help-39

[midnight haven]

@opal forge

is this what you're looking for?

[opal forge]

Tq so much @midnight haven

[midnight haven]

it's a paid app

[opal forge]

Ahhh

[midnight haven]

I think it's available for free on chrome

[opal forge]

Did you pay for it?

[midnight haven]

no

you can use it for free on their website

[midnight haven]

@opal forge

the app is paid, but it's free on their website is what I mean

[magic hornet]

it has something like premium

[midnight haven]

yeah

for step by step solutions

[opal forge]

How do we check this will form a circle?

Is there any direct formula/pdf to know about conic sections lines?

[midnight haven]

it can be decided by the equation

for circle, the equation is $|z-z_o|=r$ is of circle where r is the radius

[jolly parrot]

deltaG

[midnight haven]

for ellipse, $k>|z_1-z_2|$ where k is major axis

[jolly parrot]

deltaG

[midnight haven]

there are many, i can send you a youtube course if you want

[opal forge]

Yes. Please share

[midnight haven]

https://www.youtube.com/watch?v=gDwJK2w5NYc&list=PL_A4M5IAkMaeLzehBWWVD_EZL7EuaVP-X
its entire complex numbers, you can skip to conic sections part

i think it starts at lecture 10, i dont remember

[opal forge]

.close

[pearl pond]

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[dusty ivy]

And

$(e^{-(1)}(-(1)-1)) - (e^{-(\infty)}(-(\infty)-1))$

[jolly parrot]

SimonWin

[midnight haven]

the what

[dusty ivy]

I am trying to calculate this

[buoyant panther]

well

[dusty ivy]

Which is this

Hence F(b)-F(a)

[buoyant panther]

here you have 0 * inf

[dusty ivy]

where is 0*inf?

The e^{-inf}=0

$(e^{-(1)}(-(1)-1)) - (e^{-(\infty)}(-(\infty)-1))$

so this $(e^{-(\infty)}(-(\infty)-1))=0$

doesnt that leave me with $(e^{-(1)}(-(1)-1))$

[jolly parrot]

SimonWin

[buoyant panther]

yes and (-inf - 1) = -inf

so 0 * (-inf)

[dusty ivy]

yes and that's zerp

BBut e^(-1)

on the left most

that doesn't equal 0

[buoyant panther]

well it's zero

so

just 2/e then

[dusty ivy]

$(e^{-(1)}(-(1)-1))=-2/e$

[jolly parrot]

SimonWin

[dusty ivy]

but how can I even calculate with that lmao

do you know how I can find the integral of a function f that has two different functions in 2 intervals

so like f(x)=... when x < 0 and f(x)= ... when x>=0

[buoyant panther]

you flipped the terms

[dusty ivy]

and then integrate that

[buoyant panther]

firstly you plug upper bound

and then the lower one

you did it in reverse

this is why you got -2/e

instead of 2/e

[dusty ivy]

yes true

it's supposted to be F(b)-F(a)

sorry

[buoyant panther]

just sum of two different integrals?

for some functions it can be done simpler

especially if f(x) would be even

[dusty ivy]

i want to ask wolfram

to integrate x^2 when x < 0 and 2x when x >=0

is this possible

[buoyant panther]

yes it's just x^3/3 when x < 0 and x^2 when x >=0

[dusty ivy]

and then integral of (what you write in here)

[buoyant panther]

it's used when dealing with integration of piecewise functions, especially defined integrals

[dusty ivy]

yes, that's exactly what im doing

integral of x when x > 0 and x^2 when x <=0

doesn't work

,w integral of x when x > 0 and x^2 when x <=0

[jolly parrot]

Results provided by WolframAlpha

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Requested by simonwin

[pearl pond]

@dusty ivy Has your question been resolved?

[pearl pond]

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[pearl pond]

@dire snow Has your question been resolved?

[dire snow]

<@&286206848099549185>

[dire snow]

<@&286206848099549185>

[primal orchid]

What have you done so far?

Any first steps?

[tough horizon]

Please help

[primal orchid]

https://discordapp.com/channels/268882317391429632/488120190538743810

For virus, since I'm leaving, the key is to realize that you have only three equations but 9 variables to choose

So the problem is very underspecified. Just start making choices and you can easily find many matrices that work

[pearl pond]

@dire snow Has your question been resolved?

[pearl pond]

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[midnight haven]

how does subtraction work here, doesnt it just cancel out?

[coarse dawn]

No because it's -(x - 6) in the numerator

So you distribute that negative

[midnight haven]

ohhh

wait no

in the numerator??

