help-10

[rotund cloak]

I think I'm supposed to find the delta of each function separately and then take the min of those deltas, but I'm stuck with the epsilons

[obtuse pebble]

@rotund cloak Has your question been resolved?

[rotund cloak]

<@&286206848099549185>

[rotund cloak]

<@&286206848099549185>

[mighty geyser]

Well, remember, you can choose a different epsilon

if you add two epsilons together, you get 2epsilon

but if you add two epsilon/2 together...

[rotund cloak]

You get |f(x) + g(x) - 6| < epsilon

Which I understand, but I'm not sure where to go from there

[mighty geyser]

if epsilon > 0, then epsilon/2 > 0, and you can substitute epsilon for epsilon/2

[rotund cloak]

Would I then take the bound for delta? That is, take -epsilon/2 < x - 2 < epsilon/2?

[mighty geyser]

delta is a function of epsilon

so when you apply epsilon/2 instead, you get a different (typically smaller) value of delta

[rotund cloak]

Does that mean I'd have x - 2 < epsilon/2?

Aahhhh, I think I get it - should I replace epsilon with epsilon/2 and then find delta from min(delta1, delta2)?

[obtuse pebble]

@rotund cloak Has your question been resolved?

[novel knoll]

Wdym - you have to prove for all epsilon>0 that there exists a delta>0 such that…

You can’t just “replace epsilon with epsilon/2”

[obtuse pebble]

@rotund cloak Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

I understand this is a disguised quadratic and have reaarranged to get it all on one side

I also understand I need to take x as t^(1/3)

I got as far as factorising into (4x-3)(x+5)

but am not sure where to go from here

is it as easy as 3/4^3 and -5^3?

[obtuse pebble]

@timid silo Has your question been resolved?

[rugged kite]

don't forget the brackets, but yes

[timid silo]

.close

[obtuse pebble]

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[grave patio]

Could I get some help on these?

[obtuse pebble]

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[grave patio]

..?

.open

what.

[grave patio]

Also, with these equations, how could I solve them simultaneously? I’ve tried toying around with it but everything I try won’t work

Any ideas on where I should look first?

make the y = and solve that way, or factorise first

I assume make them equal first but it didn’t work

Figured out the simultaneous equations nvm

<@&286206848099549185>

[ancient jetty]

Hey @grave patio

You should substitute in your y value into the y values of the bottom equation

Then your equation should be easier to solve

[grave patio]

I’ve solved the bottom one, yeah that’s what I was doing but I messed it up the first time

I need help with the top two though

[finite kelp]

you've found x by solving the second one ?

well substitute x in the first one to find y

[muted flame]

Hm

Lemme show you

[muted flame]

$\left(x-3\right)^2+x\left(x-3\right)+4x=7$

[warm shale]

Captain_Mat01

[muted flame]

$2x^2-5x+9=7$

[warm shale]

Captain_Mat01

[muted flame]

$2x^2-5x+2=0$

[warm shale]

Captain_Mat01

[muted flame]

$\left(2x-4\right)\left(x-\frac{1}{2}\right)$

[warm shale]

Captain_Mat01

[muted flame]

$x=2,:x=\frac{1}{2}$

[warm shale]

Captain_Mat01

[muted flame]

@grave patio

[grave patio]

Uhh

I’m confused

oh right

I don’t need help for that one

I need help for the inequalities

[muted flame]

You said you did...

[grave patio]

As in, I solved the one at the bottom

It’s the two at the top I need help with

[muted flame]

For inequalities, just draw a graph and look at the regions to shade in

[grave patio]

That’s what I’m kinda confused on though

how do I get a graph from that

[muted flame]

This is for x>=-1

[grave patio]

it’s that simple?

[muted flame]

Yes, you are overthinking it

[unique solstice]

Sometimes

[grave patio]

ahhhh

what about for the ones that include y, just draw a line and shade in one side of it?

And sorry, final question here

if I complete the square with a coefficient, and let’s say I have 5(x+3/10)^2 > 4

Should I sqrt or divide by 5 first

[obtuse pebble]

@grave patio Has your question been resolved?

[muted flame]

Divide by 5 first

[grave patio]

Thanks

[grave patio]

.close

[obtuse pebble]

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[hoary spruce]

Hello, does anyone know the Lagrange method?

[hoary spruce]

ill post my work in a sec

[molten mason]

I can try to help, just learned it last month

[hoary spruce]

alright, my calcs are messy af

ah nvm im just looking for someone to confirm the results i got

but baseline is

i got x=13,142857...

and y=5,285714

[molten mason]

ok i'll try gimme a moment

[hoary spruce]

which adds up to 200x+600y=5799,

[molten mason]

I got x=8 and y=9

sorry y=7

and 200x + 600y = 5800

[hoary spruce]

huh

yeah i porbably got it wrong then

thank you

[molten mason]

i could be wrong

if you wanna discuss your calculations we can do that too

[hoary spruce]

sure

priv call?

[molten mason]

I'm actually at a meeting at work lmao, send me screenshots of your work

[hoary spruce]

alright, theyre suuper messy though

and in swedish

[molten mason]

i'll try

[hoary spruce]

.close

[obtuse pebble]

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[wooden trail]

I need help with the question at the top

[molten mason]

@wooden trail Can you find a right angled triangle inside the square that can help you find the side length of the square using pythagorean theorem?

[wooden trail]

Yes

I have found the right triangle

[molten mason]

send a screenshot of that drawing

[wooden trail]

Might take long because school wifi isn’t too good

[molten mason]

ok just trying to figure out where your triangle is, your triangle should have lengths that are easy to count on the grid

what are the sides of your triangle

[wooden trail]

What do you mean by that?

