help-28

[smoky wing]

And it should be 40 at the end not 50

[twin heron]

Oh, youre right

Thanks!

[smoky wing]

[twin heron]

How do you write it algebraically?
L×1.4=t^2×1.4

Or something

Should i include k or leave it out

[smoky wing]

$k\sqrt{1.4L} = \sqrt{1.4} t$

[twin heron]

For better convincy

[glossy valve]

Pure

[twin heron]

Yea, its same as what i wrote

Just taking it one step at a time

[smoky wing]

[twin heron]

I think?

No, not really

[smoky wing]

Kinda

[twin heron]

Alright so this probably is it

Actually one last question

How do i write that I have space

How do i take neat notes

Society unfortunately demands me to follow these handwriting norms to succeed

When solving equations i take up a lot of space

[smoky wing]

I really don’t know how to answer that sorry I mean im Similar to you I guess

Like this is just something I wrote earlier

Takes quite a bit of space

[twin heron]

Yea

Well thanks

Farewell mr pure

Thanks for help once again

[smoky wing]

Np

I’ll close this?

[twin heron]

I saw you typing so i was waiting for you to finish

[smoky wing]

Yeah I gave up on typing whatever I was trying to type

Lol anyway

.close

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[dire laurel]

so 3z^2 - 9z - 2z + 6 = 3z(z - 3) - 2z + 6

[dire laurel]

3z^3 and -9z share a common factor of 3z, but how is that interacting to produce 3z(z-3)?

[clear lily]

you can factor out 3z

[dire laurel]

so 3z^2 loses a 3z which makes it 3z

if -9z loses a factor of 3z doesnt that make it -3?

why is it (z-3)?

[hot herald]

3z^2/(3z) isn't 3z

[dire laurel]

?

[kind jay]

why would u think it is?

[dire laurel]

im completely lost in quadratic theory

[kind jay]

has nothing to do with that lol

how would u simplify (3z²)/3z

[dire laurel]

1z^2???

[hot herald]

no

[dire laurel]

no clue tbh

[hot herald]

clearly show work in your attempt to simplify
$$\frac{3z^2}{3z}$$
don't just write a single expression the we'd have no idea how you got

[glossy valve]

ℝamonov

[dire laurel]

9z / 3z ?

[hot herald]

no

how are you getting that

[dire laurel]

damn i really dunno sorry

[hot herald]

don't rush

[dire laurel]

gotta dip sorry class starting

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[kind jay]

watch this when ur back https://www.youtube.com/watch?v=LkhPRz7Hocg

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[full forum]

Please don't occupy multiple help channels.

[wanton flint]

My answer is s=517.856mm looking to verify

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[wanton flint]

fuckit

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[proper haven]

If a cannonball is launched at a speed of 220m/s and at a angle of 45° above the horizontal, how do you find its maximum height?

[cinder flower]

u^2/2g

[proper haven]

what does u represent? also g is gravity right

[cinder flower]

initial velocity

220

g is gravity

[proper haven]

ah ty

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[bronze dirge]

how do i start this?

any1?

;-;

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[turbid imp]

With a water pump that consumes 2 CV of power, you want to fill a 200 m tank; whose
edge is located 3 m above the water level. If the efficiency of the motor is 80%, calculate the time
which will take time to fill.

[onyx glen]

what's CV?

[turbid imp]

Horse power

[onyx glen]

,w 2 horsepower to watt

[turbid imp]

I think the fórmula is P = E/t.

[glossy valve]

Results provided by WolframAlpha

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Requested by Ann#0413

[onyx glen]

also what's a "200 m tank"?

[turbid imp]

M^3

The translator xd

[onyx glen]

ok so the tank has a capacity of 200 m^3 and its edge is located 3 meters above the water level

so that means you will need to lift 200 cubic meters of water up by 3 meters

calculate the amount of work needed to do that

[turbid imp]

How

[onyx glen]

it's just gravitational potential energy

so mass * g * height

and then calculate the time taken to do that work for an ideal 2 hp pump, and divide by the 80% efficiency...

[turbid imp]

But how can i get the time

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[torn jolt]

how is (-4)^4 positive but -4^4 is negative? are they not the same thing?

[light sonnet]

No

[scenic wren]

with -4^4 you likely mean -(4^4)

[smoky wing]

$(-4)^4 = (-4) \times (-4) \times (-4) \times (-4)$

[glossy valve]

Pure

[smoky wing]

$-4^4 = -(4^4) = -(4 \times 4 \times 4 \times 4)$

[glossy valve]

Pure

[torn jolt]

i don’t really get it? is there like an explanation why i always thought they were the same

[smoky wing]

$-4^4 = -1 \times 4^4$

[glossy valve]

Pure

[torn jolt]

that makes sense kinda

thanknyou

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[smoky wing]

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[hollow zenith]

My prof just wrote this down without explanation how do I know to choose N > 1/ε

[fast peak]

well you try out some stuff and see that it works

you know n>N, so 1/n < 1/N

so you want 1/N < eps

and that is clearly satisfied for N > 1/eps

[hollow zenith]

yeah makes sense ty

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[winter ember]

https://imgur.com/vpoR8ju

[winter ember]

How do I solve the second one?

