99 b =-100

[marble wharf]

you are not using both pieces of information

P(1)=0 and P'(1)=0

[exotic pine]

oh wait i didnt see that

[quick garnet]

Oh

Ok thanks all it worked

One last thing, how is P'(1)=0

I havent applied derivates in these kind of excecises before

[quick garnet]

I Lowkey don't understand

[marble wharf]

P(x)=(x-1)^2 g(x) for some g

differentiate that

[quick garnet]

Ok thanks 👍

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[wooden bane]

Hello

[marsh forum]

Hi, do you have a question to ask us?

[clear cloud]

We don't charge anything we swear

[wooden plover]

Nah it's 1p per word here

/j

[ancient nest]

Or you can sign up for a 1 week free trial.

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@wooden bane Has your question been resolved?

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[tardy ledge]

im getting

[tardy ledge]

$\dv{s}{t} = \frac{e^t(lnt-\frac{1}{t})}{(lnt)^2}$

anyway the answer says

idk how they are getting that t on the bottom

[solid kiln]

gamer75431

[tardy ledge]

i dont get how they get the top one either tbh

[split chasm]

multiply numerator and denominator by t

[tardy ledge]

by js t wouldn't be right tho

wait

idk

if we multiply by js t then isn't the value of -1/t changing to -1

do u mean factor 1/t

idk

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[wraith hinge]

a

[wraith hinge]

True ?

Answer is given as false

[marsh forum]

A matrix is a vector

And recall the mathematical defn of a vector

[trim lichen]

are you doing intro linalg

[wraith hinge]

Anything with magnitude and direction 😁

[trim lichen]

in that case the expected answer is false to both a AND b; vectors CAN look like this, but they don't NEED to.

[wraith hinge]

Yeah

[marsh forum]

A vector is an element of a vector space, which is a set with a list of certain properties

[trim lichen]

the more fundamental concept is not that of a vector per se, but that of a vector space.

[marsh forum]

@wraith hinge

[trim lichen]

this is a subtle but important shift in perspective.

[wraith hinge]

So they meant vector space here

[trim lichen]

no

[marsh forum]

no

[wraith hinge]

Damn

[trim lichen]

you are kinda missing my point here

[marsh forum]

read 1.21

[trim lichen]

...

[marsh forum]

Sorry, bad joke

see 1.21

[trim lichen]

bad AND poorly timed

[marsh forum]

It was a. joke on that

[wraith hinge]

"A directed line segment is a vector" is a true statement but "A vector is a directed line segment"is false

[marsh forum]

yup!

[trim lichen]

well like yeah

[marsh forum]

The converse in maths isn't necessarily true

[trim lichen]

all squares are rectangles but not all rectangles are squares

and as i said before the elements of some vector spaces definitely LOOK LIKE directed line segments

but not all of them do

[wraith hinge]

Yeah got it thanks!

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[soft pollen]

a) write down a matrix A so that R.A indicates how many sixes were scored per class.

b) Write down a matrix B so that R.B indicates how many passes were scored per class.

c) Write down a matrix C so that R.C indicates how many students there are per class.

d) Write down a matrix D so that E.R indicates how many points were scored per class in total.

e) Write down a matrix E so that E.R indicates how many fours, how many fives, how many sixes... the three classes scored together.

2nd image are the answers

But I don't know how to get there

[trim jolt]

@soft pollen Has your question been resolved?

[trim jolt]

@soft pollen Has your question been resolved?

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@soft pollen Has your question been resolved?

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@soft pollen Has your question been resolved?

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[lean lark]

Matrix
c) use a*-1 to find x

[empty orchid]

Which question do you need help with? c)?

[lean lark]

yes

[empty orchid]

Ok.

Well, what do we do when solving a linear system $AX=B$ for $X$?

[solid kiln]

;(

[empty orchid]

@lean lark

[lean lark]

yes sorry i got call by my parent

i use I?

[empty orchid]

What do you mean?

[lean lark]

identity matrix

[empty orchid]

Yeah

What about it?

[lean lark]

Since the equation is AX=B
(A*-1)Ax= (A^-1)B

[polar pewter]

this look like the things i solved in my Computer science 2nd year in Algeria

crazy

[lean lark]

Ix= (A^-1)B

[empty orchid]

Yep

[empty orchid]

So X=(A^(-1))B is your solution

Now use it!

[lean lark]

X is a 3x1

-4
7
5

[empty orchid]

Now do part d)! Verify!

[lean lark]

its not in the exercise to do nor in the extra exercise

thx again for the help

[empty orchid]

What does part d) say then

Did I self-translate it incorrectly

[lean lark]

it says verify that that the Matrix X that you have found in question c) is a solution to the equations (*)

[empty orchid]

So plug in x y and z you found into the linear system

[lean lark]

yh

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[woeful pasture]

Just started learning calc. I was working on derivative exercises and the steps in this problem left me with some questions.

