help-28

[sturdy nymph]

,, y = a(x-1)^2 + 1

[glossy valve]

rxhuls👽

[sturdy nymph]

,, 3 = a(x-1)^2 + 1

[glossy valve]

rxhuls👽

[regal copper]

okay

do you know the standard parabola equations ?

[sturdy nymph]

like y = x^2?

[regal copper]

yes

now observe

[sturdy nymph]

or ax^2 + bx + c = 0

oh ok

[regal copper]

this parabola is a shifted one

so how may we go about it ?

[pallid sorrel]

okay

personally

i would

[sturdy nymph]

well the +1 at the right of the equation means its shifted one to the right

?

[pallid sorrel]

use ax2 + bx + c = 0

[regal copper]

okay

[pallid sorrel]

then sub stuff in

[regal copper]

and what about it being shifted upwards ?

[sturdy nymph]

shifted up 1

[regal copper]

correct

[pallid sorrel]

shifted up u will get by subbing points in

[regal copper]

but here

it will be -1 not +1

now try to put

[sturdy nymph]

wait

what do you mean

[pallid sorrel]

what?

i dont think so aha

[regal copper]

the findings till now in the parabola equation

[sturdy nymph]

the +1 at the right side here means it shifted to the right

and the -1 in the brackets means its shifted upwards on the graph

[regal copper]

see look at it this way

[pallid sorrel]

wrong way

in brackets means to the right

• 1 at end means up

[sturdy nymph]

ohh

[regal copper]

since the centre is at 1,1

[sturdy nymph]

axis of symmetry is 1?

[regal copper]

we get (x-1) to be zero at 1

so we will right (x-1) not (x+1)

open desmos and try it

you will understand better visually

[sturdy nymph]

but what do i put as a

[regal copper]

we will get to that

for now assume a as 1 and look at the differences made by +1 and -1

we will figure out a as well

[regal copper]

so you see how (x+1) shifts the centre of the parabola to the left while (x-1) shifts it to the right

[sturdy nymph]

yes

[regal copper]

okay now lets rewind a little bit

[sturdy nymph]

its moving the x of the vertex

[regal copper]

yes

from here we know one point which is lying on the parabola

[sturdy nymph]

we know (0,3) and (1,1)

[regal copper]

if we consider the equation with 'a' as a constant instead of a value

[regal copper]

correct

[regal copper]

and then put (0,3) in the equation, we can find out the exact value of a

using which we can find the equation of the curve

try it

[sturdy nymph]

so i put 0 for x?

[regal copper]

yes

put 0,3 here

in the standard form of parabola

[sturdy nymph]

where did u gt the 4 from??

,, 3 = (0-1)^2 + 1

[glossy valve]

rxhuls👽

[sturdy nymph]

,, 3 = (-1)^2 + 1

[glossy valve]

rxhuls👽

[sturdy nymph]

,,3 = a(-1)^2 + 1

[glossy valve]

rxhuls👽

[vivid berry]

wowwww nice

[sturdy nymph]

,, 3 = a + 1

[glossy valve]

rxhuls👽

[sturdy nymph]

,, 2 = a

[glossy valve]

rxhuls👽

[sturdy nymph]

oh wow

[regal copper]

that is the standard form

[regal copper]

nope its a bit wrong

[sturdy nymph]

,, y = 2(x-1)^2 + 1

[regal copper]

,, x^2=4ay

[glossy valve]

rxhuls👽

yashmajor

[regal copper]

this is the standard eq

we know that the vertex is shifted so we did

,, (x-1)^2=4a(y-1)

[glossy valve]

yashmajor

[regal copper]

now in this we should put (0,3)

[sturdy nymph]

it doesnt specify which form of equation to use

so mine is fine too

[sturdy nymph]

this should be right

so i had to sub (0,3) into the vertex form since we had the vertex

[vivid berry]

yea i only use ur way

[sturdy nymph]

so we find a wiith that and put that in boom

its right

[regal copper]

congratulations (:

[sturdy nymph]

thank you sm

[regal copper]

i just prefer the standard form so tried to explain through that

[open igloo]

Are you left handed? Just asking

[regal copper]

but there are many ways to solve a problem

[regal copper]

nope

why ?

