help-13

[cedar kiln]

@waxen basalt Has your question been resolved?

[waxen basalt]

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[distant pecan]

why is it y+7 and not y+121

[cedar kiln]

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

[distant pecan]

you gonna help me?

[distant pecan]

so no ones gonna help me

dude literally asked for my work and then deleted his chat

[dire geode]

you asked only 10 minutes ago

if you can't wait, go pay for a tutor

[distant pecan]

im talking aboutthat one guy that was gonna help me andthen deleted all of his chats and left

[grand forge]

complete the square

firstly

[distant pecan]

wdym

didnt i alreadydo that

[grand forge]

Well i didnt readyour work

probably should have

[lethal jackal]

I think you have a sign error on the 8^2

because if you take a 57 and add 64 to it, oyu get 121, but if you subtract it, you get a 7

[distant pecan]

yeah but whats the sign error

[grand forge]

mhm

[slate lintel]

sometimes I don't have time to commit to actually helping someone but I can remind you to post work / say where you're stuck / etc so I delete my chat specifically so you don't think I'm actually going to keep going

[distant pecan]

i did theproblem twice and didn't see it

[lethal jackal]

I think it's tha tyou wrote -x^2 - 16x + 8^2 = -(x+8)^2? not actually sure how these two steps jive

[lethal jackal]

we have a confession now arrest her

[distant pecan]

jive?

[lethal jackal]

like not sure how you got from the first step to the second

[distant pecan]

well ur supposed to add 8 squared to both sides

[dire geode]

hayley your punishment is to be banned from having to read that handwriting

[distant pecan]

its not that bad

[grand forge]

I've seen worse

[slate lintel]

it is by far not the worst i've seen

my handwriting is pretty terrible as well

[lethal jackal]

I mean sure, you can do that, but I'm not sure how you mangaed to get -x^2 - 16x + 8^2 into what is basically -(x+8)^2

[distant pecan]

its that one formula

idk what its called

[plush warren]

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

[lethal jackal]

,w is -x^2 - 16x + 8^2 = -(x+8)^2

[wraith dagger]

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[lethal jackal]

yeah there's a sign error here

[distant pecan]

well idk

whats the sign error

[plush warren]

you can take minus out before adding things

[lethal jackal]

the way you rewrote that doesn't work

[plush warren]

-(x^2+16x+57)

and now do the formula inside the brackets

-(x^2+16x+57+7-7)

-(x^2+16x+64-7)

[distant pecan]

where did the 7 come from

[plush warren]

i just added 7 and did - seven

+7 -7 will stay the same

i just added it to make quadratic thing

[distant pecan]

uhuh

i dont understand but ok

mb

[plush warren]

so you know formula (a+b)^2=a^2+2ab+b^2

yeah?

[distant pecan]

yeah

[plush warren]

so lets talk about x^2+16x+64 first

its (x+8)^2

when you do the formula you get that right?

[distant pecan]

yeha

[plush warren]

okay

so x^2+16x+57 we have this yeah?

but we want to make it quadratic

[distant pecan]

why is it +

[distant pecan]

where did this come frm

[plush warren]

from my head

you had y=-x^2-16x-57

yeah?

[distant pecan]

mhm

[plush warren]

so y = -(x^2+16x+57)

[distant pecan]

mhm

[plush warren]

and lets talk about whats going on inside that brackets

inside brackets we have x^2+16x+57 yeah?

[distant pecan]

mhm

[plush warren]

so in this formula (a+b)^2 = a^2 + b^2 + 2ab

a^2 is x^2 in this case right?

[distant pecan]

mhm

[plush warren]

and 2ab is 16x

[distant pecan]

mhm

so we add 8x to both sides

[plush warren]

nooo

[distant pecan]

ok well keep going

oh no

[plush warren]

you need to add something to x^2+16x+57 to make it quadratic

[distant pecan]

8 squared

64

[plush warren]

okay that aint bad too

can u tell me why

[distant pecan]

uhm cuz they we can make it (x+8)^2

[plush warren]

yesssss

[distant pecan]

but then where would the 57 go

[plush warren]

ill tell ya

[distant pecan]

and wouldn't wehave to add 64 to the left side

[plush warren]

yeah

[distant pecan]

ohhh

and then here

here

this is it

[plush warren]

but youre wrong as well

a little bit wrong

[distant pecan]

y+64=-(x-8)^2+57?

[plush warren]

yes

bro i tho u didnt know shi

[distant pecan]

wait how ami wrong

[distant pecan]

well i was wondering why it was +7

[plush warren]

you were wrong when you said you would add 64 to y but you typed it right now

[distant pecan]

and not like 121

[plush warren]

and simplify this shi rn

and youre done

[distant pecan]

kk cool

[plush warren]

how do u simplify it

[distant pecan]

64-57

i have another question if thats cool

same topic

[plush warren]

yes

i dont know this topic

but i just understood it tbh

when i looked at the problem and answers

[distant pecan]

i wish i understood math easily

[plush warren]

no idea lol

maybe someone else will help u with this one

i understood other one because it was basic algebra

but this idk

[distant pecan]

ok

thanks

im struggling with basic algebra 😭

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[distant pecan]

/close

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[rancid iris]

Hi why is 330 degrees not a solution?

