help-33

[marsh citrus]

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[cobalt yacht]

How do I approach this problem?

[still temple]

,calc sqrt(2005)

[elfin berry]

Result:

44.777226354476

[still temple]

,calc sqrt(2005*2)

[elfin berry]

Result:

63.32456079595

[still temple]

ok nvm thats going to take way too long to get to 2005^2 lol

[cobalt yacht]

Yep

[still temple]

oh wait, you could probably show that every number always gets hit

[dense solar]

I don't think thats true

[still temple]

its got to be true up to some point

not sure what that point is tho

[dense solar]

(n+1)^2 = n^2 + 2n + 1

[still temple]

oh yeah

nice

[dense solar]

when the 2n + 1 part is more than 2005, you're going to be adding at least one on every step

[cobalt yacht]

So when we have

n>=1002

actually

n>=1003

So there are 1003 of these

But what about n<=1002?

No wait

It is n>=1002

So there are 1004 of these

And we have to consider n<=1001

[still temple]

,calc 1001^2/2005

[elfin berry]

Result:

499.75112219451

[cobalt yacht]

I see

500 + 1004

1504

Thanks @still temple and @dense solar

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[marsh citrus]

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[marsh citrus]

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[eternal edge]

I have to define an inner product such that a given basis is orthonormal. I've tried thinking about inner product as a linear combination of all possible products when having spaces over R or C, but the approach didn't work as the result wasn't actually an inner product. Then I tried to think about inner product with the elements of my basis as an element of the dual space, but I'm getting very close to the same result as before. What can I do to show how the inner product is?

[marsh citrus]

@eternal edge Has your question been resolved?

[eternal edge]

<@&286206848099549185>

[eternal edge]

<@&286206848099549185>

[marsh citrus]

@eternal edge Has your question been resolved?

[marsh citrus]

@eternal edge Has your question been resolved?

[marsh citrus]

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[runic temple]

1+16=17

[crystal plover]

ive been on this for an hour now - I am not understanding how they get from step 2 to step 3

[runic temple]

oh its negative nvm

[crystal plover]

np lol

[runic temple]

what line to what line

[crystal plover]

from this to this

[runic temple]

oh ok

[crystal plover]

I understand 2ln15 is the same as ln225

[runic temple]

yes

[crystal plover]

but then in the next line, where does 16 come from? is it just the lowest multiple of 4 and 4? if so, where does 2 go?

I feel dumb right now

[runic temple]

so $(2ln15)^4=2^4 * (ln15)^4$

[elfin berry]

GarlicBredFries

[runic temple]

and then in the other fraction they multiplied the top and bottom by 16

[crystal plover]

ooohhh so the power rule for logs next?

[runic temple]

no, exponent rule

[crystal plover]

oh i see, just evaluating 2^4

ahhhh

2^4 = 16, so 16*4(ln15)^4

correct?

or at least thats what they're doing

nice! i got it

tysm i am brain fried

[runic temple]

no problem, good luck

[crystal plover]

❤️

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[marsh citrus]

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[marsh citrus]

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[frosty flicker]

can i please get help with b

[ripe otter]

First part is getting the amplitude

Do you see how much it fluctuates?

[frosty flicker]

yes by 10

so 5 down 5 up no?

the centre is 7

[ripe otter]

Yeah

So amplitude is 5

[frosty flicker]

my biggest problem is knowing the pi

is it pi/3? pi/6? pi/12?

how will i know?

[ripe otter]

Well we start at when the tide is at it's highest

And at what radian value is sin at it's highest

[frosty flicker]

pi no?

[ripe otter]

Nah

Look at the unit circle

[frosty flicker]

pi/2

[ripe otter]

yeah

So you want your sine function to be offset by that

[frosty flicker]

how do you know that

[ripe otter]

Because the function should at 12 then go down to 2

Aka start at when it's the biggest

[frosty flicker]

okay so now we know the amplitude is 5

and that it starts at its maximum

meaning pi/2

[ripe otter]

Yeah so right now we have: 5 * sin(x + pi/2)

[frosty flicker]

yes

[ripe otter]

But it's also offset vertically

It needs to go up an X amount of values in the y axis

[frosty flicker]

7

[ripe otter]

yea

[frosty flicker]

so it can go down 5 to 2 and up 5 to 12

[ripe otter]

so how does your function look now?