[open igloo]

Kk

Nothin

[sturdy nymph]

u did the smiley face wrong order xd

[regal copper]

if i do it : ) this way discord converts it into the emoji 🙂

[open igloo]

I just wanted to see if my guess was right or wrong lol

[regal copper]

and then it looks creepy

[vivid berry]

:)

[sturdy nymph]

:)

[open igloo]

Lol

[regal copper]

how did you do that

[vivid berry]

idk

[sturdy nymph]

:)

[open igloo]

Anyways, have a nice day

[vivid berry]

there was a setting

anbd i turned it off

ages ago

[regal copper]

🙂

[sturdy nymph]

theres a setting

lmao

[regal copper]

lemme seeeeee

[vivid berry]

idkk i forgot

i did it in like 2019

cuz it was annoying

[regal copper]

got it

thankyou :)

[vivid berry]

np

[sturdy nymph]

anyway thank you @regal copper

[regal copper]

welcome :)

[sturdy nymph]

ill try to do this one by myself

got it ❤️

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[vivid berry]

excuse me

[vivid berry]

why is this angle 2x the one at circumference

[granite torrent]

are you familiar with Inscribed Angle Theorems?

if not, this photo will help you see:

[vivid berry]

but it doesnt look like the last one

it looks different

[vivid berry]

why cant angle DBC be the one that is x

why is it BDC

[granite torrent]

(do you mean BDA?)

[vivid berry]

yes

[granite torrent]

primary point here is that angles AOB and ADB are both constructed with two points that define the same arc (the arc here is arc AB)

[vivid berry]

can u pls draw?

i am a visual leaner

[granite torrent]

i have diagram here

[vivid berry]

what the heck 😔😔

wait

are u just connecting it to points a and b

[granite torrent]

i just colored things

[vivid berry]

so

A to O and b to O

[granite torrent]

the purple segments define the angle labeled 2x” and you see they connect at the green segment right

[vivid berry]

that angle

and then B to D and A to D

so then those 2 angles

in the middle

the one at the centre is 2x

?

[granite torrent]

i have no idea what you are saying but it seem you have some idea of what’s going on

[vivid berry]

oh

like

yk

ur making lines from

so uh

[granite torrent]

does this sentence make sense to you

[vivid berry]

angle from outside

angle at centre and angle at outside

is coming from the same point

?

i think this makes sense let me find another example

do u have another exampl;e

of this theorem?

they are coming from the same points

my 4 lines

[granite torrent]

yes that is correct

[sacred yarrow]

[granite torrent]

i see what you are referring to now yes

[vivid berry]

OK THX!

this clears up a lot for me

ALL BC OF U!!!

[vivid berry]

haave a wonderful NIGHT

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[torn jolt]

for this

[torn jolt]

in my physics class we often denote this as [
\vj F = -\grad U
]

[torn jolt]

but can the negative sign be omitted like that ?

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[pseudo cape]

@buoyant pewter we talked about differential equations earlier, so when we have a homogenous one, we just use y = e^(lambda*x) and create the characteristic polynomial, if it has multiple zeros then multiply with x

[pseudo cape]

What if we have a nonhomogenous one?

Just subtract everything from one side so it's a homogenous one?

[buoyant pewter]

then the general solution is consisted of sums of two things:

first one is:

the general solution of the homogenous equation

and seocnd part is the particular solution of non homogenous euqations

and htis secodn part we dervive

mainly with two ways

or we predict

or we use method called

[pseudo cape]

But why can we not just make a homogenous DE out of it?

[buoyant pewter]

metoda uzmienniania stałcyh , i do not know eng term

[pseudo cape]

So e.g. from y'' = y' + y''', we make y'' - y' - y''' = 0

[buoyant pewter]

i can write formual eventually if you like

[pseudo cape]

Yes please

[buoyant pewter]

yoru eq is homogenosu now

ok wait:

we discuss for simplicty such a non homogenous equation:

then it sgeneral solution is:

where:

y_1, y_2, are fundamental system of homogenous equation, means

thier wronskin is not equal to zero,

and:

first method of fidning phi is:

thsi way, you find c_1 and

c_2

to find phi(t)

in other words:

this formual you derivve when you slve this system above

iw rote

[pseudo cape]

Is that W the Lambert-W function?