[molten mason]

what are the lenghts of the sides of your right triangle

[molten mason]

you need those to apply pythagoras

[molten mason]

yup good

so what are the side lengths? I'm assuming your teacher wants you to use the grid

[wooden trail]

Yes

But I have no idea what the side length is

[wooden trail]

That’s the side lengths?

[molten mason]

you need to know that to know the third side

[wooden trail]

Sorry what?

[molten mason]

hold up

how would you use pythagoras here

[wooden trail]

I would need to find the sides

A squared + B squared = c squared ?

[molten mason]

yup

so what is A and B in this triangle?

[wooden trail]

A would be the vertical line

And B would be the horizontal line

[molten mason]

yes and you know A and B

just count the squares

[wooden trail]

The squares inside the square?

[molten mason]

yup the grid

im assuming you can count each small square grid as 1 by 1

[wooden trail]

Yes

But what about the ones that are half in half out

[molten mason]

which one

A and B are not half in half out

[wooden trail]

The one at the pointy part

[molten mason]

you mean for A?

[wooden trail]

Yes

[molten mason]

that's just 3

it looks half in half out because of drawing

technically it is all the way to the vertex

vertex = pointy part

[wooden trail]

So would it still count as a square

[molten mason]

yeah

[wooden trail]

And that’s it?

I just count the squares

[molten mason]

yeah

then pythatoras

pythagoras*

[wooden trail]

How would I do that?

[obtuse pebble]

@wooden trail Has your question been resolved?

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[rigid lodge]

ok so im trying to calculate the probability of a murderer in a set number of people

[rigid lodge]

since the probability of it is 0.0062%, how would i apply this to like 1000 people?

so like out of the 1000 people what is the chance of that

[carmine hawk]

Are you trying to calculate the probability there is exactly one murderer, or at least one murderer?

[rigid lodge]

at least one murderer

[carmine hawk]

So the opposite case of at least one murderer (since calculating that needs considering 1 murderer, 2... etc)

Would be to find the probability no one is a murderer

Can you find that probability first?

[rigid lodge]

i dont know how to

[carmine hawk]

What is the probability one person isn't a murderer?

[rigid lodge]

99.9938%

[carmine hawk]

So let's ignore percentages for now, that would be 0.999938

[rigid lodge]

yea

[carmine hawk]

What if we had two people, both of them have that same probability, so what do we do to both of those probabilities?

[rigid lodge]

would be 0.999938 x 0.999938 right?

[carmine hawk]

Precisely

So now, when you have 1000 people, it would be just 0.999938 ^ 1000

[rigid lodge]

ahhhhh

i see

[carmine hawk]

Remember, this is the probability that no one is a murderer

We need the exact opposite of this, so how do we find that?

[rigid lodge]

so would it be the opposite value

one sec

[carmine hawk]

How do we find it?

[rigid lodge]

would be 1 - whatever the answer is

[carmine hawk]

Yup, 1 - (0.999938)^1000

[rigid lodge]

so it would be 0.06012

(rounded it)

[carmine hawk]

Which is 6%, kinda scary

[rigid lodge]

thank you so much

[carmine hawk]

np (also I think the probability is a bit high because we had to take 0.000062 as our probability), not the percentage

[rigid lodge]

.close

[obtuse pebble]

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[obtuse pebble]

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[potent crest]

Can someone help me with this please. As far as I'm aware you need 5 equations to solve for 5 variables, but unless I'm missing something I can only see 4.
Thanks in advance 😄

[obtuse pebble]

@potent crest Has your question been resolved?

[potent crest]

<@&286206848099549185>

[ruby lava]

I think you cant finish it, but you still can determine values of electronics, toys and home.

[obtuse pebble]

@potent crest Has your question been resolved?

[obtuse pebble]

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[uneven sleet]

Would I factor cosine out here or change cosine^2x to 1-sin^2x?

[spiral maple]

It's a quadratic in cosx

So neither

Also what is the x=pi...?

[uneven sleet]

the answer

[spiral maple]

Ok

So yeah, it's a quadratic

[uneven sleet]

ok got it ty

.close

[obtuse pebble]

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[timid silo]

Anyone willing to help with some "basic math homework ? :)

[timid silo]

So we have a logical expression of EBA!D v B!C!A!D v !EBC!D v EBC v BCD v E!CAD, the assignment is a) using quine mcCluskys algorithm find the minimal disjunctive form of the expression b)Using quine mccluskys algorith find the minimal conjunctive form of the expression and c) Show a) and b) results using Veitch diagrams. Im not stuck I just dont know how to do it, this is my first homework and its way too hard for a first homework, and I need to do it for points , so If you could help I would appreciate it very much sir

[rugged kite]

well a good start would be to know if you know Quine McClusky's algorithm

[timid silo]

Well yeah... thats the problem I do not sir

If you do not have the time its completely understandable

I am asking for way too much

[rugged kite]

i mean

did you not study it?

i roughly know what this is about, but if your teacher didn't cover the algorithm in class that would be thoroughly surprising

cause yknow, you don't ask for something you didn't study just like this

[timid silo]

The thing is, this is my first homework and its way too hard (for me) I watched a lot of vids on yt, did some examples but then I just forget a lot of it, I will make sure to study it, but math is one of the subjects ive been lacking due to other assignments... The problem is they give out points for homeworks so yeah...

I did learn truth tables and some basic things

about axioms

I havent had the time to understand these more complex subjects tho

[rugged kite]

Alright, I've brushed up on the algorithm

Though, it might prove a bit cumbersome to explain within a text channel

well, i'll be back with some pen and paper

[timid silo]

Thank you sir!