[zenith sail]

It is, but it's already included in 5pi/6 + pi*n

5pi/6 + pi = 11pi/6

[rancid iris]

How can I find out if it is included?

Other than that

[zenith sail]

I mean, there isn't really much to it other than that

It's just that they're already pi apart

[rancid iris]

Does it have smth to do with it being tan?

[zenith sail]

well the period of tan is pi, for sin or cos it's 2pi

for sin or cos you'd do 2pi*n rather than pi*n

[rancid iris]

O

Ok ty

[zenith sail]

but you had that part right already 🤷

[rancid iris]

Well technically it's wrong

Cuz it's not on the answer key

[zenith sail]

I mean you wrote 5pi/6 + pi*n, which is correct

for sin or cos that pi would be 2pi instead

that's the only thing that's really different for tan

but you can just plug in a few whole numbers for n and see if any of your solutions overlap

[rancid iris]

Wut

Wait so

If I have a problem like

Sin(theta) = 1/2

Then one of the general solutions would be theta = pi/6 + 2pi*n?

[zenith sail]

yes

and also 5pi/6 + 2pi*n

[rancid iris]

O cuz my textbook says its only pi*n

[zenith sail]

for tan it's pi

for sin and cos it is 2pi

[rancid iris]

O ok then

Ty

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[warm tree]

I'm maybe confused about related rates, but from my understanding each term on the right side of the equation should have a dx/dt that gets factored out at the end

and 30mph is given so that is what would be plugged in

but a lot of people in my class did not include this

[cedar kiln]

@warm tree Has your question been resolved?

[warm tree]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

same question, different problem

basically do i need that dr/dt

I'm pretty sure I do, but a lot of people didn't have it so im doubting

[slate lintel]

yes you do need that dr/dt

[warm tree]

top one looks good too then?

Weird how so many people missed it

[warm tree]

.

[slate lintel]

yes top one looks good too
if you just think about units it should make sense that you need a "something / seconds" term

because to say dV/dt = 4m^2 doesn't really make any sense

[warm tree]

okay thanks

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[waxen basalt]

help

<@&286206848099549185>

wait

[knotty egret]

so

[waxen basalt]

is it everything that doesnt equal 0

lmfao

well

wait

all positive from 0?

[knotty egret]

well they both cant be negative

yea

[waxen basalt]

so (0,-oo)

[knotty egret]

no

wait

so it can only be zero or more if one or more numers are neg or 0

i hope this helps

[waxen basalt]

oh

so

(-oo,-5]

oops

[knotty egret]

well not really

[waxen basalt]

u said if they are neg or 0

right?

[knotty egret]

yea

one of the factors

[waxen basalt]

well if it went to one it wouldnt work

[knotty egret]

so find the x value where every number is positive

[waxen basalt]

ah

okay

[1,oo)

[knotty egret]

thats is the inverse

thats everything else

i think you mean (-oo, 1]

[waxen basalt]

im confused

ahhh

okay

[knotty egret]

yea

and x = 1 works

bc its zero

[waxen basalt]

wait

[knotty egret]

yea?

[waxen basalt]

any number after that doesnt tho?

[knotty egret]

ye

[waxen basalt]

so shouldnt it be until -5

bc f(x) will be positive until -5

[knotty egret]

but g(x) is neg untill 1

only 1 of them has to be neg

[waxen basalt]

• • • = -

[tame tangle]

Just to make sure I understand the problem right, f(x)g(x) means you multiple both value or I missed something?

[knotty egret]

yep

[waxen basalt]

multiply

[tame tangle]

So, for which x value the product is higher or equal to 0

Based on the graph

[waxen basalt]

so shouldnt it be (-oo,-5]U[1,oo)?

[knotty egret]

no

[1,oo) includes 2 positive numbers

which would be greater than 0

omg

im so dumb

my bad

[waxen basalt]

anywhere at or before -5

and at or after 1

[knotty egret]

i read it wrong

its just [1,oo) i belive

i thought f x g had to be less

not greater

mb

[tame tangle]

I don't understand the question xD Would be something I'd ask my teacher

[knotty egret]

alr

thats ok

[waxen basalt]

i think its this

[knotty egret]

yea

i belive

[waxen basalt]

because negative * negative will be bigger than 0 if the negatives arent 0 lol

[knotty egret]

either one or more numbers are 0, or both are + or -

[waxen basalt]

yeah

[knotty egret]

as in +2, +2 works

-2, -2 works

0, 2

works

but 2, -2 doesnt

that kinda idea

so yea

good job

i can belive i read THAT wrong lmao

mb

bye

[waxen basalt]

lol

how about this one?