[buoyant pewter]

the other method esier but of less gewnerality is

prediction

[buoyant pewter]

no

W in my textbooks

is a

Wronskin

wronskian

[pseudo cape]

Oh

[pseudo cape]

What do you mean by prediction

[buoyant pewter]

let me show you method of prediction:

thta we use only

if h(t) is a nice form of special expoential trig polynomial

i write it now:

if:

then:

if zero is not root of lambda polynomial you remind

and if zero is root of one multiplicitcy

then :

and if zero root is of multiplcity of two then:

the general form of h ( t ) when we use prediciton method is:

then you can use prediciton method

[pseudo cape]

Hm, ok, so it seems like you can't use the prediction method very often

[buoyant pewter]

no 🙂 hence i gave you formual above

so that is price

or you love

to integrate )

but it can be a pain

or you predicit

i give oyu fwe exampoel,s to solve

:

[pseudo cape]

Thank you!

[buoyant pewter]

for prediction:

and:

for integration formual :

if you sve them then yo can realluy feel

waht you liek or not )

[pseudo cape]

Ok, thank you! I will close and open a new channel if there is still something I don't know

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[pseudo cape]

.reopen

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✅

[pseudo cape]

Wait can we do one example of prediction method together please? So for example e)

$y'' + 4y = \cos(2t)$

[buoyant pewter]

sure

look:

2i, - 2i, are lambdas

hence

y_1 = cos2t

and

y+2 = sin2t

then

general of homo is:

y(t ) =c_!cos2t andC_2sin2t

now we predict:

[pseudo cape]

Yes

[buoyant pewter]

Phi(t) = t ( A cost2t + B sin2t )

that is all )

now yo mjst fidn A and B

plugign it

into non homo

then you derive A and B

and you are at home )

finally you sum all

togetehr

as you can se such euqaiotns can be solve din mind )

[pseudo cape]

Oh, thank you!

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[bitter beacon]

how to solve this ?

i think i have to use some maclaurin representation of some function but idk which

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@bitter beacon Has your question been resolved?

[bitter beacon]

<@&286206848099549185>

[pseudo cape]

I'd compare this to $f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)(x-a)^n}{n!}$

Either a = -1 or x = -1

[bitter beacon]

ok i have to go tnx for responding

[pseudo cape]

$a = 0$ and $x = -1$, then $f(-1) = \sum_{n=0}^\infty \frac{f^{(n)}(0)(-1)^n}{n!}$

We need $f^{(n)}(0) = \log^n(2)$

[bitter beacon]

what is f(x)?

what function

[pseudo cape]

We don't know yet

We have to find it

[bitter beacon]

plz let me know

i have no time toe to think

[pseudo cape]

Idk yet

[bitter beacon]

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[pseudo cape]

(You can keep it open if you like, someone else can help)

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[harsh arch]

hello

[harsh arch]

Question : find x that makes the triangle larger or equal to 20

[supple ginkgo]

hey

[harsh arch]

can anyone solve this

[wise wyvern]

Makes the triangle larger or equal to 20?

Are you sure?

[harsh arch]

idk

[supple ginkgo]

can someone help with this i dumb i in year 8 so easy work gave us easy hw this week but i still dumb how do i firgue this out

[harsh arch]

its a question from my book , the due is Tommorow

[supple ginkgo]

i mean i know kinda what to do

but degrees how to firgue that out in my head

[harsh arch]

can u read arabic?

[supple ginkgo]

nope

why

[harsh arch]

dude the question is in arabic

[supple ginkgo]

no it not

[harsh arch]

it should be like this

[supple ginkgo]

why type of angel is this x y z

[supple ginkgo]

huh

i dont even understand that

[harsh arch]

ok , thanks for the help !