[rugged kite]

The point of the Quine-McCluskey algorithm is looking for redundancies/patterns in the values that satisfy the expression

(I won't need pen and paper after all I think)

[timid silo]

makes sense

[rugged kite]

Here, I'll represent a value ABCDE in binary

So TTFFT will be 11001

[timid silo]

yep that part I understand hahaha

[rugged kite]

So to start off we will write all of the values that satisfy the expression. I will write x in the cases where the value doesn't matter

11x01
0100x
x1100
x11x1
x111x
1x011

are all the "shapes" that satisfy the expression

To calculate that I just looked at every "multiplicative" term of the expression

I put a 1 where a value needed to be true, a 0 where it needed to be false, and x if the term in question didn't depend on it

So what 0100x means is that both 01000 and 01001 satisfy the expression

Now we want to find redundancies

To do that, we can scan each value, and see if by chance, they can be simplified

For example, consider x1100, x111x and x11x1

Notice that in any case, the value of the last two bits doesn't matter

[timid silo]

how did you know which values satistfy the expression?

sorry if its a dumb question

[rugged kite]

The expression is a conjunction (everything is linked by logical ORs)

[timid silo]

yep

[rugged kite]

so for a value to satisfy the expression, it only needs to satisfy one of the terms (between the ORs)

For instance EBA!D

[timid silo]

aaa okay

makes sense

[rugged kite]

Let's reorder it to AB!DE

We need A=1,B=1,D=0,E=1 and C doesn't matter

Hence, 11x01

[rugged kite]

Back to this

[timid silo]

ahhh okay

now I understand!!

[rugged kite]

x1100 is the first one, x1101 satisfies the third one, x1110 satisfies the second one, x1111 satisfies both the last two

so really, since the last two bits don't matter in this pattern

We can merge x1100, x111x and x11x1 into x11xx

So our new list looks like

11x01
0100x
x11xx
1x011

[timid silo]

that makes sense but anither question

[rugged kite]

go ahead

[timid silo]

how?

oh nvm

I got it I got it

Wow you explain this stuff really well

I didnt expect this kind of help

[rugged kite]

you're welcome

[timid silo]

Thank you so much

continue sir

its a pleasure to learn from u

[rugged kite]

now here, scan all you want, but you're not gonna find redundancies

the patterns don't really overlap

An example of a redundancy could have been 11x01 and 11x11

Which we could have simplified to 11xx1

[timid silo]

hmmm

hiw can we simplify that the x is third in both cases

[rugged kite]

x means that this bit doesn't matter

If it's 0 it satisfies the first one

[timid silo]

yeah I understand

[rugged kite]

If it's 1 it satisfies the second one

So we can just put an x instead of the fourth bit

since it doesn't change the result

[timid silo]

ahhh okay

[rugged kite]

as for the rest of the pattern i just made up something

So if you scan you won't find redundancies

But if you really want to make sure there are none, you can do this : try to find a value that only one pattern satisfies

If you have a pattern which doesn't match a value that another pattern doesn't already match, then it's redundant

Since it adds no new values

[timid silo]

why do we reorder it

[rugged kite]

For 11x01, 11001 satisfies no other pattern (but 11101 does satisfy x11xx)
For 0100x, either 01001 or 01000 don't satisfy anything else
For x11xx, 11111 satisfies nothing else
For 1x011, 11011 satisfies nothing else

[rugged kite]

for convenience, in my world A comes before B yknow

[timid silo]

ahahaha makes sense

[rugged kite]

So now we have all our final, most general patterns

Which we can turn back to logical expressions

[timid silo]

thats awesome

[rugged kite]

11x01 becomes AB!DE (C doesn't appear because it doesn't matter)
0100x becomes !AB!C!D
x11xx becomes BC
1x011 becomes A!CDE

And since the expression has to be satisfied if at least one of these is satisfied, our equivalent logical expression looks like :

AB!DE v !AB!C!D v BC v A!CDE

This is the minimal disjunctive form

Minimal because there are no redundancies

Disjunctive because it uses logical ORs at the top level (logical OR is sometimes called a disjunction)

[timid silo]

why do you explain this better than my professor?

[rugged kite]

go figure

[timid silo]

sickk

I need to study this more

its quite interestinf

how do we get the conjunctive form?

[rugged kite]

i'm not quite sure actually

[timid silo]

Thats fine! I thank you very much sir, I will make sure to study this more and go over some of the things u mentioned!

[obtuse pebble]

@timid silo Has your question been resolved?

Channel closed

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[timid silo]

.reopen

[obtuse pebble]

✅

[timid silo]

I have another question if you dont mind

@rugged kite

[rugged kite]

go ahead

[timid silo]

so the question was actually to do the minimal disjunctive NORMAL form, is that any different?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

Are you here sir?

@rugged kite sorry for pinging

[rugged kite]

nah

the "normal" refers to the fact that each term that's OR-ed is a conjunction

some literals with AND between them

So for example A v (B(C v D)) is not DNF

but A v BC v BD is

[timid silo]

Alright, no more questions im out for real this time 😆 see ya!

.close

[obtuse pebble]

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[hexed totem]

Hey, could someone explain to me how we got 1-2sin^2alpha?