[clear lily]

$\sum_{i = 1}^{n} = \sum_{i = 1}^{n-a} + \sum_{i = a + 1}^{n}$

[glossy valve]

Gijs

[clear lily]

just rearrange this

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[winter ember]

@clear lily what's a?

would that be 4 in this case?

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[glass schooner]

I have a trigonometry (high school) problem that I've been trying to solve for 2 days.

[glass schooner]

I need to find angle D in the picture

[winter ember]

I believe the two angles on the right add up to 180

but I could be very wrong

[dire flame]

I think A is the same as C

So split in half

You have triangle

[winter ember]

i'm not too sure because the side lengths arent the same

[dire flame]

Oh I didn’t see that

[spark valve]

does b+c = 180 or na

[dire flame]

Not sure lol

Lemme see for a sec

[glass schooner]

Here what I've tried

[dire flame]

Oh bruh

Drop a line

[glass schooner]

getting the angle from the top triangle would give me the answer that I need

[dire flame]

Drop a line from C down

[dire flame]

No I thought D is the angle on the bottom right

The triangle just gives part of it no?

[glass schooner]

yeah but breaking it in two triangles seems like the way to go

[floral lava]

use the al-kashi law of cosines in triangle BCD

you'll then get angle BDC

[glass schooner]

Thanks @floral lava will try that

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[rotund blaze]

tan(pie/17)+i

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[rotund blaze]

how do i put tan(π/17)+i in the trig form

[grand yacht]

you mean pi?

π≠🥧

[rotund blaze]

yea XD

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[prisma basin]

Can anyone help me with this I have no clue how to do this

[prisma basin]

I know that secant line is between 2 points so maybe a line through A and B

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[grand yacht]

yeah draw a line through a,f(a) and b,f(b)

and then mvt

[prisma basin]

What is meant by a,f(a)?

And b,f(b)

[grand yacht]

the point (a,f(a))

and the point (b,f(b))

[prisma basin]

I got that but where would that be?

[grand yacht]

the point on the graph where x=a or x=b

[prisma basin]

Uhh I don’t really get it

But I just wanna clear something up first cuz I’m confused about f(x) stufff

So in f(x), x is the x axis value

And f(x) is the y value of that x

So for example in the equation x+2

If x value was 3, then the y value would be f(3)

So the point would be (3,f(3))

And f(3) is 5

So point (3,5)

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[spiral grail]

Quick question

[spiral grail]

A line and a plane that don’t intersect are skew T or F

I think it’s T but not sure

Wait is it sometimes true

[smoky wing]

Well if it’s true sometimes then it’s false in General

[spiral grail]

What

It says either true false or sometimes true

That doesn’t help at all….

Tysm

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[vapid meadow]

how would I solve this

[vapid meadow]

I know the answer has to be a or b

because I got 1 for the hydrogen indez

indesx

or do I sub both questions into the IHD formula

[woven sand]

This might be a question better suited to the chemistry discord in #old-network

If anyone here does happen to know then you might get answers though

[vapid meadow]

ok sounds good

should I still keep my question here?

[woven sand]

If you'd like

I'd certainly help if I could but I haven't done chem since before sliced bread was the best thing

[vapid meadow]

oh

I'm kinda bad at chem too

Just a little confused on how to approach it

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[vivid forge]

just need help on population when 5000, I know how to do it graphically but I am lost doing it algebraicaly

it has been like 6 years since I have done alg1

[clear lily]

2650e^(0.035t) = 5000

solve for t

then t + 2000 is your answer

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@vivid forge Has your question been resolved?

[vivid forge]

I forget how you solve for t

Sorry, @clear lily it’s been so long

[clear lily]

aight well

divide by 2650

such that you get

e^(0.035t) = 5000/2650

then take the natural logarithm of both sides

ln(e^(0.035t)) = ln(5000/2650)

e and ln cancel eachother out

so you get

0.035t = ln(5000/2650)

multiply by 1/0.035

which is about 28

t = ln(5000/2650) / 0.035

t is roughly 18.14

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[fleet cedar]

but that doesnt seem nearly rigorous enough

[rare dock]

what's the "for all k" for?

