help-41

[stoic ivy]

Use this as a first step I guess

See if you can reach the desired result from here

Sorry my apologies you need to put the functions in absolute values

Since the young inequality only applies to non negative real numbers

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[split sail]

why do we use inf, and sup, and not min and max for the general definition of the upper and lower riemann sum?

[indigo cloud]

min and max might not exist

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[midnight harbor]

Not sure what I did wrong here.

but its telling me that is not right

I did P = 1000

r = 6%

n = 2 x 1 = 2 (per year)

(R, n) = 0.06 / 2 = 0.03

1600 = 1000 (1+0.3)^2t

divided both sides by 1000

then 2 LN (1+0.03)

got 9.67

anyone?

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[indigo summit]

I had a question about proving greedy algorithms using an exchange proof

[indigo summit]

The problem that i'm working on is "Now let each job consist of two durations. A
job $i$ must be preprocessed for $p_i$ time on a supercomputer, and then finished for $f_i$ time on a
standard PC. There are enough PCs available to run all jobs at the same time, but there is only one
supercomputer (which can only run a single job at a time). The completion time of a schedule is
defined as the earliest time when all jobs are done running on both the supercomputer and the PCs."

The heuristic that I chose is decreasing finishing time (which i think is correct), but i'm unsure of how to go about actually proving that it's optimal

[grizzled pagoda]

Mt_Vasuveus

[indigo summit]

Let $O = {i_1 \dots i_k}$ be the optimal schedule and $A = {j_1 \dots j_k}$ be the schedule found by the algorithm
Also let $g(S)$ be the total time it takes to run a given schedule $S$ \

[grizzled pagoda]

Mt_Vasuveus

[indigo summit]

Would it be easier to prove this using a stay's ahead argument? I initially avoided it because isn't it possible that a subset of $A$ might be slower than a subset of $O$?

[grizzled pagoda]

Mt_Vasuveus

[indigo summit]

<@&286206848099549185> Any ideas on how to tackle this?

[amber wasp]

@indigo summit Has your question been resolved?

[indigo summit]

I've realized that the finishing time for a subset of jobs would be $f_k + \sum_{i=1}^k p_i$

[grizzled pagoda]

Mt_Vasuveus

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@indigo summit Has your question been resolved?

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.close

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[scenic olive]

How would I found the range?

[scenic olive]

Find*

[west phoenix]

wdym?

[scenic olive]

How would I find the range of this rational function

[elder creek]

whats the horiontal asympote

[scenic olive]

y=-1/4

I tried using that

[scenic olive]

2/-8=-1/4

[elder creek]

3/2?

where are u getting

[scenic olive]

o wait

[elder creek]

2/-8=-1/4

[scenic olive]

wrong proble

[elder creek]

?

[scenic olive]

lol

mb

[scenic olive]

yeah 3

I tried to do

(-infinity, 3) union (3, infinity)

[west phoenix]

This is your question no?

[scenic olive]

Yes yes

[west phoenix]

Could you explain to me what you mean by range, english is not my first language? But they are asking for interception and asymptotes

so you need to study when it intercepts the x-axis, and explore the asymptotes no?

[scenic olive]

like domain in range

and range

The range of y values in the function

[west phoenix]

Why do you need that?

[scenic olive]

It asks

[west phoenix]

okay

[scenic olive]

I assume I will have to know it later on

[west phoenix]

I mean typically you just study the domain and range from the answers of interception and asymptotes

So first you'd have to identify interception, and all the asymptotes

And then you look at how the graph behaves

[scenic olive]

I have the vertical and horizontal asymptopes

and the y and x intercepts

[west phoenix]

Have you searched for slanted asymptotes?

[scenic olive]

No

[west phoenix]

You find slanted asymptotes by dividing the numerator by the denominator

They don't necessarily exist, so if you know your questions are without slanted asymptotes, you don't need to do this

Secondly, you can also find other asymptotes at ''problem points'', points where the function is not determined

which in this case would be when the denominator is 0

[scenic olive]

Would the slanted asymptope be 3?

oh nvm you mean the whole numerator and denom

[west phoenix]

Yes

[scenic olive]

oh

[west phoenix]

But yes, you do have a problem point at 3

since the function is not defined at 3

[scenic olive]

Thats the horizontal asymptote

[west phoenix]

Ya, so pretty much you need to do polynomial division

for slanted asymptote

[scenic olive]

ohh

[scenic olive]

like this?

[west phoenix]

Yes

[scenic olive]

should I just

divide by x-3

and then divide that by x+1

or should I just divide by the polynomial

[west phoenix]

I would just divide by the whole polynomial

since you already have x^2 on the bottom

[west phoenix]

Ya, that's what I'd do

Now multiply by 3

to get 3x^2-6x-9

[scenic olive]

oh

does it all equal 0?

So we can just do whatever we want to it

[west phoenix]

Unfortunately no, this will not cancel out

Your solution will be

but

I just remembered something

which is good for you to know

Which I didn't think about

But you won't have a slanted asymptote in this question

You will only have slanted asymptotes, when the degree of the numerator is exactly one more than the degree of the denominator

in this case, the degree is 2 for both

which means no slanted asymptote

My bad

[scenic olive]

its alr

[west phoenix]

To find the range, you just take the limit to the point of the aymptotes

And to find the domain, you just look at when is this function defined

[scenic olive]

the vertical asymptope is 3

and -1

[west phoenix]

Yes, so now you look, how does this function behave, when it approaches 3

[scenic olive]

I assumed

[scenic olive]

goes down

[west phoenix]

It should go to infinity

[scenic olive]

I thoguht so too

I thoguht

it would be (-infinity,infinity)

[west phoenix]

Exactly

[west phoenix]

Since you look at how it behaves approaching from 3+ or 3-

[scenic olive]

But it says its this

[west phoenix]

What was the horizontal asymptote?