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[hollow grove]

In how many ways 5 boys and 3 girls can be arranged in a row so that no two girls are together?

[hollow grove]

My initial thought on this question was
(Permutations of all objects as distinct) - (Permutations when two girls are together) - (Permutations when three girls are together)

[lucid path]

The 5 boys can be arranged in 5! or 120. The 3 girls can occupy each of the 4 spaces between boys, and each of the two spaces before the first boy and after the last boy. Therefore, we can select 3 of the 6 spaces in 6C3 = 6!/((6 - 3)!3!) = 20 ways. Then the 3 girls can be arranged in 3! = 6 ways in these chosen spaces.

Therefore, the total number of ways to make the required arrangements (such that no two girls are adjacent to each other) is 120 * 20 * 6 = 14400.

[hollow grove]

Ahh that seems way easier

I just made it hard I guess

[lucid path]

maybe

[hollow grove]

Lol

What I did was say take G1G2 as one object then all its permutations

But that would only mean G3G1G2 and G1G2G3 as a possibility

[lucid path]

just google?

[hollow grove]

So I considered G2G3 and G1G3 too

Here G1G2G3 mean Girl 1,2,3 resectively

[lucid path]

That would work

i’m pretty sure

[hollow grove]

How do you know for sure

[lucid path]

how about G2G1G3

[hollow grove]

If you take G1G3 then G2G1G3 and G1G3G2 is already a possibility

[lucid path]

You lost me

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[dense nova]

This is probability and stats this wouldn’t be a discrete distribution due to the negative right?

-3

[robust slate]

You can have the variable be negative

You just need the probabilities to sum to 1

[dense nova]

Oh ok so the probabilities is the only thing that couldn’t be a negative

[robust slate]

Yeah

By definition those range from 0 to 1

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[spark granite]

I don't understand how they plot the tip of the B vector at <0,8,0> doesn't seem like that is the right point

[spark granite]

or does it work because the vecotr is parallel to that point

[full marsh]

It starts from (3, 0, 1)

and goes 8 units in the y axis

The vector that goes there is (0, 8, 0)

and its tip is at the point (3, 8, 1)

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[spark granite]

this makes sense

tyty

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@timid lotus Has your question been resolved?

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[wraith river]

why am i wrong 😢

[wraith river]

Ik i cant do it bc i have to cross out x+2

But what else do i do in this situation

[torn jolt]

Don’t overthink just use l’hôpital

[slow canopy]

Could it be that the limit doesn’t exist?

Try factoring x^6 -64 as a difference of two squares instead

You can see that -2 is a root of this so it has a factor of x+2

[dull seal]

Yes, you have x((x^3)^2-8^2) which is x(x^3-2^3)(x^3+2^3) And from here, you know x^3+2^3 is (x+2)(x^2-2x+2^2)

so you don't need more factoring

and finish simplifying

[wraith river]

Thats how they did it but i didnt know what that is

[wraith river]

Im not sure what that is my teacher never used that word

[wraith river]

Oh ok

[dull seal]

but remember that in the end i'm factoring only x^2+2^3

you have to add the rest

(x+2)(x^2-2x+2^2)(x^3-8)x

like this

from here u simplify and u subtitute by -2

[wraith river]

wait did i do it wrong/

nvm i didnt lol srry ty

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[lofty prawn]

Stuck on what to do with lambda here

[lofty prawn]

Heres my work

[gritty rose]

you used it to find x^2=y^2=z^2

lambda isn't actually part of the answer. it's just an auxiliary variable to help

[open zinc]

Use Kuhn Tucker conditions or show that if lambda =0 your constraint is not binding.
Usually you don’t need to interpret / solve for lambda though

[lofty prawn]

i thought it was split into cases. If lambda =0 is there anyway for me to find conditions s.t. x^2+y^2+z^2=0

[lofty prawn]

outside of the scope of the class

[open zinc]

The cases are essentially what kkt is doing

[lofty prawn]

or are the only points that i care about (1/sqrt3,..., ...) and (-1/sqrt3,..., ...)