[fleet cedar]

for all k in the group G

im taking an arbitrary element k of the group

then im trying to show that since hg = k, and k is an arbitrary element

hg must cover the entire group for any choice of g

i feel its too handwavy tho

[rare dock]

sounds kinda backwards

if you're saying "since hg = k" at the beginning

[fleet cedar]

idk how to phrase the next part

[rare dock]

how do you know that?

[fleet cedar]

h has an inverse since its a group element, and so i'm defining g to be the product of h inverse and k

[rare dock]

ok you can make that work

[fleet cedar]

did u have a better way in mind

i feel like my way is too clunky

[rare dock]

my way is kinda the same

it's just

let $a\in G$. Then $a=hh^{-1}a\in {hg|g\in G}$

because h^{-1}a is in G

ok actually I'll rename the things

[glossy valve]

layla💜

[rare dock]

does that make sense?

[fleet cedar]

okay yeah i see it

thank you

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[rare dock]

np ^-^

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[cedar fern]

Not too sure how to do this, please help.

[hot herald]

differentiate h(x), applying chain rule

[cedar fern]

ahh ok

thank you

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[grand parcel]

This one I've been struggling with. My thought was to run a simulation of 1000 spins, take the average and standard deviation, then determine the average needed to end with 4 or less after 12 spins, subtract that average by the sample average, divided by the sample std deviation, then use that as the z score to determine the probability?

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@grand parcel Has your question been resolved?

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@grand parcel Has your question been resolved?

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@grand parcel Has your question been resolved?

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@grand parcel Has your question been resolved?

[fallen bone]

If you can use computers you can just compute the random variable object directly

[grand parcel]

Wouldn’t that just give me the expected value after 12 spins? How would I instead find the probability to find the probability of getting 4 or less after 12 spins instead?

[fallen bone]

,calc 7^12

[glossy valve]

Result:

1.3841287201e+10

[fallen bone]

I meant iterating through the 10 billion possibilities

[grand parcel]

I could manually build that but automating that is not something in my wheel house. Looking to see if there’s a formula for this or if what I did is correct.

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[torn jolt]

31%

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[grand parcel]

Can you tell me what you did to get that number? I have many other of these so I'm trying to determine how to do them for myself.

[torn jolt]

What did you get?

[fallen bone]

SAA 'works'

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[torn jolt]

.close

[fallen bone]

@grand parcel Don't occupy a single channel for >3 days

[upbeat island]

.close

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[cunning portal]

If x^3 is divisible by 3 then x is divisible by 3, proof by contraposition.

[cunning portal]

No clue how to start, i no longer remember modules

I have written the contraposition statement as ¬x|3=>¬x^3|3

[fast peak]

if x is not divisible by 3, then it can be written as either 3k+1 or 3k+2

should be 3|x and 3|x^3 but yes

[cunning portal]

Oh yeah sry

Thank you

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[void arch]

The determinant is antisymmetric

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[void arch]

It states that det(u,v)=-det(v,u)

But isn’t antisymmetric when if a proposition R, if both R(a,b) and R(b,a) are true then a=b

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[elfin flicker]

Can anyone help me with a quation

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[elfin flicker]

$x_1 + x_2 + x_3 + x_4 + x_5 = 9$ if $ 0<x<3$

[glossy valve]

afeAlway

[elfin flicker]

I have solved it but I'm struggling with the "inclusion and exclusion" part.

The total amount of combinations is C(13,4) and we have a case called $c_1$ when we guarantee that at least one x exceeds 3. $c_1 = C(9,4) * C(5,1)$, we multiply it by C(5,1) ways since we have 5 ways of choosing which x that is gonna be guaranteed to exceed 3. Then I have another case I call $c_2$ which is the case when we guarantee that at least two x's exceed 3. We then have only 1 left to distribute to the x's. $c_1 = C(5,1) * C(5,2) $ But the problem is I don't understand if I should add or subtract

[glossy valve]

afeAlway

[elfin flicker]

My answer shit adds it to the total after subtracting c1. I kinda understand it is because I have counted one case twice somehow in c1 which means I need to add c2 to make up for it but I don't really understand this.

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[torn moon]

how did the first series become the second series are they the same?

the ones circled

[shrewd hamlet]

yes theyre the same

[torn moon]

how?

[shrewd hamlet]

(a^n) * (b^n) = (ab)^n

just a basic exponent rule

[torn moon]

ohh

ty!

[shrewd hamlet]

np

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[tepid solar]

Hi, I got the last question correct but I couldnt figure out the x's I tried answering -1/8 and 1/8

[balmy peak]

geometric series converge when -1 < r < 1

whats ur working out ?

@tepid solar

[tepid solar]

I'm here

hmm

[balmy peak]

wait did u enter (-1/8, 1/8)

?