[scenic olive]

3

[west phoenix]

So this is the correct answer?

[scenic olive]

Yes but how

Or is the homework website just dumb

which could be true

[west phoenix]

hmm

Not sure, I'm not sure if I'm missing something here?

[scenic olive]

idk either

It says the same thing when I put it into an algebra calculator

[west phoenix]

I think I know what they're doing

one moment

I mean my issue is that I'm not getting the right answer when I'm using an algebra calculator

The idea is that you check how the graph behaves when you approach the asymptotes

and when you approach infinity or minus infinity

Plotting it, it does make sense

Because if you look at it, when it approaches infinity or minus infinity

[scenic olive]

yeah its an odd question

hopefully its not on the quiz lol

[scenic olive]

Could you help me on an actual slant asymptote question?

Its fine if you only have time for 1 question

[west phoenix]

Sure

[west phoenix]

so on the top of the chair, you multiply by x

[scenic olive]

?

[west phoenix]

Yes

That means, you are multiplying your expression x(x-3)

[scenic olive]

then -3(x^2?)

[west phoenix]

That becomes x^2-3x

[scenic olive]

wait

[west phoenix]

like this

[scenic olive]

ohhhh

mb

ill try and

long division it out

[west phoenix]

Yea

Rough long division

But your final answer is that

x+5+15/(x-3)

Is your slanted asymptote

[scenic olive]

why all above 15

Oh

is that the remainder

[west phoenix]

you mean 15/ (x-3)

[scenic olive]

and then divide again

[west phoenix]

Yea exactly

reminder

[scenic olive]

kk

[scenic olive]

it wants to me to write it as an equation

[west phoenix]

y = kx+m?

[scenic olive]

y=x+5+15/x-3?

[west phoenix]

Yea

hmm, it saying wrong

we can rewrite it then

Does the quizz site say what it complains about?

type in x + 5

[scenic olive]

it sayed "write it as an equation

[west phoenix]

try this

[scenic olive]

yeah thats what

my calculator says

[scenic olive]

do we not include the last part in the slant asymptote?

[west phoenix]

You can ignore it yes

[scenic olive]

How would I know which one it is?

besides the vert asymptote

how would I know the diff

between 1 and 4

[west phoenix]

You check the line

y = x +5

where does it cross the y axis at 5?

[scenic olive]

ohh

[west phoenix]

The only one that does that, is the one to the right

[scenic olive]

so if its x+5

the y intercept

of the slant asymptote

is 5?

[west phoenix]

Yes of course, it's a normal straight line

y = x + 5

the y intercept is at 5

When you calculate for slanted asymptote, now I forgot about this since I haven't done this in like 4 years. But you ignore the remainder of the equation because it becomes irrelevant when you approach infinity

[scenic olive]

oh yeah my teacher told me that now that I remember

[west phoenix]

so when we did long division and got x + 5 + (15/x-3)

[scenic olive]

the +3 or whatever wont matter

at x= 1 gazillion

[west phoenix]

you can ignore the remainder

So we got the line y = x + 5, and yea you get now how to analyse which graph it is

[scenic olive]

well 4

only one that is right

[west phoenix]

Exactly, because it intercepts at 5

[scenic olive]

There is another one

that has two with the same y intercept

[west phoenix]

Where

[scenic olive]

the salnt asymptote is x-7

so it has to be x-7

both 1 and 3 have y ints at -7

[west phoenix]

Yes

[scenic olive]

the difference is one is negative and one isnt

[west phoenix]

x - 7

x is positive, which means the line is moving positive

[scenic olive]

So 1?

[west phoenix]

which means first one is correct

Yes

Makes sense?

[scenic olive]

Yup

I think thats all

Thanks alo

alot

[west phoenix]

No problem, hope I didn't make it complicated

[scenic olive]

Nah, very helpful

[west phoenix]

Good luck!

[scenic olive]

u to

[scenic olive]

How do I close a ticket

[west phoenix]

.close

[scenic olive]

.close

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Closed by @scenic olive

Use .reopen if this was a mistake.

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• Show your work, and if possible, explain where you are stuck.
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[scenic olive]

How could I solve A and C?

[scenic olive]

I found out A is 3 with chat GPT but it doesnt explain it properly

and I have no idea how to get C

[torn furnace]

they tell you to graph it

so presumably thats the way you find it

[scenic olive]

is there any way

to find it otherwise

[torn furnace]

you can take the derivative

but this is precalc

so maybe you don't know that yet?

[scenic olive]

is this what that is

or is that a diff thing

[torn furnace]

its a different thing

[crude kite]

Why? Why not 1 point? That's so weird

[scenic olive]

yecidk

[torn furnace]

im trying to think of how one would reason that this is the maximum without using calculus

you could use calculus reasoning

without actually take the derivative or anything

but idk how comprehensible that will be

@scenic olive the idea is like this

when the denominator is big, the fraction is small

and when the denominator is small, the fraction is big

all other things held the same

make sense so far?

👀

[scenic olive]

yes

[torn furnace]

alright

also, when x is small, then x^2 is really small

and if x is big, then x^2 is really big

we can draw a line between "small" and "big" here

it happens around x=1

because a number times a number thats less than 1 shrinks

but if you multiply something by a number bigger than 1, it grows

good to there?