[spiral maple]

cos^2(a)=1-sin^2(a)

so cos^2(a)-sin^2(a)=1-sin^2(a)-sin^2(a)

[hexed totem]

ohhhhhhhhh

i think i wrote notes on that somewhere

but just incase i didnt

[slender badge]

most classes will let you have a sheet of the trig identities during tests

[hexed totem]

what about sin^2 and tan^2? if there is an expression to them

[hexed totem]

not mine tho

our teacher wants us to remember everything we've taken so far

[hexed totem]

yeah i remember this

especially pythagorean identities

it was forced into my brain

anyway thanks @spiral maple

.close

[obtuse pebble]

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@quick mauve Has your question been resolved?

[quick mauve]

its resolved now

[obtuse pebble]

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[obtuse pebble]

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[keen owl]

hey had a question on derviatives

[keen owl]

I have 2 functions and I need to find the derivative of (dA/dw) w=1

the functions are

<@&286206848099549185>

[obtuse pebble]

@keen owl Has your question been resolved?

[keen owl]

If any helpers come please allow me 15 minutes after the hour

Ill be back in around 15 minutes From now

[atomic bear]

I am lost

Where do get help on my problems

[nocturne minnow]

Read #❓how-to-get-help

[keen owl]

Help

Helo*

<@&286206848099549185>

[hazy sedge]

What are you confused on?

also is t constant or another variable?

[keen owl]

T is time

t*

Im just not sure how to derive eoth the givens

[hazy sedge]

Are you taking a partail derivitive or is t deffined/

[keen owl]

Like i have to 1 function

[hazy sedge]

ahh sorry that was tarrible let me retype

[keen owl]

And a(w) is defined

I need to derive da/dt

One second

Going im a soyr

[hazy sedge]

wait do you need da/dt or da/dw?

[keen owl]

Store

[hazy sedge]

ok if your leaving still really need help in a bit just message me or come here again for someone else

[keen owl]

Sorry

Im back

[hazy sedge]

ok then

so are you trying to get da/dt or da/dw?

[keen owl]

The second sorry

Da/de

Dw

[hazy sedge]

really? So have you done partial derivitives?

[keen owl]

Nope

Sorry for early i was buying dinner

[hazy sedge]

And you sure your not trying to find da/dt? So your final equation would only have t's as variables?

all good

[keen owl]

Im pretty sure the question in the practice book says da/dw

I can double check if you would be so kimd to give me a moment

[hazy sedge]

if you could that would be great

OHHHH never mind i was looking t the function P not a. That would not be solviable without partail derivitive. I very sorry for the confution.

[keen owl]

The thing doesnt load on my phome sorry

I belive its da/dwtho

[hazy sedge]

All good i thing I can help you with out that. You dont need p right now.

[keen owl]

Just to recap

I need to derive this

With da/dw

Where w=1

At the derivative

[hazy sedge]

Well da refers to deriving A(w) not P. Which would be the picture below that one in the first thing you sent out.

[keen owl]

Alright

But it asked to derive the original function

I do belive

[hazy sedge]

You can derive dP/dt with that equation just not

[keen owl]

If you wsnt to help someome else ill be home in 15 mins

If you want me to go over everything just to comfirm

Otherwise we can continue eoth this

Idm either whatever’s good for you

Im getting help from ya so ill work around you

[hazy sedge]

It would probly be better to go over the problem. Mabey taking a pic and sending that if possible because I have no idea how to help without using higher level math

[keen owl]

Ok sir

Is meeting when the clock hit tjte hour fine?

So 20 minutes?

.close

[obtuse pebble]

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[swift copper]

How is this divergent? I separated the limit to the the numerator and denominator. lim/lim

I got undefined/infinity

[spiral maple]

cause it never converges

you get lim x-> inf (-1)^x by analysis

[swift copper]

oo nice

so if i ever encounter something like this i would say it diverges?

Is it because of the 8x being there?

Just testing some things out and this one converges to 0

[spiral maple]

yeah, cause you have constant/(going to inf)

[swift copper]

just to clarify if i were to draw a graph it would look like it's oscilatting and gets closer to zero

[spiral maple]

drawing the graph is near impossible

[swift copper]

o i thought it would look like

cause it alternates between negative and positive

[spiral maple]

yeah if you restrict yourself to x in N

[swift copper]

alroght thanks for the help

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[winged kettle]

my best guess would be ln(5) bht tbh no idea

[tired shell]

you're right

it's a lhopital's rule thing, and the derivative of 5^x is ln(5)5^x

[winged kettle]

ahhh okay thank you

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[vestal niche]

Can't figure this one out

[vestal niche]

Am I supposed to use cosine law?

[obtuse pebble]

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[arctic apex]

The test statistic of z=-2.25 is obtained when testing the claim that p=1/4.
a. Using a significance level of a=0.01 ​, find the critical​ value(s).
b. Should we reject Ho or should we fail to reject Ho ​?

[obtuse pebble]

@arctic apex Has your question been resolved?

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@arctic apex Has your question been resolved?

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[obtuse pebble]

@civic trellis Has your question been resolved?

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[mighty ether]

could someone maybe help explain conceptually the part that i've double highlighted here? why c = 0 for the velocity of a particle function which starts at rest?

[mighty ether]

if someone can just link up the maths with the physics stuff that would be good since i don't study physics

is this because velocity is how the position is changing, and for something which starts at rest the amount of change is just 0 anyways?

[obtuse pebble]

@mighty ether Has your question been resolved?

[mighty ether]

no @obtuse pebble

[unique solstice]

Yes

Its just because it starts at rest

[mighty ether]

okay i think tht makes sense

[prisma python]

if the c is non-zero then the velocity at x = 0 wont be zero

[mighty ether]

thanks @prisma python

[obtuse pebble]

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[obtuse pebble]

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[regal sierra]

I am currently struggling (again) with task a). What I tried is added in the following picture, but I have the feeling what I did is wrong. Can anyone of you take a look?