[frosty flicker]

perfecto

thank you

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[marsh citrus]

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[marsh citrus]

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[rigid falcon]

Hello, how do I find the domain (without graphing) in set builder notation if im given -2x^2 + xy^3 - 3 = 0

[edgy jungle]

Try to find values that won't work ig

[rigid falcon]

Well obviously, but im wondering how

[edgy jungle]

Like for sq root

Stuff inside ≥0 n stuff

Write y=f(x)

[edgy jungle]

IS That cube only on y

[rigid falcon]

huh?

yeah

[edgy jungle]

Do this ig

[rigid falcon]

the domain is supposed to be everything but 0

but what is the reasoning behind it

[edgy jungle]

XY^3 = 3+2x^2

y^3=(3+2x^2) /x

[rigid falcon]

Yeah that reasoning works

i think you can do it just by inspection now since when x=0, you'll get -3=0 but every other value of x works

[edgy jungle]

Yes

[rigid falcon]

but thank you, that's a good technique to use for my other questions

.close

[marsh citrus]

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[edgy jungle]

https://youtu.be/vwK3gs4XY6I

[edgy jungle]

Check this out if u want

[rigid falcon]

ty 🙏

[marsh citrus]

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[rustic hamlet]

Im in calc bc and curve sketching for me has been the worst. Can anyone help me understand this? Thank you.

[trim quest]

Is there any particular part you don't understand?

You have a little leeway here, there's not one specific function that fits these criteria

[rustic hamlet]

Yeah I forgot to add in the top part. It says to write what the function could look like

It’s just that we’ve only ever done curve sketching with a specific function

[trim quest]

Like sketch a graph, right?

[rustic hamlet]

Yea

That’s it just sketch a graph

I don’t know where to start

Also thank you for responding so quickly

[trim quest]

Lol no problem I just got in here 🤷

I'd start with the three specific points given, f(-2) = 8, etc

That gives you a few concrete points to work with

[rustic hamlet]

Lemme plot it rq

[trim quest]

ok

[edgy jungle]

https://youtu.be/mQdXeAY6_f0

[rustic hamlet]

Thx I’ll watch it

@trim quest

[trim quest]

Ok, looks good

The next few criteria listed deal with the derivative

so specifically f'(-2)=f'(2)=0

Do you understand what that means?

[rustic hamlet]

When f’(-2) = f’(2) ?

[trim quest]

f'(-2)=f'(2)=0 can be read as "f'(-2) and f'(2) are both equal to 0"

[rustic hamlet]

Ohhhhh ok

That makes more sense

So would I plot that as well?

[trim quest]

Do you understand what it means about the graph?

[rustic hamlet]

No unfortunately

It’s probably something simple that im forgetting

[trim quest]

It's saying the derivative is 0 at those two points

It has a "slope" of 0 there

[rustic hamlet]

Oh yeah slope is derivative

[trim quest]

Yeah, maybe that one's not as important as the others

Like the next one,

f'(x)>0 where |x|>2

[rustic hamlet]

Yeah I don’t understand that at all sorry

Derivative is greater than 0 when the absolute value of x is greater than 2?

[trim quest]

Yeah like

Where is |x| > 2 ?

[rustic hamlet]

On the graph?

[trim quest]

Like, for what values of x is that true?

[rustic hamlet]

Any value above 2?

Negative or positive?

Like -3 and 3 would work

[trim quest]

yeah, so any value greater than 2, or less than -2, right?

[rustic hamlet]

Right?

[trim quest]

yeah

[trim quest]

So this is saying, for those x-values, the derivative is positive

That is, the graph is increasing

[rustic hamlet]

Ok…

So how would a graph like that look like

It would just go up from x=2?

[trim quest]

Just based on that? We pretty much have this so far

Draw it lightly if you're going to though, we'll end up wanting something curved when we're done

But do you see how the graph is increasing on the intervals greater than 2 and less than -2?

[rustic hamlet]

Yeah

[trim quest]

So then, what about the next part

f'(x) < 0 for |x| < 2

[rustic hamlet]

So any number less than two?

[trim quest]

any number between -2 and 2

the derivative is negative there

[rustic hamlet]

Yeah that’s what I meant

Cuz the 0 is there

[trim quest]

Yeah, so the slope is negative in that interval, the graph is decreasing there, like this

[rustic hamlet]

Ahhhh ok now it’s starting to look like a curve

[trim quest]

Yep

This still isn't quite right though

Because at x=2, and x=-2, the derivative is supposed to be 0

Sharp points like this aren't really what we want

[rustic hamlet]

Yea because you can’t have a derivative at a sharp curve right?

[trim quest]

Exactly

[rustic hamlet]

So does concavity fix that?

[trim quest]

It just might

[rustic hamlet]

That’s why the double prime is there

That tells us concavity?

[trim quest]

Yep

[trim quest]

f''(x) < 0 for x < 0

[rustic hamlet]

Concave down?

[trim quest]

for x less than 0, yes

[rustic hamlet]

How would that look on a graph tho

Just a downwards curve?

[trim quest]

Concave down means the slope is decreasing as the graph goes on

Like the top half of a circle

starts with a positive slope, then gradually the slope decreases until its negative

[rustic hamlet]

Upside down parabola?

[trim quest]

yep

[trim quest]

Looks good 👍

Notice that also matches the condition we skipped

f'(-2) and f'(2) are both equal to 0

[rustic hamlet]

It equals 0 because it’s a curve and not a sharp point?