[coarse dawn]

Yes

The numerator of the second equation

$-\frac{a}{b} = \frac{-a}{b} = \frac{a}{-b}$

[jolly parrot]

dldh06

[midnight haven]

wait but its already negative

even if you do that

it becomes

(x+6) + (x-6) and (x-6) + (x+6)

which is 0 isnt it

my teacher put this though

[coarse dawn]

The numerator is (x + 6) - (x - 6)

[midnight haven]

so that becomes x + 6 -x + 6

which is 12

[coarse dawn]

Yes

[midnight haven]

then in the denominator though

thats what im trying to figure out

[coarse dawn]

What about it? It doesn't cancel out there

[midnight haven]

but each fo the terms are subtracting by the same term there

how does it not cancel out

[midnight haven]

like here ^

[coarse dawn]

Same logic with adding/subtracting fractions, the denominators don't cancel out

$\frac{1}{2} + \frac{2}{3} = \frac{3}{6} + \frac{4}{6} = \frac{3+4}{6}$

[midnight haven]

oh shit

[jolly parrot]

dldh06

[midnight haven]

im so stupid

lmao

alr thanks

.close

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[wispy comet]

To do this I would that the square root of a^2+b^2+c^2 right

and then divide the original vector by that value

[pearl pond]

@wispy comet Has your question been resolved?

[pearl pond]

@wispy comet Has your question been resolved?

[pearl pond]

@wispy comet Has your question been resolved?

[pearl pond]

@wispy comet Has your question been resolved?

[scarlet raptor]

Yes

[pearl pond]

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[plush bramble]

#old-network for physics or ee server

[pearl pond]

@dire snow Has your question been resolved?

[pearl pond]

@dire snow Has your question been resolved?

[pearl pond]

@dire snow Has your question been resolved?

[pearl pond]

@dire snow Has your question been resolved?

[pearl pond]

@dire snow Has your question been resolved?

[pearl pond]

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[frigid ravine]

Can someone help me with 4c this is what I don’t but I’m unsure if it’s correct

[blazing saffron]

Those are possible values of x

Not what the question asks for

[broken ivy]

,rccw

[pearl pond]

@frigid ravine Has your question been resolved?

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[hollow flame]

Is this where I put my question

[hollow flame]

guys what does this mean

[quiet flower]

i am getting this helper role again and again

even if i remove it

[meager parcel]

you're on thin ice buddy

[quiet flower]

WHY

[hollow flame]

WHAT HAVE I DONE

[meager parcel]

im kidding

you need to find the perimeter

[hollow flame]

I’m trying to figure it out

[meager parcel]

which is the sum of all sides

[hollow flame]

I SWEAR

THIS IS IMPOSSIBLE

I havé like 3/4 of the sides

but like

how do I get the last triangle 💔

[meager parcel]

which is the last triangle

your handwriting is a bit hard to understand

[hollow flame]

um

are you calling it messy.

[meager parcel]

no why would i do that

[hollow flame]

do u want to me to send a video

with my dilemma

[meager parcel]

no i don't think its that serious

[hollow flame]

I do.

[meager parcel]

um i'll try to solve it give me a bit

[hollow flame]

ok thx

there’s another one

with a Cartesian plane

guys I don’t know what that is

[signal tree]

Distance formula

[hollow flame]

wtf is TAGT

[signal tree]

Or by determinants

[hollow flame]

like

I count the lines

[signal tree]

Search it it up

[hollow flame]

I did

I watched the math antics video

[signal tree]

Find distance between ac you get the length of ac

[hollow flame]

ok thx

[signal tree]

Similarly find ab and bc

Then add them up

[hollow flame]

I got 12

[pearl pond]

@hollow flame Has your question been resolved?

[hollow flame]

Sure

[pearl pond]

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[thorny stream]

i got for the dot product of u * v -114, u * w = 36, v * w = -18. Did i do this correctly?

if that's the case, then none of the vectors are perpendicular

[regal herald]

seems valid enough to me

[thorny stream]

okay thank you

.close

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[finite lily]

Im not sure where to go from here

It doesnt simplify its still 0/0

[copper ingot]

Do you have to use the first principals method for this?

[finite lily]

Im not sure what that is

But we have to use the limits thing

[copper ingot]

Okay yeah then you have to use it

So this is the formula

Now $f(x)=2x^3-4x^2+1$

[jolly parrot]

Nayirus

[copper ingot]

And we want to find f'(3)

So subsitituing into the formula we get:

[finite lily]

But thats a different thing than mine

[severe quarry]

That's one of the limit definitions of the derivative

[finite lily]

hmm

but mine doesnt work?