1. Couldn't we do the exponent reduction sooner (illustrated in red)?

2. Why did the 2 in the denominator disappear(Circled in green) in the next step? Shouldn't the whole next step have 2 in the denominator, when it gets multiplied with t^(-1/2)

[tulip violet]

Your two questions are related to each other. Part 2 is basically part 1 and yes you can reduce sooner, I'm guessing that it just wants to display the full steps

Your second question, the 2 didn't disappear, because 1/2 = 2^-1

[woeful pasture]

oh! It's just a simplified version

Because 2^(1/2-1), the -1 is the (1/2) that we're multiplying initially

[tulip violet]

Yes

[woeful pasture]

Thank you, that clears it up for me

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[trim musk]

can someone help with this? i put k = 9 then used lhopital

[trim jolt]

@trim musk Has your question been resolved?

[trim musk]

<@&286206848099549185>

please ping me if you are replying

[exotic tundra]

you didn't differentiate the bottom

[trim musk]

i realised but it isnt making a difference

its still t^t

how do i deal with that

[exotic tundra]

ah, do t=e^ln(t) on the base

[trim musk]

sure i did that as well

[exotic tundra]

then put the limit into the exponent bc e^x is continuous

you should then be dealing with the limit of
t lnt

[trim musk]

tlnt?

[exotic tundra]

yea

$e^{\lim\limits_{t\to0}t \ln t}$

[solid kiln]

Sepdron

[trim musk]

Sure, how do I evaluate this now

[exotic tundra]

same method as you did on the original
$t \ln t = \frac{lnt}{\frac{1}{t}}$

[solid kiln]

Sepdron

[exotic tundra]

the L hopital

[trim musk]

Ok but this will give 1/t/-1/t^2

oh no

It was 1 this whole time

[exotic tundra]

yea

don't forget the 1/9 though

[trim musk]

yes

Thank you so much

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[exotic tundra]

np!

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[serene crane]

I have 9 questions, and would just like help working threough them

[serene crane]

hold on

[orchid wagon]

what does table of integrals say

[serene crane]

writing out my work rn

all the formulas

hold on

also is it fair to view the 17 as

sqrt(17)^2

i think i use #19

Its obviously between the second and third answers

im unsure of where i went wrong in my math for the denomiantor

[midnight vessel]

du = 3e^{3x} dx

I think you confused derivatives and integrals 😅

[serene crane]

ohhh

lol

thank you

okay onto the next

How should I view this,

obviously a^2 = 9, a = 3

I guess I could view u^2 = x^3

[midnight vessel]

Substitute u=x³

[serene crane]

i mean

oh

i see

hold on

i got it!

imma work through them rq

Should i do u sub on the whole thing?

Or how should I go about this

[midnight vessel]

Yea substitute the whole thing.

[serene crane]

or

is it that one

oh sub the whole thing

okay brb

[midnight vessel]

Wait, I got a better idea.
Just substitute u=7x

[serene crane]

okay ill try that

Im unsure of how thatll get rid of anything though

but let me try

you sure i shouldnt make u = e^7x, @midnight vessel

[midnight vessel]

Yea, I guess that looks much better 😅

[serene crane]

my only problem is

it looks like it could be this one

oh wiat

it is lol

[midnight vessel]

New integral?

[serene crane]

?

new problem yeah

thats entry 33

ig im trying to make the denom u^2

so i have to figure out what x^2 is equal to?

if so still confused

[midnight vessel]

Just factor out 5 from the numerator.

$\int\frac{\sqrt{5}\sqrt{\frac35-x^2}}{x^2}dx$

[solid kiln]

GoldBarley

[serene crane]

is it fair to say i push the sqrt 5 outside of the integral

since it’s being miktipled

multiplied

[midnight vessel]

Yea sure

$\int{cx~dx}=c\int{x~dx}$

[solid kiln]

GoldBarley

[serene crane]

wrong sadly

whered i go wrong

[midnight vessel]

Maybe simplify it?

[serene crane]

oh wait

[plush vigil]

the sqrt5

multiplies to sin^-1 too

[serene crane]

yeah i figured that

and

[plush vigil]

oh no

inside sin^-1, its x/a

here a is √3/√5

you did a^2

[serene crane]

ah

youre right

thank you

hold on

lemme try this

what would be considered u?

wait

stupid me

i tried this

[midnight vessel]

√25 = 5

√9 = 3

[plush vigil]

you divided by a instead of u

[serene crane]

i factored the 25 out

oh my

thank you

[midnight vessel]

u=x²

[serene crane]

I keep forgetting to do that

Thank you

onto the last

this one is something

make u the inside?

oh

[midnight vessel]

Yea

[serene crane]

gotta complete sqwuare first

[midnight vessel]

Not necessarily, you can do it in a later step as well.