[scenic olive]

Yes

[torn furnace]

so this tells us that a good place to look for our maximum is around t=1

because around here, t^2+1 will stop being small, and will start getting big

and, it will get bigger faster than 6t does

,w plot x^2 and x from x=0 to x=2

[grizzled pagoda]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by janniku

[torn furnace]

you can see this happen in a graph, too, x^2 starts out smaller, but eventually it catches up

and since the denominator starts getting bigger than the numerator around t=1, this is where hte function stops growing, and starts shrinking

hopefully that reasoning isnt too crazy

[scenic olive]

So basically

1 is the horizontal asymptote

Kinda

[torn furnace]

actually 0 is the horizontal asymptote

[scenic olive]

oh yeah

would it be the vertical asymptote?

[torn furnace]

so im not sure how you mean

[scenic olive]

or is it best

not to think of it as an asymptote

[torn furnace]

it has special names but they arent asymptote

[scenic olive]

so C goes up

until 1

and then it goes down?

[torn furnace]

yea

[scenic olive]

at 1 it is

6/1^2+1

which is 3

at 2 it is

[torn furnace]

and since t^2+1 never stops growing faster than 6t, the function never stops going down

[scenic olive]

12/5

which is less than three

[scenic olive]

So we know that 1 is the highest it goes for t until C goes down

We know that at T=1

C is the highest

[torn furnace]

yea, you maybe stare at the fraction, and you reason that a good place to check would be at t=1

and then you would check the graph

or try the function at some other points

like you did

and say yea, this guess seems okay

[scenic olive]

How would I go about

finding a specific T

for a specified C

like in question c

[torn furnace]

its less complicated than it seems

you want to solve $0.2 = \frac{6t}{t^2+1}$

[grizzled pagoda]

jan Niku

[scenic olive]

ye

[torn furnace]

which, you can totally solve

[scenic olive]

so do you just

do it

[torn furnace]

yea

you may run into some problems or questions

but maybe a good first step is like

$0.2(t^2+1)=6t$

[scenic olive]

6/5/.2t^2+.2

[grizzled pagoda]

jan Niku

[scenic olive]

oh tahts easier

[torn furnace]

this looks more familiar

[scenic olive]

I got

.4/6

1/15

[torn furnace]

you should get two answers

[torn furnace]

if we simplify this

$0.2t^2 - 6t + 0.2=0$

[grizzled pagoda]

jan Niku

[torn furnace]

i dont know what happened here

[scenic olive]

I dropped the t

mb

[scenic olive]

how would we figure this one out

[torn furnace]

youre in precalc

so you must have done algebra?

[scenic olive]

ye

[torn furnace]

how do we find the roots of a quadratic

[scenic olive]

o wait

ill just do quadratic formula

[torn furnace]

idk about this

,calc 4*.2*.2

[grizzled pagoda]

Result:

0.16

[torn furnace]

,calc 36-.16

[grizzled pagoda]

Result:

35.84

[scenic olive]

huh

O wait

yeah

2*.8 would be 1.16

1.6

[lusty saffron]

(i only vaguely followed all of that even though i know calculus but i think i do have an alternate actually-without-calculus way of solving it)

[torn furnace]

which one do you want

[torn furnace]

yea i figured

[scenic olive]

wdym

[torn furnace]

you should share if you have something

[torn furnace]

i mean do you want the + or - or both

[scenic olive]

as you cant have negative hours

[torn furnace]

i think they might both be positive

[scenic olive]

how

[torn furnace]

well you can figure C starts at 0

[scenic olive]

oh

[torn furnace]

and then goes up to 2

and then goes back down to 0

so it must hit 0.2 twice

[scenic olive]

because 6-sqr35.84 is also positive

[torn furnace]

it is

[torn furnace]

,calc 6(30)/(30^2+1)

[grizzled pagoda]

Result:

0.19977802441731

[torn furnace]

yea, seems about right

[scenic olive]

Ye

I'll see

if I can redo tis all from scratch

[lusty saffron]

so, we start with $\frac{6t}{t^2+1} > x$, which is true at some $t$ iff $x$ is less than the maximum value

[grizzled pagoda]

bee [it/its]

[lusty saffron]

rearrange it a bit \ $6t > x(t^2 + 1)$ \ $xt^2 - 6t + x < 0$

[scenic olive]

O no I apologize

I meant that I wanted to try myself

[grizzled pagoda]

bee [it/its]

[scenic olive]

I got the same answer

just wanted to make sure I understood what yall were saying

O mb again

[lusty saffron]

(it* but yes)

[torn furnace]

yea sorry

[scenic olive]

I am very sorry but idk if I should go into a whole nother thing

I gotta finish this last nit of homework and head to sleep

quiz tmmrw

[lusty saffron]

now, the maximum is the x which is exactly on the boundary between this equation being solvable for some t, and it not being solvable for any t

[scenic olive]

I think I generally understand how it works

Apologies

[lusty saffron]

yeah that's fair, i'm doing this more to show jan Niku

[scenic olive]

kk

I'll head out you can close the ticket when your done explaining to him

[lusty saffron]

in other words it's the point where the minimum value is exactly 0, which occurs if $xt^2 - 6t + x$ is a perfect square

[grizzled pagoda]

bee [it/its]

[lusty saffron]

so it's $x(t-1)^2$ and that expands to $xt^2 - 2xt + x$, so $2x = 6$, and $x = 3$ is the maximum value

[grizzled pagoda]

bee [it/its]