[tepid solar]

yes

[balmy peak]

thats wrong?

did u use interval notation

[tepid solar]

i believe so

i dont get why its wrong

[balmy peak]

im pretty sure its right

bec r = 8x

-1 < 8x < 1

-1/8 < x < 1/8

(-1/8, 1/8)

did u use closed brackets to indicate its not inclusive ?

[tepid solar]

I DID NOT

🤣

[balmy peak]

lol

[tepid solar]

I WROTE IT WITHOUT BRACKETS

This is embarassing

[balmy peak]

haha all good

[tepid solar]

thank you for helping me thru the tough times

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[spiral bronze]

can anyone explain why differentiating V=1/3pi r^2 h (volume of circular cone) with respect to time (t) is equal to this

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@spiral bronze Has your question been resolved?

[devout inlet]

But r and h depends on t. Use the product rule. Looks like the right term of the right hand side should be 1/3 pi r^2 dh/dt.

[spiral bronze]

youre right the solution should be $\frac{dV}{dt}=\frac{2}{3}\pi r\ \frac{dr}{dt}h+\frac{1}{3}\pi r^{2}\ \frac{dh}{dt}$

[glossy valve]

Zyraxion

[devout inlet]

That looks right now.

[spiral bronze]

yeah i got it now i actually used the correct answer

[spiral bronze]

ty

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[pliant walrus]

What formula would be used to calculate this without just counting? Asking to calculate something in game
“The price to buy 1 cow is $101360, the price to buy a second cow is $101380. The price increases by $20 each time, what is the total price of 234 cows?”

[hot herald]

look up arithmetic series

[pliant walrus]

Thx

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[gleaming fox]

I’m stuck that I don’t know what to do next

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[gleaming fox]

<@&286206848099549185>

[gleaming fox]

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[upbeat kiln]

if i input x and y, will they be in the colored region?

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Please don't occupy multiple help channels.

[hallow cedar]

? what do you mean

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[tidal reef]

The sum of three consecutive numbers in a geometric sequence is 35, the product of the first and the third term is 100. The ratio is ...

im confused. "the sum of three consecutive numbers" does it mean $a_{n} + a_{n+1} + a_{n+2} = 35$ or can i just assume it's $a_1 + a_2 +a_3$?

[glossy valve]

nichoals

[tidal reef]

if the former's the case, then how do i find the ratio?

[torn jolt]

i thinki it should be $a_n + a_{n+1} + a_{n+2}$

[glossy valve]

TheGeneralisedPerson

[tidal reef]

then how would i go through the problem?

i got $ar=10$ from the fact that the product of $a_1\times a_3=100$

[glossy valve]

nichoals

[tidal reef]

<@&286206848099549185> please :(

[abstract finch]

What is your query?

And why are you so desperate?

[tidal reef]

how would i solve this if the sum of three consecutive numbers is $a_n + a_{n+1} + a_{n+2}$?

The sum of three consecutive numbers in a geometric sequence is 35, the product of the first and the third term is 100. The ratio is ...

[glossy valve]

nichoals

[tidal reef]

i mean if i assume $a_1 + a_2 + a_3 = 35$ then it'd be easy, r = 2 or 1/2

[glossy valve]

nichoals

[abstract finch]

Does this concern Nth Term?

[tidal reef]

im sorry, i dont understand what you mean by that

[abstract finch]

Nth term is a broad subject and all you are saying right now is related to what you are saying.

Anyway...

I think this involves substitution of certain values.

I am no expert in this, however I have done some research over this subject

What is the ratio?

[tidal reef]

$a_n+a_{n+1}+a_{n+2}=ar^{n-1}+ar^n+ar^{n+1}=35$
\$ar^n(r^{-1}+1+r)=35$
\since $ar=10\implies a=10/r$
\$\frac{10r^n}{r}\left(r+1+\frac{1}{r}\right)=35$

am i heading in the correct direction?

oh wait hold on i did something wrong with the factoring

[abstract finch]

Are you getting it now?

Because I am not

[glossy valve]

nichoals

[abstract finch]

What is the full question

[tidal reef]

The sum of three consecutive numbers in a geometric sequence is 35, the product of the first and the third term is 100. The ratio is ...

that's the full question

[abstract finch]

ok

you can list the multiplier and multiplicand until it fits with the relation.

Or is this an algebraic situation?

[tidal reef]

i think it's more of an algebraic situation

[abstract finch]

Ok.

Substitution still works if you are stuck.

I don't think I am helping.

[tidal reef]

$10r^{n-1}\left(\frac{r^2+r+1}{r}\right)=35$
\$10r^{n-2}(r^2+r+1)=35$
\$10r^n + 10r^{n-1} + 10r^{n-2}=35$

[glossy valve]

nichoals

[tidal reef]

i think im heading further away from the solution

[tidal reef]

hmm.. ok i'll try it first

[abstract finch]

Can you try solving it using a power series?