[copper ingot]

$f'(3)=lim_{h\to 0} \frac{(2(x+h)^3-4(x+h)^2+1)-(2x^3-4x^2+1)}{h}$

[severe quarry]

Ah, you began with the other one, that's fine too

[jolly parrot]

Nayirus

[severe quarry]

\ before lim

[finite lily]

but what difference does it make if i use that one or the one i used?

[copper ingot]

Now we expand

[severe quarry]

No difference

[copper ingot]

Oops

[severe quarry]

The two definitions are equivalent. Sometimes, finding the derivative from one is faster than from the other

[copper ingot]

Try expanding the expression

And factoring out h so you can simplify the expression

[finite lily]

ok

[copper ingot]

Is everything going okay @finite lily

[finite lily]

yes its taking a long time XD

im almost done

[copper ingot]

Okay cool

[copper ingot]

No problem take your tiem

*time

[finite lily]

ok i got it

its 30

[severe quarry]

Yes

[finite lily]

hmm

[severe quarry]

Using the other limit definition, we have that [f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}.] So \begin{align*} f'(3) &= \lim_{x \to 3} \frac{f(x) - f(3)}{x - 3} \ &= \lim_{x \to 3} \frac{2x^3 - 4x^2 + 1 - (2(3)^3 - 4(3)^2 + 1)}{x - 3} \ &= \lim_{x \to 3} \frac{2x^3 - 4x^2 - 18}{x - 3} \ &= \lim_{x \to 3} \frac{2(x^3 - 2x^2 - 9)}{x - 3}. \end{align*} Now perform polynomial long division on $x^3 - 2x^2 - 9$ or factor it by other means. \begin{align*} \lim_{x \to 3} \frac{2(x^3 - 2x^2 - 9)}{x - 3} &= \lim_{x \to 3} \frac{2(x-3)(x^2 + x + 3)}{x - 3} \ &= \lim_{x \to 3} (2(x^2 + x + 3)) \ &= 30. \end{align*}

[finite lily]

but its weird how i couldnt do it with my original method

[severe quarry]

That's your original method

[finite lily]

oh i didnt think about using polynomial long division

ok that makes sense

[severe quarry]

With the h -> 0 way, you don't need any polynomial long division, I assume

So in this case, that way would be the faster one

[finite lily]

ah ok

thank you very much @severe quarry and @copper ingot !!!

[severe quarry]

np

[finite lily]

.close

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[copper ingot]

You are welcome @finite lily

[pearl pond]

@copper ingot Has your question been resolved?

[pearl pond]

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[short escarp]

how'd i get this wrong??

[short escarp]

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

[primal orchid]

I see some issues

You missed an interval where it's decreasing, and you should be including the endpoints as long as the endpoints satisfy the condition

Your domain might have to be in its simplest form

[short escarp]

even though the increasing is a open endpt?

[primal orchid]

This is why online hw sucks. You get 0 credit even though you have 95% of the understanding

That last point is still more than the previous points

A function can be increasing on an interval even if there's a discontinuity

If the closed point was below the open point, you would be fine

But it's above, so it's still increasing on [-9, -6]

[short escarp]

gotchaaaa

yeah i messed up on the decreasing i see where i went wrong there

but the constant????

[primal orchid]

Same issue with the endpoints I'd guess

At the point where it changes slope, the value is the same as the flat zone

So your constant interval overlaps with the decreasing interval at exactly one place

[short escarp]

okay and because the endpoint is closed its like on the point ig

brain hurt it says its wrong

everythings parenthesis

.close

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[shell moat]

.close

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[shell moat]

.close

[pearl pond]

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[wet scroll]

how would i go about solving this

where do i start etc

this isnt my paper btw

[signal tree]

You have to solve for z

Basically just like solve for x

[wet scroll]

okay
this is what i have done so far

what is the next step @signal tree

[signal tree]

That's it

Z=(3-2i)/(1-4i)

Now eliminate the denominator

[wet scroll]

wouldn’t it be Z=(3-2i)/1-4i ?

[signal tree]

Oh yes

My bad

[wet scroll]

It’s alright

Now to eliminate the denominator i need to multiply it by itself

Nvm i’m lost again

[signal tree]

Not itself

It's conjugate

[wet scroll]

Ah yes

So multiply 1+4i both sides?