[knotty egret]

wo

w

i would solve for y

[tame tangle]

I've tried to visualize the problem and I get that

[knotty egret]

thx but we kinda already solved that

thanks anyway tho

[tame tangle]

Oh well

[knotty egret]

try this one

[tame tangle]

I always hated optimization problems. Can't help there

[waxen basalt]

well i got

$$\frac{x^{\frac{3}{2}{}}(10800-x)^{\frac{1}{2}}}{100000}$$

[knotty egret]

good

[waxen basalt]

took the derivative to get

[knotty egret]

now i would square both sides to eliminate the exponential fractions

[waxen basalt]

$$\frac{32400x^{\frac{1}{2}{}}-4x^{\frac{3}{2}}}{200000(10800-x)^{\frac{1}{2}}))}$$

[knotty egret]

im not very well with derivatives

sorry

but for this

i would suare both sides

[waxen basalt]

i think i set equal to 0

[knotty egret]

yea

i think that works

[waxen basalt]

i got 8100

but idk if thats right

[knotty egret]

just make sure to avoid dips and focus on peaks

since both could have a derivative of 0

how many x values have a derivative of 0?

if just one, ig thats the answer

[cedar kiln]

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[waxen basalt]

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[tacit juniper]

Hi! Question: I'm trying to rewrite the formula for stdev with sigma notation. So far I have the following steps:

$s=\sqrt\frac{(x_1-\bar x)^2+(x_2-\bar x)^2+\dots+(x_n-\bar x)^2}n$

$s^2=\frac1n\sum^n_{i=1}(x_i - \bar{x})^2$

$s^2=\frac1n\sum^n_{i=1}(x_i^2 -2x_i\bar x + \bar x^2)$

$=\frac1n\sum^n_{i=1}x_i^2 - \frac{2\bar x}n \sum^n_{i=1}x_i + \bar x^2$

[wraith dagger]

.rafzekael

[tacit juniper]

But supposedly I'm supposed to show that it's equal to

$\left(\frac1n\sum^n_{i=1}x_i^2\right) -\bar x^2$

[wraith dagger]

.rafzekael

[tacit juniper]

And I'm not sure how the xbar squared became negative, or how the 2x_i xbar term goes away?

Is it like some case of collapsing sums?

[slate lintel]

probably

try it for two variables

and then three

and see what patterns emerge

and keep in mind what xbar means

actually that's the big thing

[tacit juniper]

Ahhh, you're right

Thats the key

Yes, ok, that resolves it all i think.

Thanks a bunch!

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[crude lagoon]

anyone know what the error means

does it want me to multiply d(x) and q(x)

[obsidian coral]

It means what it means

You gave an equation

[crude lagoon]

i gave it the form it wanted

[obsidian coral]

It wants an expression

[crude lagoon]

so no =?

does it want x

[obsidian coral]

You don't need that left part at all

[tacit juniper]

It wants an expression in the form of d(x)q(x) +r(x)

[crude lagoon]

where does it say that

its correct but idk where it says that

[obsidian coral]

In the problem itself

[tacit juniper]

At the very beginning

[obsidian coral]

It wanted you to preform long division on p(x) and wanted you to write the quotient

[upper abyss]

I'm going to guess it wants everything to the right of your equal sign

[crude lagoon]

oh

[tacit juniper]

Yeah, probably

[crude lagoon]

i see it now

its asking for the polynomial

??

x=3/4

nvm

[cedar kiln]

@crude lagoon Has your question been resolved?

[cedar kiln]

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[dreamy zenith]

how is there a difference quotient when theres no h

[muted bear]

Difference of quotients not difference quotient

$\frac{a-b}{c}=\frac ac -\frac bc$

[wraith dagger]

garlicbredfries

[rigid shoal]

not sure what the question is exactly, but this is a true equation.

[cedar kiln]

@dreamy zenith Has your question been resolved?

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[warm tree]

the graph of this function has 4 different points that has a vertical tangent line, so I did something wrong

actually, I think I just need the plus or minus

and then just plug in the y values into the function to get x?

[runic garnet]

I haven’t checked anything numerically but the thought process looks good

[livid hound]

there are two solutions to
y^2 = k

you've only written the positive ones

[runic garnet]

Yea I think they realized that here

[livid hound]

missed that

[warm tree]

look good?

[runic garnet]

Once again, the thought process looks good, but I haven’t checked it numerically

It’s just calculator work so

[warm tree]

Okay thanks

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[naive ether]

Hi

[naive ether]

Write a polynomial in factored form with the given roots: 0 (multiplicity 2), -1 (multiplicity 3)

[cedar kiln]

@naive ether Has your question been resolved?