[tidal reef]

i havent covered that with my teacher yet...

[abstract finch]

Ok.

Should I go over it?

[tidal reef]

if it's okay with you... then yes please

[abstract finch]

ok

So..

A power series is just a normal series that goes up in indices of x

And to shorten that you can use summation if you know what that is.

(if not then don't bother)

for example

,w sum(x^n)

[glossy valve]

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[abstract finch]

If it blows of to infinity then it is an infinite sum.

[tidal reef]

im familiar with the summation notation yes

[abstract finch]

Ok

[tidal reef]

aah i see

[abstract finch]

Now if you can somehow diverge the value to be around 35 then you have hit the answer.

All you need to do is then see the intersection between the graphs of the two values.

[tidal reef]

wait

what if

,w 10r^n + 10r^(n-1) + 10r^(n-2) = 35

aw

[glossy valve]

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Requested by nichoals#5812

[topaz valley]

,w 10/r + 10r = 25

[glossy valve]

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[topaz valley]

,w 10/r + 10r = 45

[glossy valve]

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@tidal reef Has your question been resolved?

[tidal reef]

i got that earlier by assuming the three consecutive terms were a_1, a_2, and a_3

r = 1/2 or r = 2

[topaz valley]

in a geometric sequence

every term is the geometric mean of its two neighbouring terms up to sign

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[steep zephyr]

what to do next?😩

[summer echo]

Foil the squared term on the left

[safe trench]

I think f(2x) is wrong

[summer echo]

You're missing a 2 in the right exponent

@steep zephyr

[steep zephyr]

ok. then i have to foil the square term right? let me try first

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@steep zephyr Has your question been resolved?

[steep zephyr]

do i need to find the x value?

[steep zephyr]

is this correct? i dont know how to continue solving it

<@&286206848099549185>

[earnest perch]

$\frac{1}{2} 10^x$ is not equal to $5^x$

[glossy valve]

Intrer

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[runic bloom]

An open rectangular box has a square base and a volume of 500 cubic inches. What dimensions minimize the amount of cardboard needed to make the box?

[runic bloom]

damn same time lmao

[hallow cedar]

so, to minimize the amount of cardboard, this would be the same as minimizing the surface area

[runic bloom]

so I made 2 equations, one is the volume, one is the surface area:
x^2 * y = 500
2x^2 + 4xy = ?

and then plugged into the second one:
2x^2 + 4x(500/x^2)
= 2x^2 + 2000x^(-1)

and then set the derivative of that to 0 to get the minimum (I think you're supposed to do that?)
4x - 2000 / x^2 = 0
4x^3 - 2000 = 0
500 = x^3
x = sqrt(5000)

but that's wrong so idk

I'm pretty new to optimization so I'm kinda guessing lol

[torn jolt]

what the!?

i did not realize that

[hallow cedar]

luanr moon, ask your quesiton in another channel :)

[torn jolt]

lol

[hallow cedar]

sir you did everything perfectly as far as I can tell

however

you didn't read the question properly D:

[runic bloom]

[hallow cedar]

lmao

[runic bloom]

I don't get it

I get that it's an open box but then?

[hallow cedar]

so the surface area

would be x^2 + 4xy

not 2x^2 + 4xy

[runic bloom]

OHH

🤦‍♂️

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[runic bloom]

i still needa solve one sec

[runic bloom]

thanks!

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[torn jolt]

Could someone check my answer please?

I'm not confident I set up the integral right.

How do you even know when to use double or triple integrals to solve a problem like this?

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@torn jolt Has your question been resolved?

[untold tree]

bc

'['

bd

dfgb

fd

df

fdb

\bfd

sfddrhbt j sh

h

[torn jolt]

<@&286206848099549185>

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[pure gate]

I got the result of p= 16 but im not entirely sure if its right can someone help

[rocky vale]

It's not

Take two steps. Add 6 to both sides, get a new equation
Then multiply by 3 on both sides and get your solution

[pure gate]

I do 16x3 on the right? Sorry im confused

[rocky vale]

Yeah, after you add 6 what does your equation say?

[pure gate]

Do i multiply anything on the left?

[rocky vale]

One thing at a time.

Add 6 to both sides. That's all

Then what does the equation say?