[signal tree]

Yessir

[wet scroll]

Ill get Z(1+4i) = 3-2i

Should i multiply z by the bracket to get rid of it

[signal tree]

No not both sides

What I mean by both sides is on the numerator and denominator

[wet scroll]

oh oh

[signal tree]

Come back to this step 😭

Now multiply conjugates

Remember i²=-1

[wet scroll]

Z= 3-2i(1+4i)

[signal tree]

After multiplication denominator will be 17

How does it cancel out bro

It will not

[wet scroll]

ok it doesnt

💀

1-4i x 1+4i

[signal tree]

Yessir

1²-(4i)²

=1+4²

[wet scroll]

Yep i got 17

Now 3-2i/17

[signal tree]

Now divide them separately

3/17-2i/17

[wet scroll]

okay

now what

[signal tree]

And you forgot to multiply the numerator

😭

[wet scroll]

oh my

Z= 3-2i(1+4i)/17

I got the rest correct

thank you very much

.close

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[polar plaza]

for this part here

why is the part from -2 to 0 not x?

if i set x^2+2x<0

it would be x(x+2)<0

and it would make -2 to 0 negative

[cursive knoll]

because x^2 +2x-6 is also negative on that region so they cancel out

assuming you're trying to solve (x^2+2x-6)(x^2+2x)<0

[polar plaza]

yeah thats what im trying to solve

wait

how do i get that end part then

i tohught

you individually solve

like

x^2+2x-6<0

and

x^2+2x<0

[cursive knoll]

If you are looking for the 0s they you sovle each inividualy

you solve for the 0s because thats where it can change form positive to negative

then you can check any number in each range to see if its positive or negative

so the zeros are -2 and 0 (from x^2+2x) and the -1+-root(7) from the x^2+2-6

so then check if its positive or negative in the ranges -infinity to -1-root(7), -1-root(7) to -2, -2 to 0, 0 to -1+root(7), and -1
+root(7) to infinity

[polar plaza]

ohh i get it

so i number line like this

then cross reference to see where is still negative

if both are negative

it will be positive

[cursive knoll]

yea

[polar plaza]

has to be 1 negative 1 positive

and also cant be -1 or -3 because it has to be <0

and these are the roots which would make it =0

.close

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[pulsar yew]

Hey I need help understanding how to find dN. I took the derivative of N and all I need to do is substitute in x and multiply by dx. In this case, we don't have dx hence the dx at the end of the answer. My question is, which x do I substitute in if x = [5, infinity)

[sharp laurel]

you dont substitute in any x

[pulsar yew]

then how would I find dy or in this case dN?

[sharp laurel]

dN = (--) dx
dN/dx = (--)

they just want the derivative

[pulsar yew]

so just 70 - 2x

[pulsar yew]

can you explain what you wrote here? I have something different in my notes

[sharp laurel]

you can divide by dx

dN/dx is just the derivative

[pulsar yew]

right

but it's dN = dx

[sharp laurel]

yeah

[pulsar yew]

which I understand it as derivative multiplied by dx

[sharp laurel]

exactly

[pulsar yew]

but shouldn't I substitute an x to find the differential?

[sharp laurel]

no

[pulsar yew]

ok

[sharp laurel]

think of it as dN(x)

like a function

[pulsar yew]

right

[sharp laurel]

just how N(x) is a function

[pulsar yew]

or f(x)

[sharp laurel]

yeah

[pulsar yew]

Thank for the help

.close

[pearl pond]

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[vivid vigil]

1/x as x-> 0+ becomes 1/.1, 1/.01, 1/.001, 1/.00…1, which starts to grow as 10, 100, 1000, 100…0, approaching infinity

log of infinity is infinity

np

[pearl pond]

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[high bronze]

How do I find the equations for these?

[signal tree]

Are these parabolas

[high bronze]

Yes so they’re quadratic equations

I know the top is positive and the bottom two are negative but that’s about all I got

[polar plaza]

use vertex form y = a(x-h)^2 + k

[high bronze]

How do I find a, h and k though?

[polar plaza]

h and k is the vertex

[high bronze]

Well actually h and k is easy

[signal tree]

Ye

[polar plaza]

and u find a by plugging in any point on the graph as a and y

as x and y*

and solve for a

[high bronze]

Ok so like if I used the point (0,3) how would I write out the equation?

In what form I mean

[polar plaza]

3 = a(0-h)^2+k with h and k being the vertex

then everytihng is filled in except a

so u can just solve it

[high bronze]

Oh ok got it

I’ll try it then tell you what I get

[high bronze]

Ok so it should be y = 1(x+1)^2+2

Ok I got 41 and 43 right but I’m having trouble with 45

[pearl pond]

@high bronze Has your question been resolved?

[pearl pond]

@high bronze Has your question been resolved?