[trim quest]

It equals 0 because a tangent line there would have a slope of 0

[rustic hamlet]

OHHHHH

yeah ok gotcha

[trim quest]

like this

[rustic hamlet]

Thanks the question above asks

[rustic hamlet]

Oh yeah I do remember this

Since it’s flat right there the derivative is zero

[trim quest]

yeah

[rustic hamlet]

But if it is flat vertically it can’t have a derivative

[trim quest]

true

That wouldn't even be a function, though

[rustic hamlet]

Yeah it doesn’t pass the vlt

[trim quest]

right

[rustic hamlet]

What can we conclude about f(3) tho

Not sure how that is relevant to this problem

[trim quest]

Not a whole lot

but not nothing either

[rustic hamlet]

There are 3 other graphs I have to draw tho

[trim quest]

Is there anything at all you can say about f(3)?

[rustic hamlet]

Isn’t it just a point?

[trim quest]

it's the y value at x=3

[rustic hamlet]

0?

No

[trim quest]

f(3) = 0?

[rustic hamlet]

Not zero

[trim quest]

why can't it be 0?

[rustic hamlet]

Cuz f(2) is 0?

[trim quest]

and

[rustic hamlet]

f’(2) =0?

[trim quest]

You're so close

[rustic hamlet]

Tangent line something? Lol

[trim quest]

what about f'(x) for x > 2 ?

[rustic hamlet]

Concave up??

No

[trim quest]

first derivative

[rustic hamlet]

Increasing my b

[trim quest]

right

[rustic hamlet]

So it can’t be 0

Cuz it’s increasing

[trim quest]

Yep

Could it be -1 ?

[rustic hamlet]

No

[trim quest]

So specifically, what do you know about it?

[rustic hamlet]

It is greater than 0

[trim quest]

Yeah, exactly

Doesn't seem like there's anything else we can say with certainty

[rustic hamlet]

Would it be the opposite for f(-3)?

Well no

[trim quest]

More or less, not with respect to 0 though

[rustic hamlet]

f(-3) isn’t even on the graph tho

Wait I’m dumb

Nvm

Ignor that lol

[trim quest]

Lol all good

[rustic hamlet]

It would have to be less than -2

[trim quest]

Not quite

It has to be less than f(-2)

which is 8

[rustic hamlet]

Oh true I was just thinking that

Cuz the graph is decreasing

It would be less than 8

[trim quest]

yeah*

You've got the right idea

but I wouldn't use the word decreasing

[rustic hamlet]

Negative?

[trim quest]

Technically its increasing there, you're just reading the graph backwards

you're going from right to left

[rustic hamlet]

Oh yeah true

It would still be less than tho right

[trim quest]

Yes

f(-3) < 8 for sure

[rustic hamlet]

Ight what about the extrema

Absolute and relative

[trim quest]

Do you know what that means?

[rustic hamlet]

Extrema is just when derivative equal 0?

No that’s critical point

Extrema is max and min?

[trim quest]

right

[rustic hamlet]

Well max would be y=8 and min would be y=0 right?

[trim quest]

relative max and min, yes

[rustic hamlet]

What’s the difference between relative and absolute

[trim quest]

absolute max would be the greatest output of the function anywhere on the domain

[rustic hamlet]

Oh would that be infinity in our case?

[trim quest]

But it increases to infinity on the right of x=2,

I'd just say there is no absolute max or min

[rustic hamlet]

Just relative?

[trim quest]

yes

[rustic hamlet]

Cool

Now for inflection points

This is when concavity changes right?

[trim quest]

yes

[rustic hamlet]

Would that be the same for relative min and max?

[trim quest]

what do you mean?

[rustic hamlet]

Since at y=8 and y=0 the concavity is changing would those be inflection points and relative mins and maxes?

[trim quest]

The concavity doesn't change at those points

[rustic hamlet]

Really?

[trim quest]

Relative extrema occur when the first derivative changes sign
Inflection points occur when the second derivative changes sign

[rustic hamlet]

So where would an infection point be on the graph?

[trim quest]

Where does the second derivative change sign?

[rustic hamlet]

X=0?

[trim quest]

right

so the inflection point is (0,4)

[rustic hamlet]

Ahhh ok

[trim quest]

its concave down before that point, concave up afterwards

[rustic hamlet]

Something like that?

[trim quest]

Yeah, I'd label (-2,8) and (2,0) as the relative max and min respectively

but you answered everything

[rustic hamlet]

Awesome

I have 3 more you think you could help me on those to?

[trim quest]

I don't think I'm gonna be on much longer, sorry

[rustic hamlet]

Alright well thank you very much for the help I greatly appreciate it

[trim quest]

No problem 👍

I'll be up later so feel free to dm them to me if you want me to check over them or something

[rustic hamlet]

Will do thanks

@trim quest if you are still on I tried another one I think it’s close but it’s missing something

<@&286206848099549185> sorry to do this and I hate to be needy but I need help

[marsh citrus]

@rustic hamlet Has your question been resolved?

[rustic hamlet]

I have more questions

Can anyone help?