[open zinc]

Yeah, those are the optima

[lofty prawn]

is it just those or is it ever differnet combo of +-

Since x^2=y^2=z^2

[open zinc]

Then you compare the levels of the objective function to classify them

[open zinc]

In this case, I believe it is every combination. You can check that every combination evaluates to 1 in the constraint, and an optimal outcome in the objective

[lofty prawn]

Looking at it thought wouldn't it just equate back to (1/sqrt3, ..., ...) since x,y,z are all squared in f(x,y,z)

[open zinc]

You have 8 optimal points in 3 dimensions, but yes they all have the same value

[lofty prawn]

looking at it though (haven't worked anything out), couldn't (0, 1/sqrt2, 1/sqrt2) also be an optimal point since it satisfies all 4 of the system of equations

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[restive hawk]

can someone help me with this

[restive hawk]

want to use lagrange multipliers

my work so far

but i am not sure what to do since i dont think ill be able to get z in terms of lambda

.close

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[acoustic drift]

How can I solve these equation: 3x + 1y = 155 and 2x + 3y = 220

[acoustic drift]

How would you go about it

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[fathom crater]

partial derivative: which one of these two is correct? I am calculating the partial derivative via chain rule for a computation graph. Is the partial derivative (Z^2+4)/ partial derivative Z^2 equal to two? or is it a constant and equal to 1?

[umbral dome]

what was the original function you're differentiating?

[fathom crater]

f(x,y,z) = x^2y + y^2z + z+2

my partials for df/dx and df/dy have the correct results

however for df/dz the top portion makes my math correct, but Idk if partial derivative (Z^2+4)/ partial derivative Z^2 is actually suppsoed to be equal to 1 or if its 2, which an online calc is saying it is

trying to determine what the actual value of this segment is

[umbral dome]

it might be easier if you imagine $z^2$ as its own variable, like $z^2=w$ for example. what would $\pdv{(w+4)}{w}$ be?

[glossy valve]

cloud

[fathom crater]

i know that thats equal to 1

but because of the exponent idk

symbolab is telling me its equal to 2

[umbral dome]

the dz^2 makes it look like a second derivative, so symbolab may be interpreting it as that

[fathom crater]

That makes sense

Is it treated as a constant in this sense? because if so it would be equal to 1

[umbral dome]

well the derivative of any variable with respect to itself is 1

[fathom crater]

chat gpt just said its equal to 2 lol, im so confused, jk its leaving out all of the constants so i am not sure what it is determining it as

pink highlighted was an instructor example, the rest is what i am trying to figure out. The value that should be placed in highlighted green, its either 1, 2 or 2z. If its 2z I do not think the stuff highlighted in grey should exist, but it does. If its 2 idk what im doing wrong, if its 1 i am correct

its a backwards propogation from a computation graph using the chain rule

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[fathom crater]

.reopen

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✅

@fathom crater Has your question been resolved?

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@fathom crater Has your question been resolved?

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[marble fable]

How do you factor number 12 and 13

[marble fable]

Ik for 12 you can factor out (x+1)½, but idk what happens to the rest

[fathom saddle]

Every term gets divided by (x+1)^(1/2), if you factor by that

[marble fable]

Um

[fathom saddle]

Give a quick check to your exponent rules

[marble fable]

½ - ½ = 0, 3/2 - ½ = 2/2 which is just one

[fathom saddle]

Yes! So you'd have (x + 1)^1

[marble fable]

Ohh

Now I have 1 - (x+1)

[fathom saddle]

Don't forget as well, you did factor by (x + 1)^1/2, so that will be factored out

[marble fable]

Here's what I have now

I just need to simplify and I believe I should be good

[fathom saddle]

I agree with that. Good work!