[torn furnace]

hrm i dont follow the leap

[lusty saffron]

yeah tbf that is the sketchiest step here

[torn furnace]

xt^2-6t+x<0 if x is less than the maximum value

[lusty saffron]

although once you've used it to identify that you should be looking at x = 3, the rest of the logic is still fine and gives you an actual proof that 3 is the maximum value, i think

well combined with finding the actual number that gives 3 as an output

[lusty saffron]

ok well the minimum of a perfect square is exactly 0, and that's the only way this can happen - if the quadratic has 0 as a value at all, but isn't a perfect square, then values in between the two roots give negative results

[torn furnace]

why is that a problem

[lusty saffron]

which part

[torn furnace]

negative results

[lusty saffron]

well then the minimum isn't exactly 0, it's some negative number

[torn furnace]

why does it have to be exactly 0

[lusty saffron]

good question

yeah i'm still thinking about why exactly that should work

[torn furnace]

i mean its the impulse right

i understand that

you stare at the set up and go yea that should be what we do

[lusty saffron]

if we're allowed whatever tools we want, then, it's continuity

the minimum of the quadratic is continuous as a function of x, we have a region where it's positive, we have a region where it's negative, and the point we're trying to find is the boundary between those two regions, therefore it's going to be 0

ok actually i got it

[torn furnace]

im not really following sorry

[lusty saffron]

with x being the maximum value, this inequality, $xt^2 - 6t + x < 0$, shouldn't have any solutions

[grizzled pagoda]

bee [it/its]

[lusty saffron]

because those correspond to solutions of the original equation $\frac{6t}{t^2 + 1} > x$ which would imply that $x$ actually isn't the maximum

[grizzled pagoda]

bee [it/its]

[torn furnace]

yea, but such an x producing that state doesnt need to be the maximum

or does that not matter

[lusty saffron]

but, because it's exactly the maximum, and therefore is actually a value of the function somewhere, we can replace all of the < with =

because there is some solution to $\frac{6t}{t^2 + 1} = x$, \ there is some solution to $xt^2 - 6t + x = 0$

[grizzled pagoda]

bee [it/its]

[lusty saffron]

so the quadratic is 0 somewhere, but never negative, therefore its minimum value is 0

[torn furnace]

no offense but I'm going to throw in the towel

the fact that my bedtime was an hour ago is compounding the problems here

that i dont think are how you are explaining it

thank you for trying

[lusty saffron]

i mean the fact that it's 5 am for me is probably not helping my explanations

[torn furnace]

i just dont understand your logic

why this roundabout loop was even necessary

[lusty saffron]

(it took me multiple attempts to actually TeX $\frac{6t}{t^2+1}$ correctly and not end up with $6t^2$ or something)

[torn furnace]

why cant we just assume that such an x exists

[grizzled pagoda]

bee [it/its]

[torn furnace]

but it doesnt matter

brains not braining

clearly $t = \frac 1 x \qty[ 3 \pm 3 \sqrt{1 - \qty( \frac x 3 ) ^2 } ]$

[grizzled pagoda]

jan Niku

[torn furnace]

annd this is good enough

[lusty saffron]

...ok my brain is just failing to comprehend this question and i genuinely can't tell at this point if it's a problem with the question or just sleep deprivation

[torn furnace]

another time maybe

https://tenor.com/view/surinoel-soul-chilled-chilling-relaxing-gif-18047038

thanks bee

we can sub in for x

$t = \frac{1+t^2}{2t} \qty[ 1 \pm \sqrt{1 - \qty( \frac{3t}{1+t^2} )^2 } ]$

[grizzled pagoda]

jan Niku

[torn furnace]

actually bee i think i understand what you mean now

about the perfect square

like here, its not enough that 3t/(1+t^2) = \pm 1

we actually need the entire quantity to only have one solution

anyways

bedtime

.close

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@hexed rampart Has your question been resolved?

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[exotic sinew]

how do I eliminate the arbitrar constants of y =ax^2 +Be^2x

[stiff jetty]

is there any context for this

u cant just eliminate it without any conditions

[exotic sinew]

Just says Elimination of Arbitrary Constants and find the differential equation

[stiff jetty]

u need conditions

i cant really help with 0 context

[amber wasp]

@exotic sinew Has your question been resolved?

[past hound]

send a picture of the problem

[exotic sinew]

The second one I think I got it

The first one is giving me trouble

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@exotic sinew Has your question been resolved?

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@exotic sinew Has your question been resolved?

[weak zinc]

Anything that preceded/followed that? There pretty much isn’t anything that will allow those constants to be eliminated/determined from what you’ve provided as is - bear in mind that the…

!original prompt wants the full context, intended to mean stuff before and after it

[amber wasp]

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

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[terse glade]

Hi does anyone know how to solve this? I'm stuck and your help will be greatly appreciated

[ancient snow]

What's exactly the problem stated here?

[terse glade]

ewll the only thing they say is crack the code which will give you three objects

and they basically leave it up to you to figure it out

[lethal steppe]

So it’s a cryptic message that has to be deciphered?

[crisp stratus]

i assume p_i refers to ith prime number

5 - 3rd
7 - 4th
11 - 5th
13 - 6th
19 - 8th

[lethal steppe]

That’s goofy

I got the solution but it’s so random

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[tender schooner]

Why is the answer (y-sqrt(y)) when the sqrt(y) is the outer function radius?

[hollow salmon]

That doesn't look correct.

[tender schooner]

Which of these is right

[hollow salmon]

The second one, sort of.

[tender schooner]

are you sure because someone smart told me the first one. is it that that’s a different method?

[hollow salmon]

Yes, I'm sure.

That's not how you use the Washer Method.

You are essentially finding the volume of the outer boundary and then subtracting the volume of the inner boundary.