[pure gate]

P= 48? My teacher told us to make the top number on the left it’s opposite so i made -6 +6 then they said add that to the other side and add it all up

[rocky vale]

48 is the solution, yeah. But it feels like you don't understand how you got there

[pure gate]

By multiplying the right side by 3

[rocky vale]

by multiplying both sides by 3

[pure gate]

Im confused on the both sides part bc all i do on the left side is make the number on top opposite of itself

[rocky vale]

$$\frac{p}{3} - 6 = 10$$
Add 6 to both sides
$$\frac{p}{3} = 16$$
Multiply both sides by 3
$$p = 48$$

[glossy valve]

tatpoj

[pure gate]

Ohhhhh i get it now i think

I got p=16 before hand

So i just multiply that by 3

[rocky vale]

Adding 6 to the left cancels with the -6
Multiplying by 3 on the left cancels the 3 in the denominator

And then whatever you do on the left, you also have to do on the right

[pure gate]

Thanks for the help!

Im sure ill be in here often

[rocky vale]

👍 No problem

[pure gate]

One last thing

Is this correct

[rocky vale]

Well, if you take your original equation, and plug 4 in for x, is it true?

[pure gate]

One sec

[pure gate]

Unless im wrong i think its right because 2x2 is 4

[rocky vale]

You're right

But I'm trying to show you how you can check these yourself

If you take your original equation and put in 4 for x, you get

$\frac{4}{2} + 3 = 5$

[glossy valve]

tatpoj

[rocky vale]

Which is true

So yeah, it's correct 👍

[pure gate]

Tysm

Probably just helped me with my quiz

[rocky vale]

Sure thing, no problem

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[short quest]

hey

[short quest]

anyone here can help pre calc

[uneven crest]

Yes?

[uneven crest]

I can.

[short quest]

9 I got A god it

for it *

im sure it’s correct and if u could help me see if it’s good i would be happy

and 10 i got E

[uneven crest]

How did you solve No. 9?

[short quest]

forget ab 9

im sure it’s A

but 10 can u check for 10

10 i literly have no clue how to do it

[uneven crest]

Ok, so how did you set up the formula?

[short quest]

isk how

can u show me how to do it :/

[uneven crest]

Ok, so.

[short quest]

ya

[uneven crest]

Do 776,000e^-1.5%t = N

Solve for t in years.

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[marble magnet]

Is there a name to pulling terms out like this? Or a tutorial on it or something? (ignore the 15n)

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[marble magnet]

<@&286206848099549185> 🥺

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@marble magnet Has your question been resolved?

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@marble magnet Has your question been resolved?

[marble magnet]

<@&286206848099549185> pls im so lost 💀 discrete midterm tomorrow

[plush egret]

looks like just exponents? what do you mean ignore

its not really clear what youre doing

[marble magnet]

cause it's not included in the right side of the equation

[plush egret]

whats in the ...

[marble magnet]

I'm jsut confused on how one goes from 4^n - 1 to (4-1)(4^(n-1)...1)

i dont know

The 4-1 is somehow pulled out

[plush egret]

i mean if its even powers, you can factor

just as difference of squares

I'm not sure what youve written here is true in general

could be wrong though

[marble magnet]

so if I had (k^n - c) how do I pull out a (k-c)

[plush egret]

idk maybe itd help to devise a case where its impossible

[marble magnet]

but it's some rule

I don't think there's a case where it's impossible

[plush egret]

how about

$2^6-5=59$

[glossy valve]

jan Niku

[plush egret]

but 2-5 is -3

and -3 doesnt divide 59

[marble magnet]

true...

My prof used it as part of a direct proof to prove this ^

and that's how he pulled out a 3...

[plush egret]

hrm

[plush egret]

well, if its part of the inductive step

he would have assumed 9 divided the 4^n+15n-1

[marble magnet]

He said to know how to prove directly and via induction

maybe I just leave it at induction and find out later 😅

[plush egret]

i dont know that the factoring you showed is true in general

there is maybe some piece of context im not seeing because im tired

[marble magnet]

that is step 1 of the proof, so there's not really prior context... I'll just hope I don't need something like that tomorrow lol

[plush egret]

it seems like something youd want to do with induction

[marble magnet]

yeah that's like the 1 way I know how

[late fable]

I would call it factoring

[marble magnet]

but my man did it like that

[plush egret]

if you are not super tired can u explain 💀

[late fable]

That’s just the general name of it

[plush egret]

no i mean in this instance

[marble magnet]

yeah it's factoring out a three, but like

[plush egret]

👀

[late fable]

When you factor expressions you find factors of the expression

I’m tbis case

[plush egret]

[late fable]

You did factor a 3

[plush egret]

but why is 3 a factor

[late fable]

You factored out a (4-1)

[marble magnet]

how though

[plush egret]

how do you know that 3 is a factor

[late fable]

But it multiplied by something else = the expression

[plush egret]

thats the question

its not

[late fable]

Think about a factor of a number

[plush egret]

its not always a factor

wait

is it

[late fable]

It is

[plush egret]

4^k-1?