[marble fable]

Idk where to start for 13

[fathom saddle]

It's a similar idea. Remember that sqrt(x) is just another way to write x^(1/2)

[fathom saddle]

Note that (a - x)^(1/2) is not a^1/2 - x^1/2

[marble fable]

idk what to do

.close

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[chrome sable]

i did this just fine when it was = to e^y but now its cos(y) i keep getting it wrong

[spice knot]

do you know what d/dx cos(y) is

[chrome sable]

-sin(y) i think

[spice knot]

do you know what d/dx cos(x^2) is

[chrome sable]

-2sin(x)

[spice knot]

uh

,,\dv{}{x}\cos(x^2)=-\sin(x^2)\cdot2x

[glossy valve]

mtt07734

[spice knot]

you didnt do chain rule correctly here nor in the other problem

[chrome sable]

do u just do power rule or also chain rule?

[spice knot]

I am doing both

I am first doing chain rule because you have one function in another

[chrome sable]

oh i didnt know i had to use the chain rule in the problem

[spice knot]

do you know what chain rule is

[chrome sable]

yea

[spice knot]

where do you use chain rule then

[chrome sable]

i have a hard time seeing it but its when a funciton is inside a funciton

[spice knot]

now take a look at cos(x^2)

[chrome sable]

so u do

-sin(x^2) * 2x

[spice knot]

yea

[chrome sable]

ok

[spice knot]

now take a look at cos(y)

you know y is also a function of x

so chain rule works here too

[chrome sable]

oh ok

uhm

[spice knot]

try this out

[chrome sable]

-sin * y?

[spice knot]

-sin(y) * y'

just like for x^2, you treat y the same way

this time though you leave y' as y'

,,\dv{}{x}\cos(y)=-\sin(y)y'

[glossy valve]

mtt07734

[spice knot]

you can also write y' as dy/dx

[chrome sable]

so -sin(y) * y

[spice knot]

hold on there

you can also write y' as dy/dx

,,\dv{}{x}\cos(y)=-\sin(y)\dv{y}{x}

[glossy valve]

mtt07734

[chrome sable]

cant we just write it like this

[spice knot]

why are you erasing the '

[chrome sable]

why do we need it

[spice knot]

when you add 1 + 1 = 2,

do you "need the 2"?

do you need to ask why you need that 2?

[chrome sable]

oh ok

so -sin(y) y' or -sin(y) dy/dx

[spice knot]

yea

and since y and y' arent the same thing you cant say -sin(y) y

[chrome sable]

yea i see it now

thank you

[spice knot]

np

also

what's $\dv{}{x}y^6$

[glossy valve]

mtt07734

[chrome sable]

sorry had to do something

uhh

6y^5

@spice knotsorry for ping

[spice knot]

,,\dv{}{x}y^6=6y^5\dv{y}{x}

[glossy valve]

mtt07734

[spice knot]

dont forget

[chrome sable]

yea

oh wait

wait its just y^6 why do we have to do chain rule there

@spice knot

[spice knot]

??????????

are you multiplying the ys again?

thats not correct

y and y' are not the same

[chrome sable]

yeaaa

6y^5 * y'

ok

-sin(y) * dy/dx

so the left side we now have 3x^2 + 6y^5 * dy/dx

on the right side we have

-sin(y) * dy/dx

we move the left dy/dx term to the right

so now we have 3x^2 = -sin(y)(dy/dx) - 6y^5(dx/dy)

@spice knotis that right so far

[spice knot]

thats not how fractions work

when you move 1/3 to the other side you dont get -3/1

so when you move dy/dx to the other side you just get -dy/dx instead of -dx/dy

[chrome sable]

oh fuck

[spice knot]

3x^2 = -sin(y)(dy/dx) - 6y^5(dy/dx) is correct instead

sounds like a typo then dw

[chrome sable]

wait what if i used y'

how would i represent that

[spice knot]

3x^2 = -sin(y)(y') - 6y^5(y')

y' and dy/dx are literally the same thing

[chrome sable]

ok and from there we do y'(-6y^5 - sin(y))

then divide

to both sides

3x^2 / (-6y^5 - sin(y))

is the answer right?

@spice knot

yessir it is

ok cool thank you

now i just have to remember that which is the hard part lol

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[obsidian fossil]

Why is this the PDF?