[naive ether]

.cloe

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[charred gust]

hi

[charred gust]

how come both of the methods (disocunt price) come to the same price of 40.3.2

[regal oak]

This is due to the commutativity of multiplication

[charred gust]

?

[regal oak]

32×1.4×0.9=32×0.9×1.4

×1.4 is the 40% markup and 0.9 is the 10% discount

[charred gust]

but is not one of the method without markup

@regal oak

[regal oak]

What do you mean?

[charred gust]

ohhhhhhhhh

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[charred gust]

tty

[cedar kiln]

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[exotic jungle]

Can somebody please help me with this question? I feel like my working makes sense but i didnt get 37%, got 29%. Where is the error in my working/logic?

[charred gust]

@exotic jungle

[exotic jungle]

ye

[charred gust]

the working looks right soo far

but

re read over this

all i can tell u is that you made a slight eroor

try reading the whole question again with a new piece of working out paper

[exotic jungle]

is it in here?

Ok i will try again

I cant see it

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[cosmic basalt]

I don't get it
I know I have the first solution down but if we go for the second solution by flipping it to a positive 9, it becomes impossible. Taking away 3 turns 9 into 6 and you can't divide 6 by 4, without getting into obtuse shit. What am I doing wrong?

[weary vessel]

Wait

[weary vessel]

Who taught you that you need to flip the sign of the 9 to look for the second solution

it's not like that!

[cosmic basalt]

The instruction videos and the teacher?

[weary vessel]

No 😛

[cosmic basalt]

Welp

What do, then?

[weary vessel]

Let's look at it in another way, follow me

$3-4 |2x-4| = -9$

[wraith dagger]

luna7427

[weary vessel]

Let's try to isolate the absolute value, the |2x-4|

What should we do first?

[cosmic basalt]

What we'd do is subtract 3 from both sides?

And then divide everything by 4 (Which really on the left side is just to cancel out said 4)

Which....leads us to the same issue

[weary vessel]

Subtract 3

$-4 |2x-4| = -12$

[wraith dagger]

luna7427

[weary vessel]

Divide by 4

$-|2x-4| = -3$

[wraith dagger]

luna7427

[weary vessel]

Multiply by -1

$|2x-4| = 3$

[wraith dagger]

luna7427

[weary vessel]

Are you okay until now?

[cosmic basalt]

Hold on lemme catch up to speed

So okay, then we'd add 4 to both sides right?

[weary vessel]

No, that 4 is inside the absolute value, you cannot add 4 to cancel it

Here is the catch

3 is the same as |3| right?

[cosmic basalt]

Correct

So -4 is the same as 4?

[weary vessel]

Then just replace 3 by |3|

$|2x-4| = |3|$

[wraith dagger]

luna7427

[weary vessel]

Therefore, what is inside | | in left hand side must be equal to what is inside | | in right hand side

2x-4 = 3

This gives you a first solution

[cosmic basalt]

Yes, which I already had

1/2

[weary vessel]

You are not focusing

It does not give you 1/2

Go do it

[cosmic basalt]

I just checked it by pressing enter

It says I'm half right

[weary vessel]

Solve 2x-4=3

The one we just arrived to

[cosmic basalt]

Add 4 to both sides

Then divide by two which leads to 7/2

[weary vessel]

Yes, nice!

7/2 is one of the answers

For the second answer, we have again

$|2x-4| = 3$

[wraith dagger]

luna7427

[cosmic basalt]

I have both answers lol

[weary vessel]

This time we do not replace 3 by |3|, but by |-3|

[weary vessel]

yeah yeah, im just showing you the correct process

you got lucky

:P

[cosmic basalt]

Now I just have a headache lol

[weary vessel]

😋

[cosmic basalt]

.closed

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[cedar kiln]

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[weary vessel]

It's very simple, you have |2x-4| = 3

For first solution you replace 3 by |3|

For second solution you replace 3 by |-3|

That's about it @cosmic basalt

[cedar kiln]

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[opal ember]

Need to solve for x. "When it comes to a triangle with two equivalent sides, the angles that they don't intercept, are the same". So C = 36°? PLEASE correct me if I'm wrong

[fluid pulsar]

u'r right

C is 36 degree

[opal ember]

36*

[fluid pulsar]

oops,sry

[opal ember]

But then I'm stuck

[fluid pulsar]

and for the same reason,∠BDC=x

[opal ember]

excuse me, but what's the formula for that?

[opal ember]

copy+pasted from someone that gave me some help, but couldn't totally understand it

give me a min, ill try it

[fluid pulsar]

humm

[opal ember]

like this?

[plush warren]

x is like 72⁰

[opal ember]

therefore
36+36+36+x=180
108+x=180
x=180-108
x=72?

[plush warren]

lol

[fluid pulsar]

how to say....∠ABD is 36,that's right,but u cant just jump to that conclusion

[opal ember]

hmm

what should i do then?