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[wary bolt]

(I changed altitude to height (h))

but

So far i have h = 7, dh/dt = 2

a = 91, dA/dt =3

The formula for the area of a triangle is A=BH/2

How would I do implict differ to BH/2?

[signal tree]

Apply product rule

Find h first then apply product rule

[wary bolt]

h = 7

@signal tree what would i do from here

[signal tree]

Findbase from original area equation

Bh/2=91

[wary bolt]

How would I isolate db/dt

@signal tree

[signal tree]

You found b?

[wary bolt]

Yes

its 26

[signal tree]

Then you just need to find db/dt

[wary bolt]

Yes I get that

how would I isolate

db/dt

In the picture above

[signal tree]

Just multiply that 2 to the other side

Then Subtract 52 on both sides

Then divide whole equation by 3

[wary bolt]

then divide by 3

Okay

Thank you

[wary bolt]

That resulted in -15.33333333

But was incorrect

[signal tree]

😭

Calculation mistake bro

Lemme check

[wary bolt]

I believe os too, I think I didnt solve for b correctly

if i were solving for b

It'd be

2A/h = B right?

As the question for A is A=BH/2

If thats correct, then B should be 26

oh wait

I put a 3 instead of 7

Simple mistake on my part

[signal tree]

Ye

-46/7 cm/min?

[wary bolt]

Yeah

Next question

@signal tree

I checked over my math and it seems correct

[signal tree]

Can I have the diagram

[wary bolt]

so funny enough that isnt actually given to us

(these questions are just randomly given to us)

(not from any book)

[signal tree]

They've asked you the rate

[wary bolt]

Yeah thats what I was finding right?

I thought in my work I just found dx/dt

Since I reorganized the problem to be x^2 = z^2 - y^2, since we're solving for x

[signal tree]

Why

[wary bolt]

Thats how we did it in class, also

We're trying to find dx/dt

[signal tree]

You are solving for dy/dx

[wary bolt]

dy/dx would be 0

[signal tree]

Dx/dt

[wary bolt]

Yes exactly solving for dx/dt

So I set it in respect to solving for x

[signal tree]

Put the values of dz/dt z and x?

[wary bolt]

heres the problem we did in class

for reference

I did the exact same thing in my problem all the way above

[signal tree]

Ok so 110 meters of rope is gone

[wary bolt]

Yes

So z would be 110

If we compare the steps taken in the class exmaple to how i did it

[signal tree]

Really?

Idk either

[wary bolt]

y = 7, z is 110 bc thats how much rope is out

[signal tree]

110 is the distance of rope let out

[wary bolt]

ye

Refering to class problem

[signal tree]

How does that correspond to length of the rope?

[wary bolt]

If 110 of the rope is out, thats the length of the rope?

[signal tree]

Ok say I have 1000 m rope

I pull and let out 1 m

Now the length of the rope is 999 meter

Right?

[wary bolt]

Yeah but

it says

"when 110 ft of rope is out"

[signal tree]

Hmm

[wary bolt]

Your example wouldnt be applicable as, we wouldnt know the starting value of rope

If the rope is pulled through the pulley at a rate of 14 ft/min, at what rate will the boat be approaching the dock when 110 ft of rope is out?

so dz/dt would be 14

oh shit

wait a second

Got it right, ive been inputting the wrong number the whole time

I should not be doing math thing late at night

Thank you Ram

[signal tree]

You had to do √110²-7²

[wary bolt]

oh i did that already

i just was doing

(110)(10)/sqrt12051

when it shouldve been 14 instead of 10

Though for right triangles

[wary bolt]

finding a side is this

but when finding z, its +?

whys that

[signal tree]

It's x

[wary bolt]

finding a side we do what you said before

or well finding x

But how come if we were to hypothetically find the z (hypotenuse) we would add the two together

[signal tree]

We're finding x

X is not the hypotenuse

[wary bolt]

Oh no i get that lol

I was giving a off topic quesifon

[signal tree]

I mean

Of course we would add

a²+b²=c²

For all real triangles

[pearl pond]

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[analog coral]

Help

Tried making a Cartesian

[pearl pond]

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[pearl pond]

@analog coral Has your question been resolved?

[pearl pond]

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[twilit harbor]

hi

[twilit harbor]

I'm lost here

-12 and 2 don't multiply to 24

nevermind it was 6 and 4

I have another question

what was wrong with the last method

[pearl pond]

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[pearl pond]

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[opal scroll]

help

[opal scroll]

(A symmetric difference B symmetric difference C) Union NOT A

I keep thinking this should have the middle part excluded from the shading too

um? bruv i think u gota pick a diff channel

[vital kelp]

oh shit my bad

was available when I was typing

[pearl pond]

@opal scroll Has your question been resolved?