3 is always a factor?

[late fable]

A factor of a number is a number that evenly divided it

[plush egret]

this is not what the question is

[late fable]

In this case 3 is a factor since it evenly divided 4^k-1

[plush egret]

theyre in discrete math they know what factors are

why does 3 always divide 4^k-1

i guess u just do that as a lemma

[marble magnet]

^ ^ ^^

[plush egret]

[marble magnet]

lol bet

[late fable]

Oh

That’s because of factoring

[marble magnet]

🤦‍♀️

[late fable]

(4-1)(4^(k-1) + 4^(k-2)…)

That’s literally what the expression is equal to

Think about expanding the expression

That I just showed

You’d get all the powers of 4 from 4^k to 4

[plush egret]

its binomial

wait

thats not the word i wanna use

[late fable]

You’d get - all the powers of four from 4^(k-1) to 1

So they cancel out

[plush egret]

ah, yea

youre right

[late fable]

And you’re left with 4^k-1

[marble magnet]

what if it were like (6^k - 3)

[plush egret]

factor a 3

3(2^k-1)

[marble magnet]

bad example

[plush egret]

yea

[marble magnet]

like (7^k - 2)

[plush egret]

factor a 2

[marble magnet]

what would factoring out a 5 look like

[late fable]

No

You can only do it with 1

Because 1 is a power

[marble magnet]

ooooooooooooooooh

[plush egret]

well

since this person is smart enough to solve

im gonna go to bed

[late fable]

1 is a really nice number

[plush egret]

thank u for the help boibored

and good luck tomorrow @marble magnet

[late fable]

I could be wrong, don’t trust me

[plush egret]

idk it seems reasonable to me

[marble magnet]

i trust you more than me

thanks guys ❣️

[late fable]

I don’t think you can always, I but me wrong though

The reason it doesn’t always work

[marble magnet]

c having to be 1 sounds correct

[late fable]

is because if you had (k-c)(k^(n-1))

You’d get k^n-ck^(n-1)

[marble magnet]

mhm

[late fable]

Wait bad example

[marble magnet]

lol

[late fable]

Ok I got it

Imagine this

[marble magnet]

imagining

[late fable]

(k-c)(k^(n-1) + k^(n-2)…)

You’d get

k^n -ck^(n-1) + k^(n-1) -ck^(n-2)…

So in order for it to cancel out

[marble magnet]

ahah

I see

[late fable]

You’d have to have k^(n-1) - ck^(n-1) = 0

C = 1

Hopefully I’m right

[marble magnet]

yeyeye

looks right

[late fable]

Looks right

You should probably wait till you get confirmation from a more experienced person

[marble magnet]

midterm tomorrow morning so you're the best i got lol

I'll stick to induction, and keep this in my back pocket

but i feel like you're right

[late fable]

What class is this

[marble magnet]

dsicrete

discrete

[late fable]

They don’t offer that in my school

[marble magnet]

really?

hmm

[late fable]

They only offer to Calc bc

[marble magnet]

ohhhh

I'm in college

[late fable]

oh

Lol

My bad

[marble magnet]

lol

proof that you dont wanna forget the basics lol

[late fable]

This isn’t basic

[marble magnet]

fundamental then

[late fable]

1% of my class could solve this

Ok maybe 2%

[marble magnet]

lol

stay in that 1% then 😅

2%*

[late fable]

Math isn’t the biggest

Subject of competition

[marble magnet]

true

but maybe you'll need it later

anyway, i gtg, so thank you @late fable and @plush egret

[late fable]

No

No

Np

[marble magnet]

YES

.close

[late fable]

Sorry autocorrect

[full forum]

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[late fable]

Good luck

[marble magnet]

thanks

🫡

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[worn cedar]

am i missing soemthing?

[onyx glen]

symmetric matrix A

[worn cedar]

yk how to do that

[onyx glen]

do what

[worn cedar]

make a symmetric matrix

[onyx glen]

do you know what a symmetric matrix is?

[worn cedar]

yea

[onyx glen]

ok then tell me

what is a symmetric matrix?

[worn cedar]

A^T = A

[onyx glen]

right

[worn cedar]

idk how to make it tho

[onyx glen]

or, speaking in terms of entries, $A$ is symmetric if $a_{ij} = a_{ji}$ for all $i$ and $j$.

...the bot is down, apparently

a_ij = a_ji

this is what i meant

so for example you're told a_12 = 4, so this along with A being symmetric also means a_21 = 4

also i think you screwed up the entries you did put. a_12 means row 1 col 2, not row 2 col 1 as you wrote

[worn cedar]

wait so would this be correct

yes it is

thank

i also dont know how to do this

and this one as well

[onyx glen]

@worn cedar please stay in this channel and do not open any new ones

[onyx glen]

show part (a) from here.