[main idol]

Just close this and open a new channel

[rustic hamlet]

.close

[marsh citrus]

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[wise jackal]

find (f^-1)'(a) (inverse derivative)

[wise jackal]

okay so I did 6 = x^3 - 4/x to get the inverse value

into my calculator

but it has 2 values, -0.64 and 2

so does that mean this question has 2 answers?

I feel like it's obviously yes but I haven't had a question like this so I'm kinda worried one is extraneous or something

[tulip idol]

uhh

you know it's asking for derivative of the inverse function right?

at a = 6..

so evaluating the inverse function at x = 6 doesn't help whatsoever (@_@;)

[wise jackal]

uhh

yeah

I'm trying to find (f^-1)'(6) which equals 1 / f'(f^-1(6))

and I'm finding f^-1(6) rn

so why wouldn't I evaluate the inverse function at x = 6?

@tulip idol

[tulip idol]

wut?

you need to find the derivative of inverse function... and whatever derivative you get.. you plug x = 6 in that...

so like if f(x) = x²/5 the inverse is f^-1 (x) = √(5x), the derivative is (f^-1)'(x) = √5/2√x and so (f^-1)'(5) would be 1/2

... that's how it works

[wise jackal]

I know that but my teacher said you can't get the inverse of these kind of functions that easily

so you have to use the formula that she taught us

g'(x) = 1/f'(g(x)) where g(x) = f^-1(x)

I'm pretty sure I have this part right, I've been doing other problems abt this

my question rn is just if there can be 2 answers

but I'm pretty sure you can for these weird kinda functions now that I graphed it

i gotta sleep

so thanks ig

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[marsh citrus]

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[still temple]

it seems to me that it is not right

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[upper gust]

can someone pls explain me the intuition between these 2
m is element of the natural numbers

[gritty pagoda]

https://en.wikipedia.org/wiki/Geometric_series if you scroll down to "sum formula" there's a picture and a description for the geometric derivation of this

might be what you are looking for

[upper gust]

thanks

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[hallow kettle]

For (a) I'm not fully sure on how to go about the question

[hallow kettle]

I was considering using the inverse cdf transformation but I'm thrown off by the X-1

Should I let U be a random variable equal to (X-1) and then Y =U**2 and then find du/dy to be 1/(2sqrt(y)) and then multiply that by the pdf of u which is equal to the pdf of X with x-1 subbed in?

[solid echo]

Not too sure if this works, but you can case-base the different parts of (X-1)^2:
https://www.probabilitycourse.com/chapter4/4_1_3_functions_continuous_var.php

[marsh citrus]

@hallow kettle Has your question been resolved?

[hallow kettle]

Thank you

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[marsh citrus]

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[cobalt yacht]

What’s the fastest way to evaluate 8^2007 in mod 13?

[devout mauve]

fermats little theorem

[cobalt yacht]

Hm

[hushed egret]

8^2 mod 13 = -1

[cobalt yacht]

But 2007 is not prim?

[cobalt yacht]

Oh

(-1)^1003

x 8

[hushed egret]

a breeze

[cobalt yacht]

So it’s 5 mod 13

Aight

Thx

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[marsh citrus]

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[still temple]

using pigeonhole principle

[still temple]

.close

[marsh citrus]

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[inner acorn]

on the second line, how is it that we don't have to multiply by 2x at every instance we substitute x^2 +5? Is this correct?

[tiny abyss]

pretty sure this is correct and that you only have to multiply the derivative outside once

[inner acorn]

wow

welp

that settles it

thanks

.close

[marsh citrus]

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[tiny abyss]

np

[marsh citrus]

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[west sigil]

Hi

[west sigil]

I am currently on my first semester in uni

and starting calculus

[bright jay]

No need for a backstory

[west sigil]

sure

[bright jay]

• Show your work, and if possible, explain where you are stuck.```

[west sigil]

prepping right now what i am unsure about give me a minute

[dusk coral]

I am expecting a full anime-style intro tbh 🫡

[west sigil]

I'm looking for the horizontal asymptote

[high pine]

limit at +/- inf

(in fact in +inf, since we have x inside log)

[west sigil]

Yeah and now I'm getting that, since I have inf/inf, I should apply De L'Hospital

[high pine]

so what's the problem, derivatives?

[west sigil]

Yh I'm messing it up the xlnx

[high pine]

$\ProductRule$

[elfin berry]

Modus

[high pine]

where f(x) = x and g(x) = ln(x)

[marsh citrus]

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[viral lily]

,calc 25+10

[elfin berry]

Result:

35

[split ridge]

how?

i got 29

[viral lily]

[marsh citrus]

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[hollow dock]

Could you please check if these answers are correct?