Because of integral rules, you can combine the two integrals into one integral.

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[velvet junco]

Am i messing something up?

[median granite]

so with quadratic inequalities

the easiest way to solve it is to factor the quadratic

are you able to do that?

[patent raptor]

looks good so far

[velvet junco]

Hmm

[patent raptor]

now factor

[velvet junco]

Its not matching with this

[median granite]

is this the answer?

[velvet junco]

Yep

[median granite]

the answer is half right

the x values are correct but it should be an inequality

[median granite]

ok this is good

so answer is x<=-3 and x>=1/2

and they obtained the x values from the factored form

do you understand their table of sign method?

[velvet junco]

This is the same way of doing the table right?

[patent raptor]

-5x = -6x+x might work

[median granite]

your factored form looks different

[velvet junco]

This is from a different question

[median granite]

oh

ic

looks about right

were you able to factor the quadratic?

[velvet junco]

Not working

[velvet junco]

Nah uh

[patent raptor]

what are you doing

-2x²-6x+(x+3) = -2(x+3)+(x+3)

[velvet junco]

Of

I got it

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[tranquil pasture]

in this triangle problem, when we solve for z, why do we do 12^2 = z^2 + 6^2, and not 9^2 + x^2 = z^2, i have a test today please anyone help!

[tranquil pasture]

and how will i know when to add the entire side and not just one part

[patent raptor]

12^2 = z^2 + 6^2 refers to the right triangle GHK

[tranquil pasture]

so why do we use the entire triangle to find z?

[patent raptor]

I suppose y is given and x is not?

[tranquil pasture]

yeah its not given

[stiff jetty]

where does it say y is given 6 diretly

[tranquil pasture]

but if we can solve to find it

i solved to find y is 6

[patent raptor]

But both ways are correct, it's just you don't know what's x is as opposed to y

[tranquil pasture]

ohhh okay

the problem tells me to solve for x, y, and z

[patent raptor]

Well you can get z and by that x

[tranquil pasture]

ohhh true

thanks

[wanton sentinel]

How exactly? If you managed to find y I don't understand how you didn't also find x and z at the same time

[tranquil pasture]

first i found x by doing this

then i found y because it was a hyoptenuse of the smaller triangle

[wanton sentinel]

Oh then you can find z the same way

With 9^2 + x^2 = z^2 like you said, except you know x

[tranquil pasture]

when my teacher did it in class she used z^2 + x^2 = 12^2 , so i was wondering why she did that bc thats what i was thinking too

[wanton sentinel]

As long as you get the same answer...

[tranquil pasture]

wait ill check rn if its the same

[wanton sentinel]

I would have just solved a system of equations because I don't have that h=sqrt(nm) formula in mind

[tranquil pasture]

yes i got the same answer!

[tranquil pasture]

ohhhh smart

[wanton sentinel]

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[amber wasp]

@valid cobalt Has your question been resolved?

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@valid cobalt Has your question been resolved?

[patent raptor]

Looks to me you are supposed to find such a and b so that you get rid of the constant -3 and -1

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[split sail]

How do I solve this?

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[split sail]

I need some help

[astral olive]

!1c

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[slender verge]

Back at it again ( i got help to derivate this), but he told me that the a is a constant, but can you assume that really?

In the excercise the x is “about 0”, is it fair to assume that the a is a constant then if it does not say anything about it ?

[astral olive]

Can you send the original question as well please

No context otherwise

[astral olive]

Okay, now I'm really confused why you thought a may not be constant

a = e, and e is a constant

[slender verge]

Well that is the “next question” if so to say

The first is to compute the Taylor polynomial

From that first sentence, can we assume that the a is constant?

[astral olive]

I mean, you are told to assume a = e, and e is a constant, so yes

[shadow stump]

a can be anything but it doesn't depend on x so you treat it as a constant

[astral olive]

Anything other than an x term yeah

[slender verge]

Like this

This should be the correct way right ?

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[hallow gull]

I have to get this final answer (i.e. 3623) into the form x ≡ a_k(mod m_k). So I know that I need to get it into mod330

In my textbook I have this but not sure how they got it

Where the first formula equation is

[amber wasp]

@hallow gull Has your question been resolved?

[hallow gull]

wait nvm ! I got it

[weak zinc]

Just as I started typing

[hallow gull]

I still have one question though just for clarification! So we essentially are doing this? 3623 ≡ rmod330.

[weak zinc]

Pretty much, as before, you can add/subtract any multiple of 330 and you’ll have something equivalent, so, say, 323, or -7 are both “good” choices (the first one as it’s between 0 and 329, the second as it’s a number that’s smaller in absolute value so easier to do arithmetic with)

[hallow gull]

Okay great thank you : )

[hallow gull]

Also, I used the method of back substitution for a question. I always get confused by the inverse step. Do I need to do the whole extended euclidean algorithm to find it or is there a quicker way?

[weak zinc]

There is a quicker way (this was something I wanted to comment on before, but thought better of it)

Remember how I said that you can add/subtract numbers, and how I alluded to previously that -7 is smaller in absolute value?

[hallow gull]

yeah

[weak zinc]

Here, you know that 6 is equivalent to -1, so then you're reduced to thinking about what to multiply -1 with, such that mod 7, you get 1, but then, of course, -1 would be a good choice to use

You could try the whole Euclidean algorithm, and find that 6's inverse mod7 is 6, cause of course 6* 6 = 36 = 7(5) + 1, but that saves a bit of time

[weak zinc]

In your working too, using that same idea, 165 is equivalent to 1 mod 2, so its own inverse, 110 is -1 mod 3, 66 is 1 mod 5, 30 is -3 mod 11...