[serene crane]

just for the aske of the problem i have to do it now but i do want the otehr way explaiend

[midnight vessel]

Oh

[serene crane]

confused on this part now

[midnight vessel]

Lemme also solve it rq

[split chasm]

that step is distribution

[serene crane]

the red box?

[split chasm]

(p - q) * r = ?

[serene crane]

oh

[split chasm]

the box marked with x, yes

[serene crane]

oh i think

i got it now

no i dont

lol

[midnight vessel]

I think you just repeat what you wrote in the last box for the next box as well.

Then finally write the solved expression in terms of u.

[serene crane]

id guess

youd put the two things together

[midnight vessel]

Yea, in terms of x.

[serene crane]

oh thats confusing the hell outta me rn

[midnight vessel]

u=x+4

Just substitute that in place of u.

[serene crane]

oh youre right

okay thank you all

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[little egret]

Can i pls have skme help with thsi question

(2-6i)z^2 - 2z^2

How do i do

[trim lichen]

well what is 2 - 6i - 2?

[little egret]

-6i

[trim lichen]

the fact that the coefficients are complex doesn't really change any of the underlying process.

you were able to do the subtraction just fine right now.

[little egret]

Wat

[trim lichen]

$(2-6i)z^2 - 2z^2 = -6iz^2$ ...

[solid kiln]

ann.in.a.teacup

[little egret]

Oh i didnt know u could expand it lol

Ok thx

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[fossil cobalt]

can someone guide me

[trim lichen]

it means you are going backwards

relative to whichever direction is chosen to run the x axis along

[trim lichen]

on a speed vs time graph, area represents distance

[round imp]

Velocity is vector

Depends on direction

[trim lichen]

well in this case velocity would be the correct word to use

[fossil cobalt]

yes i know this

but they say the distance between them

so i got confused

[trim lichen]

find the position of each car at time 3t

using the graph

brb

[fossil cobalt]

how if time = 0 , it travels a distance

the question probably means velocity by speed

it is a prep book

so its not accurate

[trim lichen]

i have a hunch they meant from 0 to t

the question is rife with typos

[fossil cobalt]

probably

wait

let me show u the answe

because i didnt understand it

where did the t/2 come from

[carmine skiff]

it's a triangle

[fossil cobalt]

where did the d come from

[trim lichen]

should have been v for sure

this book looks like hot garbage

[fossil cobalt]

it is

ok so velocity * t/2 = x

what do i do next?

i find the area under the curve between 2t-3t of K/

?

[trim lichen]

actually theres an easier way

you just find the area between the velocity curves for K and L, and not from 2t to 3t but from 0 to 3t

[fossil cobalt]

let me try to understand what you did

how did we get 2x?

it is a square

t * v

oh so if t/2 = x

then t = x

?

[carmine skiff]

the book defines the triangle as x

[fossil cobalt]

and then we subtract

(x+2x+x+x+2x) - (2x+x)

4x

[carmine skiff]

you don't subtract though

[fossil cobalt]

why

we have negative area

[carmine skiff]

the part underneath is negative

[fossil cobalt]

oh they want the distance between them

so -(-

10x

[trim lichen]

yes

[fossil cobalt]

thanks

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[brisk forum]

In Poisson Distribution, when modelling it, events have to occur singly in time. My question is what does this mean?

[brisk forum]

When patients arrive at a rate of 6 per hour , does singly mean that these patients just dont arrive at the exact time which is unlikely isn't it because patients arriving at the same time can occur

[marble wharf]

well thats always a problem with these models. they dont necessarily completely reflect reality

but if you only have 6 patients an hour then it is quite likely that probably they wont arrive at the same time

[brisk forum]

Then how can you say that it occurs singly?

[brisk forum]

I get that

[marble wharf]

its just an extra assumption you are placing on the situation

to be able to model it

[brisk forum]

So my take on it that its about whether the patients arrive at different times within the hour, is that correct?

[marble wharf]

yes

[brisk forum]

Okay thanks for clearing that up a bit

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[proud thistle]

i'm stucked

[shell apex]

show ur working out? which part are you stuck on?

[proud thistle]

i dont know where to begin to attack the problem

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[carmine skiff]

You can try using the square triangle that contains bah

And isoceles triangle that contains cao

They have the same arc

[proud thistle]

ok thanks i try

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[pliant lava]

Pls Explain how they got the two numerators equal

In zhe bottom most equation

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[pliant lava]

<@&286206848099549185>

If there is something I've not explained properly about this problem then please say that, I'll do my best to explain

<@&286206848099549185>

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[wraith hinge]

why does

[wraith hinge]

plugging in

15,-5

give u the other points

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