[plush warren]

well

[fluid pulsar]

BC=DC=b

[plush warren]

U know that BDC is x as well

[fluid pulsar]

so,∠CDB = X

that is the right process

[opal ember]

wait so ∠BDC isnt the same as ∠DCB or ∠CBD?

or are they all the same?

sorry

[plush warren]

BDC is isosceles triangle

BC=DC=b

[opal ember]

yes, same sides

[plush warren]

so angle CDB = angle CBD

yeah

[plush warren]

u understand this yeah?

and they are both equal to x

And than( 180-36)/2

because in triangle BDC sum of angles are 180

so x+x+36=180

x=72

That's it

[opal ember]

So my answer was right but the process was wrong?

[plush warren]

Yes

do you understand solution?

[opal ember]

ohh so because D can be represented as x that's why we do 2x+36=180?

then we combine like terms

2x=180-36
2x=144
x=72

tysm!

one question though

Let's forget about the previous problem for a moment. What's the difference between ∠ABC, ∠BCA, and ∠CAB? don't they all have like the same measures and angles?

[plush warren]

oh whats different?

everything basically

well not really everything since ABC is isoscles triangle but angle BAC and angle BCA are equal and they are 36

and angle ABC is 108 so it isnt same measurement

[opal ember]

Ohh so the letter that's in the middle represents the measure of the angle?

∠BAD = ∠DAB = ∠BAC = ∠CAB = 36° ?

and so on

[plush warren]

they are all the same angle... but yes

[opal ember]

i get that

but can you just make clear why ∠BDC = x?

is it because BC = DC = x or for another reason

[cedar kiln]

@opal ember Has your question been resolved?

[cedar kiln]

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[crimson sedge]

just to make sure

[crimson sedge]

the axioms of an inner product say that <a,a> >= 0 IFF a> 0 correct?

[short blade]

no

<a,a> >= 0

usually a > 0 doesn’t make sense

[crimson sedge]

hes saying for C(0,1), <f,f> = integral of f*f on [a,b]

[short blade]

who is he

[crimson sedge]

Koffi Enakoutsa

my optimization professor

[short blade]

right

what are you confused about

[crimson sedge]

so i think of an inner product as like a metric or just a function or what

[short blade]

it gives a notion of length and direction yes

[crimson sedge]

ok sounds good

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[boreal trail]

why do lim h-->0

like i get 1a)
but how come finding gradiant of tangent at P = the secant but as h = 0

[dire bronze]

Revise on what a derivative is

[boreal trail]

derivate is function which is the average rate of a specific point/instantenous

isnt it that making limit h-0 its making it so there basicly isnt a 'h' as in ^ and then make the tangent

[cedar kiln]

@boreal trail Has your question been resolved?

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[thin roost]

Seems right, but idk where I went wrong...

The ans is 53 idk how they got that....

[muted bear]

the constraint a+b+c=0 is more restrictive on the inequality

Because you cant have e.g. a=b=c=1/3

Easy fix, square both sides of the equality

Then do your amgmshenanigans

[thin roost]

Hmm

How's this working out...

[muted bear]

Uhh

1+2(ab+bc+ac)=0 right?

[thin roost]

Yes

[muted bear]

(ab+bc+ac)/3=-1/6

[thin roost]

Alr

[muted bear]

-1/6>=abc^(2/3)

Hmmm

[thin roost]

How's that happening??

[muted bear]

Amgm

[thin roost]

a²+b²+c²/3 >= (abc)^(2/3)
...

Oh ye

[muted bear]

Wait hold on

[thin roost]

a²+b²+c²=-2(ab+bc+ca)

[muted bear]

2+2(ab+bc+ac)=1

(1+ab+bc+ac)/4=1/8

1/8>abc^(1/2)

Hmm no

[thin roost]

How's 2 getting in action here... ??

[muted bear]

From this

[thin roost]

Alr

[muted bear]

The consequence of this is (abc)^2<1/4096

But thats not true

Also imforgetting >=

[thin roost]

[muted bear]

Man i thought it was gonna be an easy fix

[thin roost]

[vague rapids]

so seems all qs of IOQM

[thin roost]

It's not even from ioqm, it from the module...

[vague rapids]

I mean it is for IOQM

[thin roost]

Ye yen

It is

For that

[muted bear]

<- Failed oly competitor

[thin roost]

This is from number theory, so there's gotta be more than inequality ...

[muted bear]

Have you tried squaring the second equation

And/or cubing it

[thin roost]

a²+b²+c²=1??

This??

[muted bear]

Idk maybe it leads to something good

[thin roost]

,w expand (a^2+b^2+c^2)^3

[muted bear]

Actually you know what

c=-a-b

[wraith dagger]

Results provided by WolframAlpha

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

Requested by _basudev#0

[muted bear]

Have you tried that

[thin roost]

Oh ye

Shit

[vague rapids]

a+b+c=0
So (a+b)=-c and similar stuff would help (maybe)

[thin roost]

Perhaps...