[obtuse pebble]

@regal sierra Has your question been resolved?

[regal sierra]

<@&286206848099549185>

[obtuse pebble]

@regal sierra Has your question been resolved?

[tribal stump]

lol

[obtuse pebble]

@regal sierra Has your question been resolved?

[obtuse pebble]

@regal sierra Has your question been resolved?

[fallen charm]

@regal sierra if you just want to check if your taylorpolynomial is correct i'd suggest checking it on wolframalpha or something similar

and if its wrong you can come back and pose a more specific question

[obtuse pebble]

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[regal sierra]

Okay thank you!!!

[fallen charm]

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[regal sierra]

.close

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[charred solar]

I need to write this in to one log

[charred solar]

thinking about doing this but not sure what to do next

^with all of them so they are all 10th log

not sure what to do now

[north acorn]

why did you choose the 10th log in particular

[charred solar]

not sure would it help to get another?

[north acorn]

how about you try converting them to the 2nd long

it should simplify a lot nicer

[charred solar]

let me see

[north acorn]

there you go

[charred solar]

I just multiplied the first with 3 and the second with 3/2

but got a different answer

[charred solar]

if i do this i can add them all up right?

[north acorn]

yeah but by multiplying by 3 you mean 3/3

if so

then it should work

[charred solar]

yes the "upper part" and the "lower part" not sure what it is called in English

[north acorn]

me neither but i got what you mean

[charred solar]

ill try again

[north acorn]

what did you mean by got a different answer

[charred solar]

well i have the answers next to me but probably made a small calculation mistake gonna try it again and be more secure (if that's the right word)

[floral void]

=lnx/ln2 + lnx/ln4 + lnx/ln8
=(1/ln2 + 1/ln4 + 1/ln8)lnx
=11lnx/ln(64)
=11 log_64 x

or (11/6)(log_2 x)

[charred solar]

thanks but I'm not looking for the answer trying to understand this subject 🙂

[floral void]

well…i showed all the step tho

[charred solar]

I'm so confused after this i can do this

thats log rules right?

if you log(x) + log(a) = log(xa)

[floral void]

nvm…

[charred solar]

wait i should have it now

should be 11/2 no?

11/2^1/3 = 11/6

so will be log_2 (x^11/6)

which is correct

thanks!

It's so amazing how a discord server like this actually works such an awesome server

.close

[obtuse pebble]

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[lavish kite]

x = 2|y| - 1

Find four points contained in the inverse. Express your values as an integer or simplified fraction.

{( , ), ( , ), ( , ), ( , )}''.

[obtuse pebble]

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[timid silo]

Hi i'm new to this math topic. What is the method for solving these? :0

[drifting wraith]

2. is just an identity you memorize

[timid silo]

i seeee :0

[drifting wraith]

and 1 is like, (1.5x + 3 - (0.5x+1))(1.5x + 3 + (0.5x+1))

so same thing

but actually you're probably supposed to multiply

without that

x(2x+4) + 2(2x+4)

[timid silo]

I figured that, i was a bit confused at the time

[timid silo]

ty

[fervent cradle]

if you have (a+b) * e, that's the same as ae + be
so now let e = (c + d)
then (a+b) * (c+d) = (a+b) * e = ae + be = a(c+d) + b(c+d) = ac + ad + bc + bd

hopefully that makes some sense

[timid silo]

YES :]
and ty guys

[obtuse pebble]

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[warm siren]

can someone tell me what i did wrong

do i just have to state more than one ?

[quaint warren]

what is local maximum then?

ah ye

i though the left of c was closed

zoomed the pic

it was open

🙃

should we close the channel

[obtuse pebble]

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[warm siren]

those were not correct unfortunately (this is a different problem)

[obtuse pebble]

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[warm siren]

.close

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[toxic bridge]

@pine smelt

[pine smelt]

hello

[toxic bridge]

yes

ok

so

[pine smelt]

the last part

it was supposed to be a minus

instead of a plus

[toxic bridge]

ohhhhhh

so it does equal 1

[pine smelt]

yes

[toxic bridge]

ok

[pine smelt]

so is one of the things (x-1)

[toxic bridge]

so that means that it is (x - 1)(x - b)(x - c)

yes

[pine smelt]

ok

[toxic bridge]

do you know how to do polynomial division

[pine smelt]

uh

not anymore

[toxic bridge]

oh

so it's like

this

$\frac{x^3 - 9x^2 - x - 9}{x - 1}$

[warm shale]

Tra-Guy

[toxic bridge]

do you know how to simplify that

[pine smelt]

uh

one sec

[toxic bridge]

ok

[pine smelt]

nope

[toxic bridge]

ah

ok well it's kinda hard to explain

how to do it

we're basically factoring out an 'x - 1' from x^3 - 9x^2 - x - 9

[pine smelt]

oh

wait

i think i did polynomial long divison a couple weeks ago

im gonna go through notes for like

3 minutes

[toxic bridge]

wait is it +9x^2 or -9x^2

[toxic bridge]

ah ok

[pine smelt]

ok nvm im dumb

no clue

[toxic bridge]

ok so when we factor out an x-1

it becomes (x - 1)(x^2 +10x + 9)

[pine smelt]

ok

[toxic bridge]

do you know how to factor quadratics

[pine smelt]

yes

[toxic bridge]

ok

so you can factor that quadratic

[pine smelt]

(x-1)(x+9)(x+1)

right

[toxic bridge]

yes

[pine smelt]

ok

[toxic bridge]

and youre done

[pine smelt]

oh shoot

thank you

[toxic bridge]

yeah is that all

did you just want to factor it

[pine smelt]

yeah

thanks

[toxic bridge]

all g then

.close

[obtuse pebble]

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[obtuse pebble]

@gusty wren Has your question been resolved?