[obsidian fossil]

Like I get that all the computers need to have completed their task

but why are we multiplying by the PDF once and then raising the CDF to the power of n-1

[gritty rose]

That's not the CDF, that's the complement probability

set q = exp(-lam * z) and rewrite the density

f = lam * n * ...

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[copper ridge]

i cant do math rn

[digital quarry]

Hi, do you know that the property that come with diameter and 2 other line?

[copper ridge]

wha

[digital quarry]

Scratch that, how much do you know about circle property

[copper ridge]

i started learning this around a week ago but then missed a weeks worth of lecture after so almost nothing

[digital quarry]

So in circle, there are some property that relate to the angle inside

how far back can you remember?

btw, here's some web that might be a help, it contain most of the things you need to know about circle.
https://e-gmat.com/blogs/circle-properties-circle-formulas-circle-area-circle-circumference/#:~:text=1.-,Angles formed by the same arc on the circumference of,semi-circle is always 90.&text=The angle formed by an,formed by the same arc.

one of the property you need to know for this question is this

[copper ridge]

tangents

but even then im still not 100% confident at it

[digital quarry]

I see, perhaps read the web that I sent you and ask anything you don't understand would be for the best.

[full forum]

@copper ridge Has your question been resolved?

[full forum]

@copper ridge Has your question been resolved?

[full forum]

@copper ridge Has your question been resolved?

[trail rose]

hello

[silk nacelle]

hii

[trail rose]

@digital quarry

[digital quarry]

I'm not the owner of question

sorry

@copper ridge

is the owner of the question

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[wise otter]

could i have help with this question please

[tender cloak]

what have you tried

[wise otter]

i dont know how to approach any of them

[steel veldt]

So

@wise otter

[wise otter]

what is this?

[tender cloak]

do you know some properties of differentiation?

about the four fundamental arithmetic operationos

[steel veldt]

?

[wise otter]

oh yes, this is first principle type?

[tender cloak]

do you know those

[wise otter]

like the rules? product, chain quotient?

[tender cloak]

yes

[wise otter]

yes

i do

[tender cloak]

then can you differentiate g(x) in a?

find the derivative with terms of x and f'(x)

[wise otter]

ill try give me a sec

would it not just be 6x-5f'(x)

[tender cloak]

it's right

g'(x) = 6x - 5f'(x)

[wise otter]

so would i then do g'(3) = 6*3-5(5)?

[tender cloak]

yes

[wise otter]

so g'(3) = -7, is this all the question is asking for?

[tender cloak]

yes

[wise otter]

and for the second i'd do the same but use quotient for the deriviatve and chain for the last?

[tender cloak]

yes

[wise otter]

appreciate your help, it makes sense now

[tender cloak]

no problem

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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[pale cargo]

okay hai

[pale cargo]

i need help

so i have an equation

$cos 2a=-\frac{sqrt{3}}{2}$ for the domain 0 equal/less than a equal/lessthan 360

[glossy valve]

sin

[pale cargo]

and so, i got the 2 values of a as

75, 105

but my textbook gives 4 values

how the hail do u get 4 values im so stuck

😭

[glossy valve]

NEON

[pale cargo]

woahhh

How did u do that

[sharp flame]

do you know the general solution

[pale cargo]

yea, i got the 2 values as 75 degrees and 105 degrees

[sharp flame]

LaTeX code, pretty simple, recommend learning if you're doing anything math related on a computer

[pale cargo]

fair

[sharp flame]

Hmm

[pale cargo]

yes

[glossy valve]

NEON

[sharp flame]

This is what you had?

Are you familiar with radians btw

[pale cargo]

yes

[pale cargo]

wtf is that

help

whats n-

and 5 pie/6

[sharp flame]

okay

[pale cargo]

How much is that in degrees again

[sharp flame]

nvm

[pale cargo]

150

[sharp flame]

I thought you just said you were familair with radians

[pale cargo]

Degrees

Yes i did have

150 and 210

Initially

[sharp flame]

uh

[pale cargo]

what?

[sharp flame]

okay lets just stick with this

[sharp flame]

ignore the second image

[pale cargo]

gah damn

i quit india in 10th

🙏🏻

wtf

[glossy valve]

NEON

[sharp flame]

do you agree?