[marsh citrus]

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[honest perch]

just want to be sure if this is ok?

specifically if i calculated the cos-1 correctly with the negative fraction i got as my answer

[marsh citrus]

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[honest perch]

<@&286206848099549185>

[leaden minnow]

yes that looks correct

[honest perch]

lets gooo thx bro

[leaden minnow]

no problem

[honest perch]

hey so in this problem... they r wanting to compare the distances... which i marked in blue??/

also the red line with the length between the two stations... is that accurate as the arrows seem to be a bit longer than the corner of the triangle itself

[leaden minnow]

yeah thats just like a small error in the drawing but ignore that, its just meant to mark the longest side of the triangle

[honest perch]

k

[leaden minnow]

i think what you have to do is find the 3rd angle

and then law of sine from there

[honest perch]

yup doing exactly that rn

thank god pythagoream theorem is easy lol

i mean whoops

i forget what it's called but the rule that the angles have to add up to 180 in a triangle lol

[leaden minnow]

mhmm

[marsh citrus]

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[still temple]

Let f(x) = x + 1/x + 5 for x>0

If I only apply AM-GM inequality on x + 1/x, I get min{f(x)} = 2 + 5 = 7

But, if I apply AM-GM inequality on the whole f(x), then I get min{f(x)} = 3 × (5)^(1/3)

Why does this happen?

[night orchid]

can anyone can an answer for 1b and 1c? i two answers

60% for b and 45% for c

[still temple]

This channel is occupied........

[night orchid]

yeah mistake sorry

[still temple]

Sure wait a sec

[still temple]

Edited a minor error here ^

Please read my message again I have done some changes in it. Sorry for the inconvenience caused

I do understand that. But, why am I getting different results in both cases?

How come?

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[elfin berry]

kangaroo rat
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[marsh citrus]

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[deft apex]

This was in our midterm today and we were allowed to take the questions home to practice them if we didn't understand them during. Anyone have any ideas? the hardest part for me was picturing what is actually happening.

[deft apex]

i've posted in the physics discord as well but if anyone here has any ideas I would love to hear them 😄

for 3.

[still temple]

Let the initial velocity of the bullet be V.
Using conservation of energy we get
MV^2/2 = mgh + (M+m)gR/2
Where h is the maximum height attained by the block of wood, R is the length of the rope and g is the acceleration due to gravity.

Solving for V, we get
V = sqrt[2gh/(M+m) + 2gR/(M+m)]
This is the minimum velocity required for the block of wood to hit the bell.

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[smoky oracle]

I need ti siplify it

[worthy trellis]

Is that $2a^{-7}$ or $2a^{-1}$

[elfin berry]

Deol

[smoky oracle]

Why 2 answeres

[gusty fossil]

@smoky oracle did you write $2a^{-7}$ or $2a^{-1}$ it's hard to read

[elfin berry]

johanna

[marsh citrus]

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[still temple]

I need help with this

[marble glacier]

hm?

[still temple]

idk what to type for print thing

[marble glacier]

well

what do u want to display?

[still temple]

a machine depreciating at the rate of 3 percent over the course of 25 years and the starting price is 5000

[marble glacier]

firstly for count in range (25) is exclusive

it does not include 25

it goes from 0 to 24

so u have to do 26

[still temple]

ok

whats next

[marble glacier]

ok

now u want to times 5000

by 97/100

25

times

[still temple]

how am i supposed write it out

[marble glacier]

hm

ill write it out

[still temple]

damn

[marble glacier]

cost = 5000
for count in range (1, 26):
cost = cost * (97 / 100)
print(cost)

what IDE is that

spyder?

[still temple]

ok

thanks

[still temple]

ide?

[marble glacier]

like

nvm

the code should work

[still temple]

yeah it does

thanks

[marsh citrus]

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[cinder ether]

n=5

delta x = 5

lower should be from f(10) to f(26)

and upper should be f(14) to f(30)

like the sum of all of those f(x)s multiplied by 5

why it's not correct?

[worn nimbus]

draw it

might help

[cinder ether]

ok i'll try

but is there anything wrong with my numbers?

[worn nimbus]

i dont know

[cinder ether]

how does that make it different

[worn nimbus]

so you choose the right value to make the rectangle with

because thats probably where youre going wrong

remember the upper sum is the sum of suprema

and the lower sum the sum of infima

[cinder ether]

yeah i got the delta x wrong

it's 4 not 5

.close

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[digital field]

$x = r \sin\theta \cos \phi$

[digital field]

why is

[elfin berry]

Xetrov

[digital field]

$\frac{\partial r}{\partial x} = \sin\theta \cos\phi$

[elfin berry]

Xetrov

[digital field]

why can the partials be flipped like this?

[still temple]

flipped?

phi and theta are held constant

[digital field]

shouldn't it be dr/dx?