[dawn arch]

I don't think this question is written correctly. The way this is worded it implies that expansion of exponents is equivalent to finding the derivative and that is false.

[high lily]

the question is fine

[dawn arch]

Not really, hence = therefore

What I believe the question is trying to say is expand the given functions THEN find the derivative and compare it.

[high lily]

hence is different from therefore

[dawn arch]

Exactly

[high lily]

hence is appropriate here

[dawn arch]

https://www.dictionary.com/browse/hence

Definition 1, lists "therefore"

[high lily]

there's a distinction when used in math

[unique solstice]

Theres multiple uses

Its super olden to use it that way

But its technically correct

[dawn arch]

But we're not in olden times...

[unique solstice]

Shrug

Maybe its not olden in math like hes saying

[dawn arch]

I dunno, I've never come across it outside of it being used in the case of it being interchangeable with "therefore"

[high lily]

hence is used in questions when they want you to explicitly use what was just obtained to do something

[dawn arch]

Interesting, I wonder how many students taking Calc 1 truly understand that distinction

[unique solstice]

I mean the hence is more of a flow word

All you have to read is "find dy/dx"

[dawn arch]

yeah true, but I'm thinking for average students it could be a confusing flow word to use and they might confound things a bit

[high lily]

dont think I've seen that many people that had an issue with "hence"

[dawn arch]

Well, I rarely see it being used in this fashion, but that would be a good question to ask! But thanks for letting me know, I learned something new!

[high lily]

depends a bit on context I guess

[dawn arch]

Yeah, that's what was confusing me.

[high lily]

perhaps in stuff like proofs some people may prefer to use hence

[dawn arch]

Yeah, that's where I see it often

Not in basic Calc 1 questions haha

[high lily]

they use it when they want you to solve a question in a specific way

[dawn arch]

ah ok, cool

[spiral maple]

Hence in a question is just saying "Using the previous answer, do this"

[dawn arch]

Yeah, I looked it up in the "math" sense

[spiral maple]

Hence or otherwise is "Use the previous answer, or don't, up to you"

[high lily]

some writers may offer an alternative with
hence or otherwise

eep

[spiral maple]

Usually you just follow the hence

[dawn arch]

Right, I was just musing over how many students would read it that way in this question without it being explained?

[obtuse pebble]

@gusty wren Has your question been resolved?

[dawn arch]

oh god, I just realized I read this too quickly and left out the word "and"... gah, that's why I was SO confused. I'm really sorry.

[unique solstice]

The and I think is the weirdest part

[dawn arch]

really? To me, this makes it more like "Do this AND then do this..."

[obtuse pebble]

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[small knoll]

Why this and not the second one?

[small knoll]

are the two end results the same?

there fixed it

[obtuse pebble]

@small knoll Has your question been resolved?

[small knoll]

??

<@&286206848099549185>

[obtuse pebble]

@small knoll Has your question been resolved?

[obtuse pebble]

@small knoll Has your question been resolved?

[mighty geyser]

my guess is that it'll probably be equivalent, remember your change of base formula

[obtuse pebble]

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[versed bison]

Can someone help me these 5 questions with steps so that I know how to solve them? Thank you so much!

[obtuse pebble]

@versed bison Has your question been resolved?

[versed bison]

<@&286206848099549185>

[small knoll]

For the first one, you just plug it in a graphing calculator and you get this:

Making the solution (0,2)

for the second one, you subtract the second equation from the first one to eliminate one of the variables:

x+3y = 7

• x+y=3

∴ 2y = 4
∴ y = 2
then plug into one of the original equations to find the x
x+y = 3
∴ x+2=3
∴ x = 1
thus giving the answer (1,2)

for the third one, isolate one of the variables and substitute it into the other equation.

2x - y = 3
x + y = 0

the second one you can make it x = -y and plug it as x for the first one giving you
2(-y) - y = 3
-2y - y = 3
-3y = 3
y = -1
then plug it into one of the equations to find x.
x + y = 0
x + (-1) = 0
x = 1
thus giving the answer of (1,-1)

fourth one part (a) you just isolate y so that it is in standard equation form

y - 2x - 5 = 0
y - 2x = 5
y = 2x + 5

then you can plug in the given values for x to find y for each ordered pair in the table
y = 2(-2) + 5
y = -4 + 5
y = 1
(-2,1)

y = 2(0) + 5
y = 0 + 5
y = 5
(0,5)

y = 2(2) +5
y = 4 + 5
y = 9
(2,9)

number 4 part b is just drawing the graph from x = -2 to x = 2 you already have the values for -2,0, and 2, so you just draw a mini graph to the scale they want, put in the dots representing the 3 ordered pairs you got from part a and connect them

for part c, you use the graph you made to find the value of k in the ordered pair (1.5,k)
alternatively, although they didnt tell you to, you can just plug it into the original formula since the x value is given as 1.5
y = 2x + 5
y = 2(1.5) + 5
y = 3 + 5
y = 8
thus the ordered pair would be (1.5 , 8) making k = 8

for 4 part d part i, you just draw a horizontal line in your graph you made in part b at y = 3

for 4 part d part ii you see the point where the horizontal line you drew intersects with the graph of y = 2x + 5 and state the x-coordinate of said point.
Alternatively again, not asked but you can just plug it into the original equation given of y = 2x + 5 since the y-value is given already as 3.

y = 2x + 5
3 = 2x +5
-2 = 2x
-1 = x
x = -1
∴ the answer would be -1

as for part 5, the x solutions are given as 3 and -4, as such, you can assume the polynomial factors were (x-3)(x+4) = 0 making either (x - 3) = 0 or (x + 4) = 0 making x = {3 , -4}. As you found the factors are x-3 and x+4 you can use the FOIL method (Front, Outside, Inside, Last) to find it in ax^2 + bx + c = 0 form. or in this case, it's asking for the values of the second and third so x^2 + ax + b = 0.