Ye ye I'ma sleep now

Thanks

.close

[cedar kiln]

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[thin roost]

I'm tried

[muted bear]

Same

[cedar kiln]

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[vague rapids]

I'm tried ok

[vague rapids]

.close

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[mint oxide]

what is the total number of distinct divisors of 2^9 x 3^19

[mint oxide]

<@&286206848099549185>

[dusty hazel]

Hello.

[mint oxide]

hy

[dusty hazel]

Have you tried anything?

[mint oxide]

no

[dusty hazel]

Do you want to?

[mint oxide]

yes

help /

[dusty hazel]

The factors to this number will be of the form 2^x * 3^y; 0=<x=<9 and 0=<y=<19

Do you realise that?

[mint oxide]

yes

[dusty hazel]

Do you think x and y are dependent, or are they independent?

[mint oxide]

independent ?

[dusty hazel]

That's right.

If they are independent, how many choices for x do you have?

[mint oxide]

9

with integers 19

[short blade]

how did you get 9

where did 9 come from

[mint oxide]

i just don't get it

[short blade]

that didn’t answer my question

how did you get 9

[mint oxide]

if they r independent, 2^9 then it can be distictly divisible by 9 numbers 2 4 8 16 32 64 128 256 512

[short blade]

you’re missing 1

2^0 is also an option

[mint oxide]

oh

ok

[short blade]

so you have 10 options for x

now how about for y

[mint oxide]

20

[short blade]

yes

so for each value of x you have 20 options for y

[mint oxide]

yes

[short blade]

how many total combinations

[mint oxide]

200

thank u

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[hazy flicker]

HEllo! Please help one more time!!...

[hazy flicker]

I mean it's just disk method right?...

like I thought I did it correctly, integral from 0 to 4 with (sqrt(4-x)-7)^2 ?

<@&286206848099549185> ;(

[dire geode]

draw a picture

[hazy flicker]

alright

Top part

@dire geode I'm hoping im right but it said I was wrong obviously so

[dire geode]

,rotate

[dire geode]

draw a picture of the region being revolved

[hazy flicker]

I don't really have time mane </3

[smoky idol]

then you don't really have time to learn. Do it right or don't do it

[hazy flicker]

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[crimson sedge]

Let's Consider the conservation law: [
\pdv[\map c{x,t}]t + \pdv[\map j{x,t}]t = 0
]
where [
\map j{x,t} = -\map D x \pdv[\map c{x,t}]x
]
The diffusion coefficient $\map D x$ depends as follows on $x$ [
\map D x =\map {D_0}{1+gx} \quad g \ge 0
]
Thus $\dv[\map D x]x = D_0g$. Inserting the current $j$ into the conservation law we obtain a \emph{drift diffusion equation}[
\pdv[\map c{x,t}]t = D_0g\pdv[\map c{x,t}]x +\map {D_0}{1+gx}\pdnv{2}[\map c{x,t}]x]
The initial conditions for the partial differential equation is [
\map c{x,t = 0} M_0\map \delta x
]
where $\delta$ denotes the delta function. The boundary conditions are [
\map j{x=x_0,t}=\map j{x=+\infty, t} = 0
]
Solve the one-dimensional drift-diffusion partial differential equation for these initial and boundary conditions

[crimson sedge]

I included all the physics for context. I want help with how to tackle the PDE though

I have to use a product ansatz: $\map c{x,t} = \map T x \map X x$

\vs{3 mm}
I inserted the product ansatz into the one-dimensional drift-diffusion PDE and got [
\f1{\map T t}\pdv[\map T t]t = D_0g \f1{\map X x} \dv[\map X x]x +\map {D_0}{1+gx}\f{1}{\map X x}\dnv{2}[\map X x]{x}
]

[crimson sedge]

unsure on how to proceed with this though

[crimson sedge]

oh frick, forgot that lol

[calm sierra]

if f(x)=g(t) for all x,t then f,g equal a common constant

[crimson sedge]

the seperation constant, right?

[calm sierra]

so just tack on to the above equation
[\dots=\dots=-\lambda]
for some constant $\lambda$

the minus sign is usually done for convenience

[crimson sedge]

i guess like

solve for T(t) from that?

[dire geode]

you get 2 ODEs from that

[calm sierra]

yeah can you write the eqn for T from that?

[crimson sedge]

good question

[wraith dagger]

rokemily

[crimson sedge]

you are bound to get a second order linear ODE right

from what i am seeing

yeah okay i think i know how to go from here

ty you two

.close

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[rotund jasper]

need help with limits

[rotund jasper]

,rotate

the answer i am getting is 1/3

where as it's given as -1/3

[south tundra]

Because x = -sqrt(x^2) in this case

Since x is negative

[rotund jasper]

x tends to - infinity

that's the reason ?