[opal scroll]

is anyone out there?

[craggy lake]

The middle part should be included, so your diagram is right

[pearl pond]

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[pearl pond]

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[brisk mulch]

Solve system by graphing.

Please @ me because I was absent Friday and Ion even know what to do fr

[lilac quartz]

can you graph eq 1?

[brisk mulch]

Yeah if I put it into vertex form

[lilac quartz]

can you try?

[brisk mulch]

sure

[lilac quartz]

,w plot -3x^2-30x -71

[jolly parrot]

Results provided by WolframAlpha

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

Requested by dotdoc.

[lilac quartz]

Keep doing

[brisk mulch]

y=-3(x+5)+4

Now what

[lilac quartz]

ie your vertex is at?

[brisk mulch]

Wtf is a vector

Oh

(-5,4)

[lilac quartz]

roots?

[brisk mulch]

Wym roots

[lilac quartz]

the x intercepts

[brisk mulch]

-6 and -4

but its some weird decimal

[lilac quartz]

Not irrational right?

roughly you can sketch them

-3x-17 is a straight line

[brisk mulch]

i mean yeah

[lilac quartz]

If you draw both graphs

you can see they intersect nicely

[brisk mulch]

Ok

So basically i just have to find where they intersect

thats it

[lilac quartz]

yeah

[brisk mulch]

Thanks

[lilac quartz]

That’s the solution

For the system

[brisk mulch]

Thanks thanks thanks

.close

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[midnight haven]

Question: Suppose we are flipping a fair coin 5 times. Find the expected number of heads we obtain given that the first and last flips are of opposite parity (i.e. one is heads and one is tails).

Not sure how to start with this one, could you do (5C0+5C1+5C2+...5C5)*1/2^5?

[toxic lichen]

could you do (5C0+5C1+5C2+...5C5)*1/2^5?
that's just 1

[midnight haven]

right 🗿

[toxic lichen]

the first and last flips are a heads and tails in some order, so that gives you 1 heads guaranteed

what's the expected number of heads from the middle 3 flips, which are not constrained by any conditions?

[midnight haven]

no its 1/32 smh

oh wait

never mind im stupid

ignore me

i go to my corner of shame now

[midnight haven]

Is it 1.5 H just from 0.5*3 (probability of heads * number of flips)?

[toxic lichen]

indeed

[midnight haven]

So then it's 1.5 + 0.5 =2 expected heads

from 5 tosses where the first and last are oppositee

[midnight haven]

ah sorry not 0.5 but 1

so 1.5+1=2.5

just like 5 * 0.5 🧍

Thank you!

.close

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[runic glacier]

Taking the derivative of g3(x) it’s absolute value is not less than one near the root x=3 yet it converges to a solution when I put it into a fixed point iteration algorithm

Is there a specific reason why?

[pearl pond]

@runic glacier Has your question been resolved?

[pearl pond]

@runic glacier Has your question been resolved?

[pearl pond]

@runic glacier Has your question been resolved?

[plush bramble]

!show

[pearl pond]

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

[pearl pond]

@runic glacier Has your question been resolved?

[runic glacier]

Derivative of g3(x) is 6lnx

Which is not less than one near the root x=3

[plush bramble]

that's not the correct derivative

use power rule

[runic glacier]

Oh right I was I integrating instead

Thanks for the help

.close

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[timid path]

Need help with this number theory/linear recursion exercise: Prove that there exists a fibonacci number which is divisible by 100

[pearl pond]

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@timid path Has your question been resolved?

[pearl pond]

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[midnight haven]

hello can someone help me solve this equation on gradients and y intercepts

[midnight haven]

this is the question

i thought i had it but whern it tried to help me the videos couldnt load so im stuck

[mild bronze]

Oh

So

Y=mx+c

Where m is the gradient and c is y intercept

Gradient = (-2,-8) (2,4)

[midnight haven]

ok

[mild bronze]

So

Equation=

[midnight haven]

im starting to get irt

[mild bronze]

Y2-Y1/X2-X1

-8-4/-2-2

-12/-4

=3

Gradient=3

So we get

3x+c=y

C is y intercept

So y intercept is y's interceot when x=0

Which is -2

So

3x+-2=y

Y=3x-2

[midnight haven]

ah ok

ty

You helped me understand

ty

[mild bronze]

Np

[midnight haven]

so is it 3x+-2 or 3x - 2

sorry but you put both of those

[pearl pond]

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[pearl pond]

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[meager elm]

need help

[dense goblet]

what don't you understand

[meager elm]

how to get the coordinates

i mean i know the coordinates for vkgo

but which one isnthe reflection

[dense goblet]

reflection across y=-x

[meager elm]

which means -x,-y right

[dense goblet]

not really

just try to imagine flipping the shape across that line

[meager elm]

i mean y-,-x

[dense goblet]

that's not going to work

[meager elm]

but isnt y=-x -y,-x?