The first three, it's pretty immediate to see their inverses in the new form

[hallow gull]

Hmm yeah see idk why but my mind has trouble seeing this all in my head. Maybe you can help explain with this example that my prof showed? I am not really sure where the conclusion that we needed to find the inverse of 3mod7 came from in this

but it is a similar question

I just feel like seeing it already completed will help me

Sorry if I am basically ask you to repeat the saem thing again T_T . I am a bit slow

[weak zinc]

In this one, you're dealing with quite small numbers already, so for 3's inverse, you wanna find some number such that when you multiply it by 3, you get 1 more than some multiple of 7

Of course, 3 * 5 is 15, and that's 1 more than 14, a multiple of 7 (times tables )

So you know that 3's inverse mod 7 is 5 (or any integer equivalent to 5 mod 7)

[weak zinc]

For your case, finding some integer such that if you multiply 110 by it, you get 1 more than a multiple of 3 is a bit pain, but finding a number such that if you multiply -1 by it, you get 1 more than a multiple of 3 is nicer to deal with (also remembering all the modular arithmetic properties)

"ah, if i multiply -1 by -1, I clearly get 1, so -1's inverse must be -1 mod 3(!)"

[hallow gull]

Okay that makes sense, I think I get most of, I just don't know how I know that I want "1 more than some multiple of 7". Is that just by the theorem of an inverse that it a times a^-1 is equivalent to 1(modn)?

[weak zinc]

Yep, that's it, a * a^{-1} = 1 mod(n) is saying e.g. that a * a^{-1} = 1 + kn for some integer k, i.e. "the product is 1 more than some multiple of n"

[hallow gull]

Okay so then going back to this one. On the side there am I showing it correctly?

hmm I do't think so

[weak zinc]

It could be a bit clearer: taking from here, you know $6 \equiv -1 \mod(7)$, so your equation is now $(-1)s = 1 \mod(7)$ (from where, it should be instant that -1's inverse mod 7 is -1)

As for why 6 is -1 mod(7), if you really need to demonstrate it, you could of course say that 6 = 6 - 7 = -1 mod(7)

[grizzled pagoda]

@weak zinc

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[hallow gull]

Can anyone help me find my error in this? It should be 53mod60 not 25(mod60)

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[past island]

im really stuck on this bruh

conjugate axis on the line y=1
one vertex at (-2,5)
asymptotes with equations 4x-5y+13=0 and 4x+5y+13=0

i managed to find the center using the interescting point of the asymptotes which should be at (-13/4,0)

but the given vertex is at (-2,5)
shouldn't atleast one of their values be common (center and vertex) so that we can tell if its either horizontally-aligned or vertically-aligned?

i also tried plotting the given in desmos and i got this. WHICH DOESNT MAKE SENSE AT ALL 😭

1.) the conjugate axis passes through the center right? so based on this, the center would somewhere at (?,1)

2.) the intersection point of the asymptotes is where the center is, based on this, the center should be at (-13/4,0) or (-3.25,0) in decimal form, which already contradicts with my first point 😭

3.) shouldnt the center be alligned with one of the points of the vertex? so if the vertex is at (-2,5) then the center would be at (-2,?) or (?,5) right? WHICH ALSO CONTRADICTS MY FIRST TWO POINTS

WHERE IS THE CENTER BRAHH

[past island]

Pl

Pls

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[deft hound]

Good evening, how are you?
I have a question about Fermat's theorem.

Fermat's theorem says that if a function f has a relative extrema at x=a and f is differentiable at x=a, then f'(a)=0.

To apply Fermat's theorem, do I need to ensure that the point where f has a local extremum is a right and left accumulation point of the domain? Or is it enough that it is a relative extreme?

As I understand it, if there is a relative extreme in a, by definition, there must exist a neighborhood with center a and radius delta such that every x that is in this neighborhood has an image greater than/equal to f(a) (as appropriate, if relative minimum or maximum). With this, that point a must be inside the domain and therefore it is an accumulation point to the right and left, without the need to clarify that.

Furthermore, for f to be differentiable at x=a, that is, for f'(a) to exist, point a must belong to the domain and also be an accumulation point; and if the lateral derivatives are not clarified, it should be understood that it is the derivative at that point and that it is the point of accumulation to the left and to the right.

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.close

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[proven palm]

ill send a picture of myw ork in a second awk

im pretty sure i got the slope right?

but then the issue is how do i get the x in sinx = 2/π

and theres no calculators or anything like that allowed

just like

paper and pencil

[elder creek]

uh

just arcsin(2/pi)

and thats ur answer

[proven palm]

i can just say that,,, as an x coordinate,,??

[elder creek]

ye

[proven palm]

oh my god

[elder creek]

x=arcsin(2/pi)

arcsin(2/pi) is a value

[proven palm]

wait

your right sorry i keep thinking it needs to be like a number nunmber

youre

fhoijk

thakn you

.close

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[quartz hornet]

This is getting me

It’s either first or second

<@&286206848099549185>

[inland pulsar]

O has 8 elec and F has 9

Should be second

@quartz hornet

[quartz hornet]

Do you see the 2-

[inland pulsar]

8+2 isn't 9

O- and F would have been the same

[quartz hornet]

But Florine is a flouride

So it’s a -

[inland pulsar]

No

F means F

[white sundial]

Nope, F- is fluoride

[inland pulsar]

If they meant F- they would have said so

[quartz hornet]

So it isn’t the first?

[inland pulsar]

.