[vagrant elbow]

You need to write it as -|x| and only take the |x| part inside the square root if you want to be fancy about it

[south tundra]

Pretty much

[rotund jasper]

is there a reason for that ?

[rotund jasper]

ok ty much appreciated

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[gleaming summit]

can someone help me go through a paper i have sat

[crimson sedge]

don't ask to ask, just ask!

[gleaming summit]

wdym?

[tropic oxide]

send the paper here along with all relevant info e.g. your work, the marks you've received if any, etc

[sick rose]

send the paper

[gleaming summit]

i havent go tmy mark but i wanna see what i got ish

[gentle flower]

[sick rose]

[gleaming summit]

the marks i gave is the marks i think i got

the black writing is what i wrote in the exam

i wanted to go over the paper with someone if possible

<@&286206848099549185>

[dire geode]

are you supposed to show any work

[gleaming summit]

we were given an answer booklet the uni takes in

[dire geode]

or did you like do all that first order linear differential equation stuff with integrating factor in your head

[gleaming summit]

we can take question paper home

]i just wrote final answer on the question paper

[dire geode]

then just plug your y into the differential equation and see if it's right

[gleaming summit]

how?

[dire geode]

you know y

find dy/dx

and then plug in dy/dx and y into that equation

[gleaming summit]

but theres a +C

[dire geode]

so?

$\frac{d}{dx} \frac{C}{x^4} = ?$

[wraith dagger]

rie.mann

[dire geode]

you just use power rule

,tex .diff rules

[wraith dagger]

rie.mann

[dire geode]

and constant multiple rule

[gleaming summit]

im confused

are you saying to diffferentiate the formual i got

and put x and y values in and compare it to the orignal ode?

[dire geode]

yes. that verifies if your answer's correct

[gleaming summit]

but the +C would mean i would get two different answers

theres an answer with the orignal formula

and theres an answer with the formula i got

[dire geode]

show this

[gleaming summit]

intergating factor

[dire geode]

show the answer

in fact show all context for that problem

[gleaming summit]

i think its correct i just checked online and i got my answer

its just that for the problem

would you be able to help me with the second problem

third and the sixth problem

[dire geode]

just reopen a new channel with just one problem at a time

[cedar kiln]

@gleaming summit Has your question been resolved?

[cedar kiln]

@gleaming summit Has your question been resolved?

[cedar kiln]

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[stark breach]

I need help with 48

[rare vault]

is Ln natural log

[stark breach]

yeah

[cedar kiln]

@stark breach Has your question been resolved?

[eternal nymph]

dunno if the exercise got a problem or anything but there isn't solutions bigger than 2

[stark breach]

there is an answer sheet and it says 44.7 but i dont know how

[dawn junco]

and it's prolly not even solvable exactly

so even if there was one solution >= 2, idk how you would find it by hand

[stark breach]

i plugged in the answer it gave and they dont match there might have been an error on the answer key

[dawn junco]

or even the question itself idk

¯_(ツ)_/¯

[stark breach]

yeah that could be it too

[worldly chasm]

There should be a solution bigger than two. That's an exponential vs a quartic

x ≈ 44.69112...

@stark breach

https://www.wolframalpha.com/input?i=solve[{x^4+%3D%3D+6+exp(0.3+x)+%2B+log(x^3)%2C+x+>+40}%2C+x]

In order to solve this by hand you will likely need to choose a large enough seed value and do newton's method for a few rounds

You won't be able to get an exact value in all likelihood, even using common special functions, such as Lambert W.

[cedar kiln]

@stark breach Has your question been resolved?

[cedar kiln]

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[rancid iris]

Can someone pls check if a is right?

[open nacelle]

Looks good i think

But

You don't have to include 2pi for any of these

It's already in the problem description that theta is strictly less than 2pi

Theta also cant be 0

For a

[cedar kiln]

@rancid iris Has your question been resolved?

[rancid iris]

O ok cool

Ty

.close

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[shadow kite]

Hi I really need help with this question

[crimson sedge]

you asked this yesterday

[gritty galleon]

Ig u-sub u =x^2 + 1

[crimson sedge]

this looks like integration by parts

with u-sub as x^2 + 1 we still have the x inside the second logarithm

[gritty galleon]

u yes

[stray otter]

just make x in terms of u

[gritty galleon]

but couldnt u use 2 log x = log x^2

x^2 is is u -1

[cedar kiln]

@shadow kite Has your question been resolved?

[cedar kiln]

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[gritty galleon]

Problem on a practice test paper on my tution

[cedar kiln]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[gritty galleon]

4

I think its (c)

[tropic oxide]

let's see here

[gritty galleon]

my confusion lies in the fact we have no restrictions on q(x), whatsoever

[tropic oxide]

by the looks of it you're expected to recall the defn of euclidean division

you didn't do it correctly

should bee (d), p = gq + r

q is the quotient

[gritty galleon]

hmm

[gritty galleon]

how can u be sure tho?