[meager elm]

is whck the reflection i use

[pearl pond]

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[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

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[heady oak]

how do i solve this

.close

[pearl pond]

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[vagrant ingot]

∀x(P(x)∨B(x))
∃x¬P(x)
Therefore, ∃xB(x)
Can I use disjunctive syllogism here?

[solid ivy]

Need to use disjunctive syllogism on instances of x, so universal instantiation first, then syllogism, then generalization

[pearl pond]

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[woven crow]

i need help with 2v

2b

[pearl pond]

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[manic sundial]

Fully factor 8m^3-125n^3

[coarse dawn]

Hint: difference of cubes

[manic sundial]

Got it what about 54xm^9n^12+16xp^3?

[unborn abyss]

do GCD first

[pearl pond]

@manic sundial Has your question been resolved?

[manic sundial]

2?

[unborn abyss]

oh yeah probably

[pearl pond]

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[shy ginkgo]

So I have a few points and I’m trying to find the slope

[shy ginkgo]

But I’m unsure if I’m supposed to include a point or not

It wants me to graph the points and it says “set y-int to zero”

So should (0, 0) be a point?

[midnight haven]

full question please

[shy ginkgo]

So I have the points (5.09E-5, .096) (1.02E-5, .214) (1.53E-5, .3) (2.04E-5, .403) (2.55E-5, .516) (3.05E-5, .61)

[shy ginkgo]

And it says set y int to 0 so I’m unsure if I should include (0, 0) as a point when calculating slope etc

Because the answer drastically changes

[midnight haven]

i mean

!original

[pearl pond]

Please show the original problem, exactly as it was stated to you. A picture or screenshot is best.

If the original problem is not in English, then post it anyway! The additional context might still help helpers help you. Do your best to translate.

[shy ginkgo]

Oh

[midnight haven]

you can use the slope formula

[shy ginkgo]

Ik how to find the slope I just don’t know whether or not to include (0, 0) in the data points

Lemme take a photo

@midnight haven

I calculated the absorbance and concentration so that gives me my points

Since the problem is chemistry related I’m not sure context will give you much help

[midnight haven]

if (0,0) and those data points lie on the same line then you can include it in your calculation of the slope

[midnight haven]

oof yeah

[shy ginkgo]

I’m just unsure what setting the y int to zero means for me

It’s not in my data points but it says to set y int to 0

So if there’s a y-int of 0 then… (0.0) is a point?

Idk I’ll figure it out

.close

[pearl pond]

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[fresh nymph]

how do we know two planes are coincident and is cut by a third plane

as opposed to (7)

like in this example

[rough stream]

It's case (7)

As given at the bottom there

[fresh nymph]

no the first photo i sent

e)

it says it is case 2

[rough stream]

Oh mb missed it.

I disagree with them. It's case (7) as well.

[fresh nymph]

hmm

how do you know if its one of those other cases then

[rough stream]

Note that case 2 talks about two planes

[fresh nymph]

idk how i can find which case just by looking at the matrix

only whether its parallel, meeting at a line or meeting at a point

[rough stream]

OH I didn't realize case 2 had coincident planes

[fresh nymph]

yeah

[rough stream]

Fml I'm leading you on a roller coaster here let me get my bearings

[fresh nymph]

its fine lol

[rough stream]

It's the original matrix. One row is a multiple of another

Those two rows represent coincident planes.

The matrix reduction hides this. That is, one of the rows immediately goes to 0.

[fresh nymph]

ohh i missed that

does that mean it doesnt matter which row has the 0000

[rough stream]

You're allowed to rearrange the rows during a reduction, so it never matters

[fresh nymph]

i see

and also just out of curiosity

is this smth you would find in a linear algebra course

like is this what matrices are or is it smth else

[rough stream]

Matrix reductions and interpretations of it? Yeah you'd find that

[fresh nymph]

ok thanks a lot!

.close

[pearl pond]

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[sharp crater]

Probably easy, but how to do this?

[unborn abyss]

draw a sketch

[sharp crater]

Ah yeah that made it easy lol

Would this be just zero? @unborn abyss

[unborn abyss]

how could y be 0?

what value could x be to make that happen?

[sharp crater]

1?