[still temple]

well yes

they can't be flipped

[digital field]

I'm currently looking at the derivation of Laplacian from rect to spherical coordinates

looks a bit off to me

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[past eagle]

assistance understanding this question please

[worn nimbus]

8 choose 3

how did you get that

[past eagle]

i used the combination formula

which I thought was the right way to go but was wrong

[worn nimbus]

i dont even know what that is

[past eagle]

if you know an easier way to get the answer 56 I would greatly appreciate the lesson

[worn nimbus]

8 choose 3 is the easy way

i think your intuition on n choose k is behind

[past eagle]

not sure what that means

[worn nimbus]

well i cant do it justice in chat

especially not if theres videos explaining n choose k

ill find a video for you

https://www.youtube.com/watch?v=WHztDZECzlM

[past eagle]

and this should show me how to the answer 56 was gained

so what i got from that is it should be 8 x 7 x 6 / 3 x 2 x 1

but that does not give me 56

ok nevermind I figured it out

thank you

.close

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[still temple]

[marsh citrus]

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[atomic solar]

If I have those polynomials and I want to see if they generate P_2[X] (the set of polynomials with degree<=2), is my resolution ok?

[marsh citrus]

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@atomic solar Has your question been resolved?

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@atomic solar Has your question been resolved?

[normal moss]

So, you're trying to show they span P_2[x] in the linear algebra sense right?

If so I'd probably try to do something similar. Take any element b in P_2[x] and let b=Ap_1+Bp_2+Cp_3+Dp_4 (for A,B,C,D) scalars then try and solve for A,B,C,D via row reducing, which is similar to what you did. The rref form of a matrix spans the same column space as the original matrix iirc (I think) so I'm pretty sure your method is even faster than how I would try to do it.

At the end though, in the last line, idk why you wrote things in terms of tuples rather than in polynomials.

[marsh citrus]

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[atomic solar]

oh I though it would look better in that way

but ty, needed some confirmation

.close

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[nimble hound]

Please explain question 14

Please don't give the solution

<@&286206848099549185>

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[twilit arrow]

what will be the amount of fluid left when it is all used up?

[midnight gazelle]

3742.5?

[twilit arrow]

no

if we use up all of the saline solution, there should be 0 left

so we want to solve for the time such that 3750 - 7.5t = 0

[midnight gazelle]

Ohhh right

[earnest ginkgo]

8 hours 20 minutes

[twilit arrow]

don’t give out answers

[earnest ginkgo]

7.5 ml per minute is being depleted until the entire 3750 ml is empty. Divide 3750 by 7.5

[midnight gazelle]

Thank you!

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[frank oriole]

How do i solve this integral

[twilit arrow]

you could do u substitution on the bottom. then x^2 = u-1

scratch that

don’t listen to me

[crimson frost]

I'd probably try partial fractions

[marsh peak]

Integration by parts?

[glass silo]

Use the fact that $x^2 = x^2 + 1 - 1$?

[elfin berry]

chartbit

[marsh peak]

Ah right

[crimson frost]

Oh that's a good idea

[marsh peak]

Yeah x^2/(x^2 + 1) is just 1 - 1/(x^2 + 1)

Integral of 1 is x and integral of 1/(x^2 + 1) is arctanx, that's it

[crimson frost]

Oh that's an even better idea

[marsh peak]

Don't credit me

[marsh citrus]

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[obtuse heath]

Stuck after rationalizing

[proven timber]

Show your work

[obtuse heath]

1/sqrt(x)-1 * 1/sqrt(x)+1 / x-1(1/sqrt(x)+1)

= 1/x - 1 / (x-1/sqrt(x) + x-1)

$1/x - 1 / (x-1/sqrt(x) + x-1)$

[elfin berry]

Trolleroof

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[still temple]

<@&286206848099549185>

[marsh citrus]

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@still temple Has your question been resolved?

[gentle owl]

use equation of a cone and find derivative

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[marsh citrus]

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[marsh citrus]

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[still temple]

done (got wrong)

.close

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[orchid oracle]

(derivatives) How do I simplify this expression so I can solve for y' ?

[marsh citrus]

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[copper wraith]

I have no idea how to do this

<@&286206848099549185>

[marsh citrus]

@orchid oracle Has your question been resolved?

[orchid oracle]

Use other channel

[marsh citrus]

@orchid oracle Has your question been resolved?

[marsh citrus]

@orchid oracle Has your question been resolved?

[proud ice]

What have you tried?

[orchid oracle]

Maybe im doing this step wrong.

Or can't I do this...

[proud ice]

You can solve it.

The algebra is a little ugly, but you can.

I would suggest expanding the product on both sides, then putting everything with y' on one side and everything else on the other

[orchid oracle]

Wym 'solve it'

[proud ice]

You said "Or can't I do this...", I thought "do this" meant "solve the equation"

To which I said, yes. You could

[orchid oracle]

Ill try this again when I've got time. In the meanwhile hopefully this channel can stay open

[tulip idol]

you're worrying too much, if anything
start by splitting the 4 on the left and sending one of the 2s within the second parenthesis:

2(y² + x² - 2Rx)(y² + x² - 2Rx)' = R²(y² + x²)'
=> [(y² + x² - 2Rx)²]' = R²(y² + x²)'

wait 😶 do you mean you wish to isolate y' ?

if that's the case, expand everything and pull all terms with y' on one side and then divide both sides by the coefficient of y'

[shadow relic]

Do you know what the vertex formula of a parabola is?