[tropic oxide]

all the answer options are in the same form

[gritty galleon]

oh wait 💀

[tropic oxide]

just with the letters shuffled around

[gritty galleon]

hmm u r right

[earnest socket]

isn't (c) trivially true

just set r(x) = 0

[gritty galleon]

but if its division then we must have the resstriction deg(p(x)) = deg(g(x)) + deg(q(x))

[gritty galleon]

wdym..?

[earnest socket]

"then we can find polynomials q(x) and r(x) ... such that:"

if you set q(x) = p(x)g(x) and r(x) = 0, then (c) holds

obviously they want you to choose (d) though

[gritty galleon]

well r(x) = 0 is given

[earnest socket]

huh

[gritty galleon]

bruh 😐

well tysm @tropic oxide and @earnest socket

.close

[cedar kiln]

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[earnest socket]

rip

premature closure

[cedar kiln]

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[thin roost]

Q 28)

I could only conclude this...

[vague rapids]

try to find the value of 3^3+4^3+5^3

[gritty galleon]

,w 3^3 + 4^3 + 5^3

[wraith dagger]

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Requested by itzkraken.#0

[thin roost]

216

[tough aurora]

at the very least, 3^3+4^3+5^3 is an upper bound...

[thin roost]

Yep...

[vague rapids]

216 can be expressed as?

[thin roost]

6^3

[vague rapids]

Yes

and K is a positive integer

so smallest K value is?

[thin roost]

1?

[vague rapids]

Yes

and there you go

[thin roost]

Alr

Damn

That's was ezz idk where I went to think on

.close

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[ashen anchor]

Is there a difference when we write $\sqrt[3]{-5}$ and $-\sqrt[3]{5}$ as a solution?

[wraith dagger]

bennxy

[crude blaze]

Yes

The answer of √-5
Would be i√5

[thin roost]

[ashen anchor]

Hmmm ok, but $\sqrt[3]{-5}$ could also be -1,709 right?

[wraith dagger]

bennxy

[crude blaze]

Yes

[thin roost]

$\sqrt[3]{-5} = \sqrt[3]{-1×5} = \sqrt[3]{5} × \sqrt[3]{-1} = -\sqrt[3]{5}$

[wraith dagger]

_basudev

[thin roost]

,w (-1)^3

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[thin roost]

So difference...

$i\sqrt{5} \neq \sqrt[3]{-5}$

[wraith dagger]

_basudev

[ashen anchor]

Okok got it

Thanks guys

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[magic solar]

how do you find the horizontal range?

[magic solar]

do i just solve for x and the non 0 is the answer

[tropic oxide]

it's the distance between the two x-intercepts, yes

[magic solar]

ah ty

for iv) what does direction refer to?

ok assuming its the angle with the horizontal, im still not quite sure how to do it

[tropic oxide]

it is the angle with the horizontal

it's the angle between the tangent line and the horizontal

[magic solar]

its been a while, how would you find it again?

it was probably something along the lines of tantheta= somethnig

[tropic oxide]

tan(θ) = y'(120) in your case

[magic solar]

ah thanks

wait y'(120) means dy/dx when x=120 right

which is 265/4956, which when tantheta= that, theta is 3 degrees. But the answer is 6 degrees ;-;

[tropic oxide]

yes]

hold up

,w 2 * (-5/2304) + 5/12

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[tropic oxide]

wait uh. no

,w 2 * (-5/2304) * 120 + 5/12

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[magic solar]

-5/1652x + 5/12 if your wondering

[tropic oxide]

1652?

[magic solar]

the derivative

it should be

[tropic oxide]

i think your denom's messed up there

2304/2 = 1152 not 1652

,calc atan(-5/48) * 180/pi

[wraith dagger]

Result:

-5.9468630539735

[magic solar]

oh right

[tropic oxide]

yeah -6 degrees as it should be

so fraction bullshit

[magic solar]

yeah you're right sorry

oops

to find the greatest heigh, i used -b/2a

which was 96 and the wrong answer

so i tried when x=96, which is half its horizontal range, and got the corerct answer

oh wait i need to sub in

im so dummbb

.close

[cedar kiln]

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[teal magnet]

Most of the pages on I found on trigonometric angle addition identities just list them and I can't remember that all straight through. I found 2 pages that seem to go through the material slower, do you know of something even slower? https://math.libretexts.org/Bookshelves/Precalculus/Book%3A_Precalculus__An_Investigation_of_Functions_(Lippman_and_Rasmussen)/07%3A_Trigonometric_Equations_and_Identities/7.02%3A_Addition_and_Subtraction_Identities and Section 7.2 of https://resources.saylor.org/wwwresources/archived/site/wp-content/uploads/2011/11/MA003-3.1.pdf

[cedar kiln]

@teal magnet Has your question been resolved?

[cedar kiln]

@teal magnet Has your question been resolved?