[quartz hornet]

Badass

[inland pulsar]

Lol

[quartz hornet]

What’s it asking for here

[white sundial]

Second second

[quartz hornet]

Almost done couple more questions

[inland pulsar]

I hope you are not cheating on a test

[quartz hornet]

This is a assignment

Who has a test at 8pm 😭😭😭

[inland pulsar]

It's 8 in the morning for me lol

[quartz hornet]

It’s 7:50PM

[quartz hornet]

What’s this asking

[inland pulsar]

8:20 am

If you have Na+ and SO4 2-

And someone asks you to write the compound

Then you write Na2SO4

Does this ring a bell?

[quartz hornet]

To

Yup

[inland pulsar]

Doe the same for ammonia and phosphate

Do you know their symbols?

[quartz hornet]

NH_4

Is ammonium

Right?

Wait

[white sundial]

NH4+ is ammonium

[quartz hornet]

Shouldn’t it be the first

Cause of the 3

Since PO^3

So there needs to be a 3

@inland pulsar

[white sundial]

$\ce{PO_4^{3-}}$

[grizzled pagoda]

Kex

[quartz hornet]

Yup

So therefore

The first one is correct

[white sundial]

So yes, you need 3 ammonium

indeed

[quartz hornet]

Potassium is 19

Chloride is 17

So should ie be the last one

Cl2-

[white sundial]

Which period is chlorine?

[quartz hornet]

-1

Potassium is +1

So it would be

The second

One combines with one

[white sundial]

Yes

[gloomy tide]

https://discord.gg/eexdsFw

[inland pulsar]

Yeah first is correct

[quartz hornet]

They suck

2 more questions and I’m out ur hair

Hw due tonight

So for this one

X Must be greater by 3

Since Y needs 3 to make it even

[white sundial]

Yes, so?

[quartz hornet]

So it should be the second

Why can’t It be the 4th

1 to 1/3

[white sundial]

Check the signs

[quartz hornet]

Ahhh both positive

[gloomy tide]

not really our problem. to everyone else, if you want to help him with this, join that server as well

.close

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[quartz hornet]

.reopen

[amber wasp]

✅

[quartz hornet]

@white sundial @inland pulsar dm me

.close

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[alpine lodge]

I need help with a math algorithm to be put in code. This is for a computer network operation and a few neural net that is behide the decider bus logic making the choices. This math problem sits at the kernel level,
This is a story base problem for me. I do not know how to do the math that goes into the neural net. but I want someone to solve my math problem so that it to be uses in this niche neural nets. This is about computer governance.

[pallid canopy]

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@alpine lodge Has your question been resolved?

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@alpine lodge Has your question been resolved?

[alpine lodge]

what the the 64 numbers of pi?

What is the 64 numbers of pi?

what is pi?

[pallid canopy]

This is an incoherent question

pi is a number defined to be the ratio of a circle's circumference to its diameter

There are no "64 numbers" of pi without more context

[white sundial]

Maybe he meant digit(s).

In that case, you could use the BBP formula.

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@alpine lodge Has your question been resolved?

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[nocturne sierra]

Can anyone help me pleasee

[nocturne sierra]

I've found the sides but I cant seem to find the height of the pyramid thats in the cylinder

[hearty sleet]

Right triangle no?

[nocturne sierra]

No i think its an equilateral

Im pretty sure so

[hearty sleet]

My visualization skills aren't good enough for this holy

[white sundial]

Top one is equilateral, others are right

[hearty sleet]

Can't you find height if yk sides + equilateral?

[nocturne sierra]

What?

Uhh please explain

Hello?

[white sundial]

$V=\frac{1}{3}Ah$

[grizzled pagoda]

Kex

[white sundial]

With A being the area of the equilateral triangle and h the height of the tetrahedron. I would start with the easier part, which is the area of the equilateral triangle that inscribes the circle.

[nocturne sierra]

I have, its 12 sqrt 3

How do I do that?

$A=12\sqrt{3}$

[grizzled pagoda]

Reefus

[nocturne sierra]

Oh ok yeah

That

[alpine lodge]

What are the numbers of pi to the 64 number?

[white sundial]

!occupied

[amber wasp]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[alpine lodge]

....

[white sundial]

As for the height, you could start determining the sides of the other triangles. They are isosceles and rectangular.

[shadow stump]

at this point you need to get a new help channel. yours timed out due to lack of timely response and has been claimed by someone else

[alpine lodge]

I opened this channel to ask help so I can get my story based questions answers and solve buy someone else.

[alpine lodge]

Oh...

👋

[nocturne sierra]

Wait...

Is it possible that it might be an equilateral pyramid?

Or no?

[white sundial]

no, it's not a regular tetrahedron

the top triangle is equilateral, but the other three are only isosceles.

[nocturne sierra]

Okay

[white sundial]

keep in mind that the tip that points to the bottom of the barrel has right angles, since it's a cube.

[nocturne sierra]

Okay i found it

The area of the top is 12 sqrt 3 and the sides are 12

Now what?

[white sundial]

Sides of the equilateral triangle or the other equal sides of the isosceles triangles?

[nocturne sierra]

The other ones

The three identical sides

[white sundial]

You need the equal sides of the isosceles triangles.

Since those form a rectangular triangle, with the other sides being the radius of the barrel and the height of your tetrahedron, which is what you ultimately want.

[nocturne sierra]

Oh you mean the diagonal lines?