[marsh citrus]

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[umbral kraken]

If the distance of point C from the center of the circle is 17 cm and BC = 15 cm,
Find the diameter of the circle.

[umbral kraken]

can you help me locate the diameter?

[late geode]

do you know properties related to tangents to a circle

[umbral kraken]

yes, i think i will be using the pythagorean theorem but i am not sure

[late geode]

i think i will be using the pythagorean theorem
yes

[umbral kraken]

3. AB is the radius,

AB = √ (17²-15²) = 8cm

Diameter is twice radius = 16cm

this is my work but i think its wrong

[late geode]

why do you think its wrong

which part of what you did are you unsure about and/or think was illegal

[umbral kraken]

i think i am wrong with placing the hypotenuse

[late geode]

the hyp is the side opposite your right angle

you applied pythag correctly

[umbral kraken]

so is this correct or wrong?

[late geode]

its right

[umbral kraken]

thanks

[marsh citrus]

@umbral kraken Has your question been resolved?

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[sleek crater]

Sorry guys im posting this here and not in the physics server because it takes a long time to get a response there and this is kinda related to Mathematics in the form of algebra

So Newtons = kg * m / s^2

How is kg/s^2 * m/s^2 the same thing? Or am I reading it wrong?

[wind palm]

Where did you get that from?

[sleek crater]

A formative assessment

[wind palm]

Is it $\frac{kg}{s^2} * \frac{m}{s^2}$ instead of $\frac{kg}{s} * \frac{m}{s}$ ?

[elfin berry]

Potet

[sleek crater]

Yes that's whats written and what I wrote above

[wind palm]

Well, weird.

[sleek crater]

Did you learn it different?

[wind palm]

Nope, the newton is the same.

You should maybe ask your teacher about that.

[sleek crater]

How is kg / s * m / s the same as this?

[wind palm]

Since upper part is kg * m. And the lower part is s * s = s^2

[marsh citrus]

@sleek crater Has your question been resolved?

[marsh citrus]

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[mild tendon]

ah by chance

[mild tendon]

what would you do in situations where the equation is like!

x^2(3) + x = 0?

like!

if an equation were

x^2 + x = 12

then how would i find x?

[still temple]

You factor an x

To have

x(3x+1) =0

Here u have two things that can make this be 0

x = 0
3x+1 = 0

Obviously x = 0 is just that

for 3x+1 = 0 isolate the x to have

x = -1/3

[still temple]

Take 12 to the left to have

x^2+x-12 = 0 and then realise that this is just a quad that u can factor

[marsh citrus]

@mild tendon Has your question been resolved?

[marsh citrus]

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[umbral hearth]

find a circle equation if its tangent to 2 lines which are parallel to 4y+3x+17 = 0 and 4y+3x-33 = 0 one of them at (-3,-2)

[umbral hearth]

I've found that one of the line its tangent to is y = -3x/4 - 17/4

and the other one is prob impossible to find

but idrk

.close

[marsh citrus]

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[hasty osprey]

The length of the smallest side (or leg) of a right triangle is 16. The lengths of the other two sides are consecutive odd integers. Use the Pythagorean theorem to solve for the the larger of the two missing sides (the hypotenuse).

Me too

[ionic owl]

Since you're given the latus rectum you know your p value, in this case -7. And you're given the vertex so you know your h and k values. h=0, k=0, p=-7. If you plug this into your equation you get (y-0)^2 = -4(-7)(x-0) or y^2 = 28x

MyLab says -28 is the correct coefficient. So did I mess up with the p value? Or did I use the wrong equation for the parabola?

<@&286206848099549185>

Just sending this to keep channel from timing out

Again keeping channel open

[marsh citrus]

@ionic owl Has your question been resolved?

[spark berry]

You did something wrong at = -4(-7)

It's not supposed to be -4

It's just 4

[spark berry]

See, you're not supposed to multiply it with negative 4

[marsh citrus]

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[ionic owl]

@spark berry Thanks

[marsh citrus]

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[rotund walrus]

hey i need someoe to check my work

[rotund walrus]

someone plz help

im in a hurry

<@&286206848099549185>

[marsh citrus]

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[marsh citrus]

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[wraith trench]

Hello, how can I solve this?

[wraith trench]

I tried getting it to standard quadratic form but was unsuccesful

not sure how I can get the x values

[fossil osprey]

You can do that i think

$$ -{(x+11)}^{2} +1 $$

[wind palm]

Expand -(x+11)^2

[elfin berry]

jafar/جعفر

[wraith trench]

what is expand?

and how do I do it

[wraith trench]

but I'm not sure how

[wind palm]

Like, what do you get from -(x+11)^2

[wraith trench]

I don't know

[wind palm]

What is (x+2)^2

[wraith trench]

x^2 + 4x + 4

[wind palm]

What is -(x+11)^2

[wraith trench]

I don't know how to calculate that

[wind palm]

How do you do with (x+2)^2?

[wraith trench]

I just do xx + 2x + 2x + 22

discord removed the "*"

but I just mulitply and add

[fossil osprey]

Hint: ||It's the same as (-1)(x+11)(x+11)||

[wind palm]

This is the same.