[unborn abyss]

well there's a couple problems there

for one, x can't be 1 since it says x > 1

[sharp crater]

True

[unborn abyss]

for two, if we plug 1 in for x

we get

[sharp crater]

Undefined?

[unborn abyss]

$\f x{x-1} = \f{\blue1}{\blue1 - 1}$

[jolly parrot]

hayley.jpg

[unborn abyss]

which yeah is undefined

[sharp crater]

Yeah

So would it be -1.9 since plugging that in would be the only way to have a value more than one?

@unborn abyss

[unborn abyss]

would it? what value for x causes y to be -1.9?

[sharp crater]

Or no -0.9?

Cause -1.9 makes it less than 1

[pearl pond]

@sharp crater Has your question been resolved?

[pearl pond]

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[oblique yacht]

does anyone see how the circled substitution is made

[oblique yacht]

oh maybe z' is dz/dx

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[pearl pond]

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[wild fable]

Find the region bounded by the curves $x = y^2 + y$ and $x + y = 3$. I've found intersection points (y-1)(y+3)=0 and so got y = 1 and y = -3. So my integral will be $\int_{-3}^{1}\left(\left(3-y\right)-\left(y^{2}+y\right)\right)dy$. Is this correct?

[jolly parrot]

water beam

[wild fable]

and I guess im finding this area ?

[toxic lichen]

appears correct yes

[wild fable]

appears

ok cool

[pearl pond]

@wild fable Has your question been resolved?

[wild fable]

If I have 3 lines, y = x, y =2x, y = 6 - x

is it illegal for me to do this

x - 2x = -x

then

-x = 6 - x

0 = 6

ok

so it is illegal

lol

cuz that dont make sense

@toxic lichen how do I set all three of them equal?

[toxic lichen]

i really want to xyproblem you rn

you mean you want to find the AREA of the region bounded by 3 lines?

[wild fable]

yes

[toxic lichen]

graph it, then split vertically at an appropriate point.

[wild fable]

what is considered "appropriate"

[toxic lichen]

graph it and you will see

if you don't see then show me the graph and i will tell you

[wild fable]

i dont see where to split it here

[toxic lichen]

zoom in on the triangle

[toxic lichen]

right

you split vertically at the purple-black intersection

[wild fable]

so what does this do

[wild fable]

whats the purpose of this?

[toxic lichen]

so that you can do the left half and right half as two different integrals

of (purple - green) and (black - green) resp.

[wild fable]

so im trying to find this?

[toxic lichen]

yes

[wild fable]

thats gonna be confusing

so i cant just randomly draw the line I need to find out its x coordinates then

[toxic lichen]

coordinate, singular

but yes

[wild fable]

to find it i will do 2x = 6 - x

i assume

3x = 6 so x = 2

[toxic lichen]

why assume

[wild fable]

ok

i definitely 99% know

okay yeah i think i got this

.close

[pearl pond]

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[split dirge]

Doing linear algebra here, where do I start? Why are matrices and vectors in a denominator?

[split dirge]

I suppose g and h transpose "cancel" to give a number in R¹ but other than that, i'm stuck

[solid ivy]

yeah $B^{-1} g$ gives a nx1 column vector, then $h^T$ times that is the dot product of a row and column vector

[jolly parrot]

992qqoloy

[split dirge]

yes

[solid ivy]

$gh^T$on the other hand gives an nxn matrix

[jolly parrot]

992qqoloy

[split dirge]

oh i got my dimensions confused

[solid ivy]

Sooo yeee idk how it'll workout exactly but I think you're supposed to multiply C by D and show it's the identity matrix

[split dirge]

ok

thanks!

[solid ivy]

Np

[split dirge]

are g and h necessarily column vectors of B or is it arbitrary?

or just ones that satisfy that inequality

[solid ivy]

just arbitrary I think

Such that they don't satisfy the equality

like such that that isn't equal to 0

[split dirge]

so does 1 + (h^T)(B^-1)g just end up giving a number

ok it does

[solid ivy]

mhm

[split dirge]

g(h^T) produce a matrix of same size as B, but not necessarily B, but I don't know how I would get further multiplying the inverse of B by it

unless g(h^T) = B

which i dont want to assume

[solid ivy]

Hmm

[pearl pond]

@split dirge Has your question been resolved?

[pearl pond]

@split dirge Has your question been resolved?

[split dirge]

.close

[pearl pond]

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[rigid mist]

$\sum_{r=1}^{r} {2r}^3=\left(\sum_{r=1}^{r} {2r}\right)^{3}$

[jolly parrot]

yajatk07

[rigid mist]

can we do this?

[foggy mason]

no