(x-3)(x+4)
Front: x * x = x^2
Outside: x * 4 = 4x
Inside: 3 * x = 3x
Last: 3 * 4 = 12

Combined:
x^2 + 4x + 3x + 12
∴ x^2 +7x + 12

making your a and b values 7 and 12

@versed bison

[versed bison]

thank you so much 😭😭🙏🙏

[obtuse pebble]

@versed bison Has your question been resolved?

[obtuse pebble]

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[languid grail]

i'm kinda stuck on how to find the sum of this, i've proven that it is convergent using the alternating series test. to find the sum, i made it into a telescoping series but it's alternating and that's been messing with me.

[obtuse pebble]

@languid grail Has your question been resolved?

[languid grail]

<@&286206848099549185> any kind souls wanna nudge me in the right direction here?

[royal basin]

may i suggest grouping the terms into pairs?

your series is absolutely convergent, so grouping will not mess anything up

the general idea is that $\sum_{n=2}^{\infty} (-1)^n a_n = \sum_{k=1}^{\infty} (a_{2k} - a_{2k+1})$

[warm shale]

Ann

[royal basin]

it might be a good idea to decompose 3/(n^2 + n - 2) into partial fractions first, just in case

[languid grail]

ah ok, so i should group the even and odd terms and then see what i get? i'll try that

even with the grouping, either i'm doing something severely wrong, or using the wrong method, none of my terms cancel out

$\frac{3}{n^2+n-2}=\frac{1}{n-1}-\frac{1}{n+2}$

[warm shale]

KentyMcDclr8

[languid grail]

i get this but can't seem to go further

[royal basin]

well now you have $$\sum_{k=1}^{\infty} \frac{1}{2k-1} - \frac{1}{2k+2} - \paren{\frac{1}{2k+1-1} - \frac{1}{2k+1+2}}$$ in accordance with that formula i wrote earlier

[warm shale]

Ann

[languid grail]

what would be the continuation of that because i'm still stuck even after making the last 2 denominators 2k and 2k + 3?

[royal basin]

$\sum_{k=1}^{\infty} \frac{1}{2k-1} + \frac{1}{2k+3} - \frac{1}{2k+2} - \frac{1}{2k}$

[warm shale]

Ann

[royal basin]

hmm

i think i might've painted us into a bit of a corner here

[languid grail]

nothing cancels out =/

[royal basin]

hence my last remark.

full disclosure, i did not quite see this coming.

[languid grail]

haha, all good

[royal basin]

let's throw this into WA

,w sum[n=2, infty] (-1)^n * 3/(n^2 + n - 2)

[warm shale]

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[royal basin]

log

hm

what if...

what if we consider $f(x) = \sum_{n=2}^{\infty} \frac{3x^n}{(n+2)(n-1)}$?

[warm shale]

Ann

[royal basin]

then our series will be the value of f(-1), and maybe we can glean some information about the function this way...

[languid grail]

hmm

[obtuse pebble]

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[obtuse pebble]

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[chilly ledge]

Hey hi, i'm having trouble with question C

[chilly ledge]

i have the answer for this, but i don't understand it.

why can i not just multiply the sum of Xi and Yi?

and why is the formula for sample covariance relevant here?

[obtuse pebble]

@chilly ledge Has your question been resolved?

[chilly ledge]

<@&286206848099549185>

plz halp

someone? anyone?

[molten mason]

think about a simpler example of (x1+x2+x3)(y1+y2+y3) - if you multiply those you get a bunch of extra terms

instead of just x1y1 + x2y2+x3y3

[chilly ledge]

ok yes, that makes sense

so the method they are using here is to just rewrite the covariance formula

[molten mason]

yup so that you can find the sum of cross prod with the given info

[chilly ledge]

is there another way to do this? since i know xi, yi and what n is

[molten mason]

you could calculate it by hand...which is annoying

[chilly ledge]

ooh wait i'd have to do each xi*yi manually

times 20

yeah that's REAL annoying

[molten mason]

yup XD

[chilly ledge]

so the answer is just a bit of trickery really

to save time

[molten mason]

and its also sometimes impossible with real data

yeah it's just trying to get you to see the relationship between all the things you could possibly manually calculate

by manipulating formulas

[chilly ledge]

yeah, actually a pretty good question in that regard

[molten mason]

[chilly ledge]

i was just confused by "cross product"

because i immediately thought vectors

[molten mason]

yeah same

xD

and the way we're calculating it ironicaly reminds me of dot product instead of cross product

[chilly ledge]

also that other form of the covariance formula... i think i've seen it before but not sure

so it never occurred to me

our professor really likes us to rewrite shit, i should've known really

alright i actually understand this now.

thanks for the help.

[molten mason]

[chilly ledge]

.close

[obtuse pebble]

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[timid silo]

Hi

[timid silo]

What does it mean to put an interval or a set into a function?

[lethal crest]

$f^{-1}(A) = { x\in \mbb{R} | f(x) \in A }$

[warm shale]

Ryuzaki

[lethal crest]

you can replace R by the domain of the function

[timid silo]

Sorry, what is the domain? I have only had maths in german so I'm not familiar with the international words yet

@lethal crest

Ah wait I got it

what is A?