[wraith trench]

yeah I know

but I can't calculate that

[wind palm]

(a+b)^2 = a^2 +2ab +b^2

[wind palm]

Are you unsure of how you do it?

[wraith trench]

yes I am confused by the -1

if I convert it to standard form

what do I do with the -1

[fossil osprey]

Since (x+11)(x+11) is the same as (x²+22x+121)

You can replace (x+11)(x+11) with that

(-1)(x+11)(x+11)

(-1)(x²+22x+121)

[wraith trench]

yes

[fossil osprey]

And then can you simplify this?

[wraith trench]

how?

[fossil osprey]

The same method

[wraith trench]

so far I understand

but I don't know what to do with the -1

[fossil osprey]

Distribution

[wraith trench]

ok so I need to multiply the entire expression with -1?

[fossil osprey]

Yup

[wraith trench]

alright

[wraith trench]

what about the +1 I have

[fossil osprey]

Leave it alone

The -1 is only acting on the squared term in the bracket

Not on the whole equation

[wraith trench]

ok

[fossil osprey]

What do you have now?

[wraith trench]

-x^2 -22x -121 + 1

[fossil osprey]

You're on the right track

[wraith trench]

ok so do I add the one to the 121

then factor

[fossil osprey]

yes you can do that

Now you have
h(x) = -x² -22x -120

And you want to know what that is equal to zero

And then you know how to do the rest

[wraith trench]

but I have a negative 2nd degree term

so how can I factor that?

[fossil osprey]

When you have
-x² -22x -120 = 0

You can multiply both sides by -1

[wraith trench]

how did you get 0?

[fossil osprey]

The zero is what we want to find out

We want to know when does h(x) = 0

[fossil osprey]

That's what it means when it says "What are the zeros of the function?"
It want you to find the x values such that h(x) = 0

[wraith trench]

okay yes I see it now

thank you for the help

much appreciated

[fossil osprey]

np

[wraith trench]

.close

[marsh citrus]

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[raven olive]

If I had a function and I wanted to mathematically exclude some values in a function for example a trig function and I wanted to exclude anywhere where y = 0 and just skips to the next number how would I do that?

[raven olive]

Oh an just adding a rule like sin(x){sin(x)≠0} won’t work

[dry spindle]

is this desmos?

and wdym by exclude

like not display?

[raven olive]

Like changing the formula so that it’s no longer an element of that equation

[dry spindle]

x or sinx

[raven olive]

Anywhere that sin(x) = 0

[dry spindle]

alr you could mess with the output set of the function

gimme 1 sec

ill latex it for ya

$f:\mathbb{R}\to{\mathbb{R}-\{0\}}, f(x)=sin(x)$

tf

[raven olive]

The entire equation I’m working on is but I’m simplifying it to isolate the problem

[elfin berry]

Kel.plush

[dry spindle]

there we go

[dry spindle]

you could do piecewise notation

you could take all values of f(x)=sin(x), {-0.01<sin(x)<0.01}

wait what is the problem itself

[raven olive]

So the best way to explain it is a table gimmie a sec

[dry spindle]

mkay

[raven olive]

So say I have a table like this that goes on forever with 0s popping up after each number I want to remove those

So that it would end up like that

[dry spindle]

so you want to merge all the values of sinx being not 0 into a liniar function

this table thing is confusing me

like why

[raven olive]

I’m using sinx as an example

[raven olive]

I would have to explain a lot for that to make sense

[dry spindle]

im listening

[raven olive]

Alright

So that actually formulas look like this

And basically the idea was to display all odd numbers excluding the multiples of 3 5 7 and 9 which would leave only prime numbers

[dry spindle]

and you want to excude 0 out of that?

[raven olive]

I got that part working but it left my with some 0s in place of where the multiples were excluded

[dry spindle]

how exactly are you displaying the numbers

could you maybe share the graph?

[raven olive]

https://www.desmos.com/calculator/dveje0rfam

[dry spindle]

hmm

wait you can put piecewise functions in tables

can you

[raven olive]

Something like this

?

[dry spindle]

id just put it as greater than 0

i did it in the function itself and it seemed to work

[raven olive]

Wait where?

[marsh citrus]

@raven olive Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[hot mantle]

I feel like I overcomplicated this problem?

[hot mantle]

Was there an easier way to solve it?

[proud ice]

Are you trying to find a value for b such that the parabolic equation has a minimum of 32?

Or such that minimum occurs at x=-32?

[hot mantle]

the vertex would be (x,-32)

So the former

[flat raft]

well, you know that the vertex happens at -b/2, so I assume you get something like (b/2)^2 - b^2/2 + 9 = 0

,w solve (b/2)^2 - b^2/2 + 9 = 0

how did you get from this step to the next step?

[hot mantle]

Those are separate equations probably should’ve specified that

[flat raft]

oh

[hot mantle]

I just set up the equation in vertex form