I suppose its 2 sqrt 6

[white sundial]

They should be equal one side of the equilateral triangle divided by sqrt(2)

Yes, essentially the relation between the side and diagonal of a square

[nocturne sierra]

Yeah, 4 sqrt 3 divided by sqrt 2 would be 2 sqrt 6

$4\sqrt{3}/\sqrt{2}=2\sqrt{6}$

Oh wait

I meant 4 sqrt 3

My bad

[grizzled pagoda]

Reefus

[nocturne sierra]

Better

[white sundial]

And now the height

[nocturne sierra]

How

[white sundial]

#help-41 message

You can try to draw it for yourself if it helps. I'm not good at it 😅

[nocturne sierra]

I still dont get it ;-;

[white sundial]

Well, I tried, maybe it does help

x being what you just calculated

[nocturne sierra]

Ohhh

Oh right I completely forgot the corner to the centre of the triangle is the same as the radius of the circle

[white sundial]

Well, I did say inscribe earlier 😄

[nocturne sierra]

Sorry I dont quite get the words from earlier

Not so good with words

So uhh

$V=6\sqrt{6}$

[grizzled pagoda]

Reefus

[nocturne sierra]

?

Oh, also ft³

Thanks a lot though!

Byee

.close

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[nocturne sierra]

.reopen

[amber wasp]

✅

[nocturne sierra]

A bit miscalculated

I meant $V=8\sqrt{6}$

[grizzled pagoda]

Reefus

[white sundial]

Ye, I didn't really calculate it myself, just wanted to guide you from a conceptual point of view.

[nocturne sierra]

Yeah

Again, thanks a lot!

Byeee

.close

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[exotic scroll]

Hey guys
Can I get help with math ?

[amber wasp]

Please don't occupy multiple help channels.

[inner prawn]

just ask brotha

[exotic scroll]

Card game

A deck contains 52 cards, 13 of each of the four suits hearts, spades, clubs and clubs. The cards are numbered by denomination in descending order, Ace, King, Queen, King, 10, 9, 2, so Ace is the most valuable, but can count as 1 in straights

Ida and Matte play a variant of poker where at the beginning they have two hidden cards in each hand and three cards are face up on the table. The best poker hand that can be formed from the two cards on the hand and the three on the table is of interest for further play and betting. The most valuable poker hand is straight (five cards in a row and of the same suit). Next, in descending order, fours, house (three and pair), suit (all of the same suit), straight (five cards in a row), three, two pair, one pair, and highest card. If two players have the same type of hand, the one with the highest number on the trick wins if both have a house, on the highest pair if both have two different pairs, on the second highest pair if both have the same highest pair, or on the highest card if both have the same two pairs If they also have the same denominations (eg two pairs with the same two pairs and the same other cards) it is a tie and no one wins. No color is worth more than any other

Task 1: Assume that there are hearts Ace, hearts King and spades Queen on the table and that Ida has diamonds Ace and clubs Queen in her hand. Ida thus has two pairs.

(a) How many different pairs of cards can Matt have in his hand? (Note that the expression "pair of cards" in this and the following questions refers to two cards and not to "pair" in the sense of two cards of the same denomination.)
(b) How many different pairs of cards can Matt have that gives a better poker hand than Ida's?

I solved a

Now it’s b

We have 6 different ways

Three of same
5 in order
5 Same colour
Thee of same kind and pair
4 of same kind
5 Same colour and in order

@inner prawn

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@exotic scroll Has your question been resolved?

[quartz hornet]

https://tenor.com/view/wesker-albert-albert-wesker-resident-evil-resident-evil-4-gif-5143688443782660001

[cursive ermine]

woah, words

[amber wasp]

@exotic scroll Has your question been resolved?

[exotic scroll]

Its a game, just a descriotion 🥲

[cursive ermine]

terrifying

[amber wasp]

@exotic scroll Has your question been resolved?

[amber wasp]

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[shadow galleon]

$\int \frac{r^2 \cdot e^{2r \sin(\theta)}}{r\sin(\theta) + 2} dr$

[oblique marlin]

HELP RN

[cunning birch]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[grizzled pagoda]

alee

[shadow galleon]

Can this integral be solved?

[ruby dome]

Context?

[cunning birch]

if I had to guess, I'm pretty sure there's some exponential integral in there

e^x/x has Ei(x) as its primitive

and it's not pretty

[shadow galleon]

What could I do?

[cunning birch]

maybe add more context

where does this come from

are there bounds?

[shadow galleon]

$\iint_D \frac{xe^{2y}}{y+2} dxdy$

[grizzled pagoda]

alee

[shadow galleon]

where D is the plane domain between the unit circle with center at the origin and the ellipse with semi-axes a = 4 and b = 2 and contained in the first quadrant

[grizzled pagoda]

alee

alee

[ruby dome]

what should be the result?

[grizzled pagoda]

alee

alee

[shadow galleon]

$\int_0^{\pi/2} \int^{\frac{4}{\sqrt{\cos^2\theta + 4 \sin^2 \theta}}}_1 \frac{r^2 \cos(\theta) \cdot e^{2r\cos(\theta)}}{r\sin(\theta) + 2} dr d\theta$

[ruby dome]

Mmm

Why are we missing an extreme?

[grizzled pagoda]

alee

[shadow galleon]

@cunning birch

[ruby dome]

Where did you get this exercise from?

[shadow galleon]

So taking the cos outside since I will have to integrate in dr I get this

[shadow galleon]

.

Does anyone have any ideas on how to calculate this?

[patent raptor]

interesting

[patent raptor]

Maybe do a substitution to get rid of the y+2

y = rsin(theta)-2

then it basically becomes a xe^x problem

[grizzled pagoda]

bacc the sigma😔🤞

[patent raptor]

I am just afraid if the 2nd integral with theta will give something neat

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@shadow galleon Has your question been resolved?

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