[lethal crest]

$f:\underbrace{D}{\text{domain}} \to \underbrace{R}{\text{range}}$

[warm shale]

Ryuzaki

[timid silo]

How do you solve that then?

[lethal crest]

what's your f here?

[timid silo]

I will invent something similiar because I don't want to use the one from the assignment:
f(x,y) = xy + 3x

[lethal crest]

so you need the find the points s.t. f(x, y) lie in the interval [0, 1]

that

that's what $f^{-1}([0, 1])$ means

[warm shale]

Ryuzaki

[timid silo]

so [0,1] is the domain?

[lethal crest]

subset of the range,

you are inverting f

I'll give you an example if that helps

[timid silo]

wait how can you invert a function with two unknowns?

[lethal crest]

take f(x) = x^2 then f^{-1} ( [0, 1]) is the set where f(x) lie in 0 to 1

i.e. 0 <= x^2 <=1 gives -1 <= x <= 1

so in this case $f^{-1}([0, 1]) = [-1, 1]$

[warm shale]

Ryuzaki

[timid silo]

ah, so you put in 0 and 1 and use equal or greater than?

[lethal crest]

that's a specific example but mostly yes

[timid silo]

okay

[timid silo]

this

[lethal crest]

find all the points that maps to the interval you are trying to invert

[timid silo]

Imma be real I have no clue

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[lethal crest]

hint1 || sin(x) is odd function, so that part doesn't matter ||

hint2 || u substitution u=x^5 ||

$\int_{-2}^2 \frac{\sin(x)}{x^{10}+4} \dd x = 0$

[warm shale]

Ryuzaki

[obtuse pebble]

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[obtuse pebble]

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[tired shell]

not sure what g is for but everything you said is correct

that's right

standard basis is assumed

uh without that negative sign at the front yeah https://en.wikipedia.org/wiki/Rank–nullity_theorem

oh that's just your way of bullet-pointing

[obtuse pebble]

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[slow jackal]

Let's assume that we are faithing against some hydra with red, blue and black heads..
We are starting with 6 red and 8 blue heads...
You will be cutting only red heads..
When you cut red head... Then.. Hydra gets 6 red and 3 black heads insted of one cutted red head...

You will get this sequence: 14, 62, 350
What is nth term formula(or recursive function) and can you prove it?

[tired shell]

I don't understand what the sequence is counting

oh it's cutting all the red heads sure

every red head cut is 1 head to 9, so it's +8

so 14->62 is 14 + 6*8, since you start with 6 red

try to write out how many red heads are at each step first

[slow jackal]

Hydra have 6 red and 8 blue heads...
After first round 62 = she will loose her 6 red heads... But she'll get 6 * 6 red and 6 * 3 black heads.... Plus.. She already had 8 blue heads.

[obtuse pebble]

@slow jackal Has your question been resolved?

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[sour crypt]

i need help
so if tan(t)=-5 what does tan(-t) equal

[sour crypt]

would it be tan(-t)=5 ?

[tired shell]

yup

[sour crypt]

k thanks just making sure

[tired shell]

tan has that 180 symmetry

[obtuse pebble]

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[timid silo]

how do i get the answer

@worldly forum

can you help me

please

.help

[obtuse pebble]

Commands:
clopen: .close, .reopen
factoids: .tag
help: .help

Type .help <command name> for more info on a command.

[spiral maple]

dont ping random users.

[timid silo]

but it said feel free to ping helpers

and then i looked in helpers and then i just saw the first one that was on online

have i misunderstood anything?

[green compass]

isolate X

[timid silo]

how

i really dont know anything about this subject in math

[green compass]

x is over 2 right

[timid silo]

i think

[green compass]

what do you have to do to get x on its own

it's an equation

whatever you do to one side happens on both

[timid silo]

idk

no idea

thats why im in help

[green compass]

have you tried

multiplying both sides by 2

[timid silo]

no because i dont know what i have to do

[timid silo]

so 2 * x *2

[green compass]

(x/2)*2

multiply everything by 2

[timid silo]

so the (x/2) is just the x over 2?

[green compass]

yes

[timid silo]

can u use the bot to make that picture

[green compass]

no

idk how to use it

[timid silo]

ok

[timid silo]

ok so x over 2 times 2

so that 4?

[green compass]

x over 2 times 2, 4 times 2, 3 times 2

[timid silo]

huh

[green compass]

no it's x

it's an equation and you're trying to isolate X, so the first thing you do is get rid of the 2 beneath the x, hence multiplying everything by 2

[timid silo]

man i have no idea

i think im about to have a mental breakdown soon

its just we have a test soon so have to study

wait can you do this

[green compass]

do what

[timid silo]

x/2 + 3 = 9
x/2 = 9 - 3
x/2 = 6
2 * x/2 = 2 * 6
x = 12

[green compass]

Lol seems like you already know what you have to do

that's the exact same as the question you have, just with different numbers

[green compass]

Show me what you'd do for this. that example and this are fundamentally the same question

[timid silo]

x/2 + 4 = 3
x/2 = 7 - 4
x/2 = 3
2 * x/2 = 2 * 4
x = 6

?

i think i did something wrong

[green compass]

x/2 + 4 = 3
x/2 = 3 - 4
x/2 = -1
2 * x/2 = -1 * 2
x = -2

[timid silo]

why do you do x/2 = 3 - 4?

[green compass]

there is no 7

[timid silo]

how

[green compass]

because there isn't???

[timid silo]

then how was this right

[green compass]

It's right