[onyx glen]

right so you have found A^-1 great

notice that your linear system is actually Ax = b where b is the vector [1; 1; -4]

[worn cedar]

i dont understand the last part

[onyx glen]

$b = \bmqty{1 \ 1 \ -4}$

augh

it's a column vector with entries 1, 1 and -4 in that order

[glossy valve]

Ann

[onyx glen]

ah there we go

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[signal storm]

help

[onyx glen]

have you made any progress so far?

[signal storm]

eh

not rlly

[full forum]

@signal storm Has your question been resolved?

[signal storm]

**

<@&286206848099549185>

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@signal storm Has your question been resolved?

[onyx glen]

when you replace x with (x+2), by how much does the expression for A change?

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[torn jolt]

How do we Evaluate the expression

[atomic crater]

just plug those values in

so 9 is (1/5)(1/5-6/5)

[torn jolt]

wait

how do we do that..

WAIT WHAT..

[onyx glen]

is it the fractions that are scaring you?

or is it the concept of replacing variables with numbers

[torn jolt]

I don't get how do we put it together

[onyx glen]

put what together?

[torn jolt]

like this..

[onyx glen]

the problem says, "Let's say the letter x stands for the number 1/5 and the letter y stands for the number 6/5. What is x * (x-y)?"

[torn jolt]

umm..

1/5 ( 1/5 - 6/5)

?..

[onyx glen]

yes

and now you have an arithmetic problem

it's gonna involve some fractions and some negative numbers, but it is an arithmetic problem all the same.

[torn jolt]

I mean how does this go in to question 11?

[torn jolt]

It looks so different..

[onyx glen]

what do you mean?

these two questions are independent of each other.

they're the same kind of question but they are not related in any way aside from that.

[torn jolt]

ohh

[onyx glen]

but it is still the same principle of "plug in the numbers where the letters are and do the arithmetic"

[torn jolt]

how do we plug it if it is that long..?

do we have to do something more?

[onyx glen]

you just do...? the length of the expression doesn't matter much

you simply replace all instances of b with the number you're told b stands for, and replace all instances of c with the number you're told c stands for...

the fact that the expression is a little bit longer does not make a difference

[torn jolt]

so its...4(2+1)2 - 6(7-b)4 ..?

[onyx glen]

4 * (2+1)^2 - 6 * (7-2)^4

^ for exponents when you're writing plain text

btu yes

but yes*

[torn jolt]

thanks!!!

.close

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[loud hull]

[r = \sqrt{x^2 + y^2}]
[5^2 = \sqrt{2^2 + x^2}]
[x=\sqrt{21}]

[csc = \frac{r}{y} = \frac {5}{\sqrt{21}} = \frac{5\sqrt{21}}{21}]

[loud hull]

What am I doing wrong here?

[humble chasm]

so cos theta is opposite over adjacent

[glossy valve]

dopediscorduser

[humble chasm]

since tan theta is less than 0

it must be sine is negative

sorry I mean adjacent over hypotenuse

LOL

[torn jolt]

no

[humble chasm]

cos is adjacent over hypotenus mb

xd

[torn jolt]

2nd quadrant

sin is positive

cos is negative

so tan is negative

[humble chasm]

really now?

[torn jolt]

so how is this true

[humble chasm]

cos theta = 2/5

[loud hull]

How did you figure out it was in Q2?

[hot herald]

its not in q2

[loud hull]

Oh

[torn jolt]

oh

[humble chasm]

not in q2

must be in q4

tan theta is negative

cos theta is positive

[loud hull]

Well how do you figure out what quadrant it's in?

[humble chasm]

cos (60) in quadrant 1 is 1/2

since 2/5 is less than 1/2

and greater than 0

[loud hull]

I was honestly hoping I just had the sign of my answer wrong, does this mean I did the math wrong somewhere?

[humble chasm]

it's either in q1 or q4

yeah the sign of your answer for

csc is wrong

because if tan is sin/cos be less than 0

and you KNOW cos is positive

[loud hull]

Right because sin is negative?

[humble chasm]

then sin is negative

and so is it's reciprocal

[loud hull]

Okay but wait

That's assuming it's in Q4

What's stopping it from being in Q1 and sin being positive?

[humble chasm]

Tan theta being negative

Assume that it was in q1

you know tan theta is negative

but that's impossible for tan to be negative in q1 because in q1 you have positive divided by positive

That's a contradiction

cos theta = 1/2 in q1 and q4

since tan is less than 0 you have a negative divided by a positive

give that cos is positive

it must be that sin is negative

therefore your answer lies in q4

@loud hull

[loud hull]

Ohh

Gotcha

Alright thanks!

.close

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