[pale cargo]

How da hail

Did u get that

[sharp flame]

well you said a is in between 0 and 360 degrees

[pale cargo]

yea

[sharp flame]

I simply multiplied each part of the inequality with 2

[pale cargo]

why

[sharp flame]

because your question is cos 2a

[pale cargo]

yea

[sharp flame]

lets let 2a = b

fine with that?

[glossy valve]

NEON

[pale cargo]

how da hail did b come between 0 and 720

[sharp flame]

because b = 2a

[pale cargo]

oh

real

so how do ik when to

MULTIPLY

thr

inequality

HELP

[sharp flame]

basically you want a simple term inside the trig function

[pale cargo]

mhm

[sharp flame]

we had 2a

[pale cargo]

yes

[sharp flame]

we made it b

if we had like

2a + 60 degrees

[pale cargo]

yes

[sharp flame]

we would have also made that b

[pale cargo]

okay

[sharp flame]

because then its easily solvable

[sharp flame]

now how many solutions?

[pale cargo]

uh

well

how do i find out solutions-

if its more than 360

[sharp flame]

are you not familair with the fundamentals of trigonometry

[pale cargo]

i am...?

why

[sharp flame]

it repeats after 360 degrees

basically 361 degrees = 1 degree

367 degrees = 7 degrees

you get that?

[pale cargo]

ohhh

so like

360 then uh

Idk

410?

wait

Nah

uh

360+90

450

then 540

and 630?

😭

[sharp flame]

between 0 and 360 what are the values of b that satisfy?

ignore the 720 for now

[pale cargo]

uh

so

it will be in

2nd and third

So im assuming

540 - 30 and 540 + 30

Giving us

510 and 570?

[rugged void]

SINNNN

[pale cargo]

HIIIIIIIII

[rugged void]

HYD

[pale cargo]

GOOD WBU

[rugged void]

gr8

can you send the question again
and where y'all at

[sharp flame]

no no

between 0 and 360

don't go beyond that

you're right about it being in the second and third quadrant

[pale cargo]

im talking about

values of

2 a

not a

[rugged void]

2a = arccos -sqrt 0.75, right

[pale cargo]

^

[rugged void]

yea yea

[pale cargo]

finding all values of a

[rugged void]

i know i saw

[pale cargo]

correct

[rugged void]

so $a=\pi \pm \frac{5\pi}{12}$, correct?

wait

sorry one sec

[pale cargo]

it has 4 values

[rugged void]

one second, mb

[pale cargo]

allg

[rugged void]

sorry, back

my bad

[glossy valve]

ren

[pale cargo]

yes

i guess

[rugged void]

?

[pale cargo]

i solved it in degrees

[rugged void]

oh

alr

[pale cargo]

not radians

[rugged void]

that works too

[pale cargo]

yes

[rugged void]

yea we can convert back

[pale cargo]

mhm

[rugged void]

so 2a = 150 or 210

right

?/

[pale cargo]

yes

but we need 4 values

[rugged void]

BUT also remember that cos 510 and cos 570 r the same

because rotation by 360 deg

??

@pale cargo

[pale cargo]

mhm

OH

UM

YEAH

wait how are cos 510 and cos 630 the same-

OH WAIT

Do u mean

510 and 570?

Or-

[rugged void]

oh yes

mb

sorry

i'm dumb aren't i?

[pale cargo]

all good

kinda

JOKING LMAO

KIDDING

[rugged void]

ik ik

so anyway

[pale cargo]

ur good everyone makes mistakes

yuh

[rugged void]

2a = 150, 210, 510, or 570

so just divide and convert into radians

[pale cargo]

yes

[pale cargo]

why radians

why would i make my life harder

[rugged void]

because they're easier to work with

[rugged void]

i agree lmao

[pale cargo]

😭

[rugged void]

i like degs too but radians r useful smtimes

[pale cargo]

sometimes

yea

i mean i didnt like fuckin scrub my ass in 10th using degrees as a measure in trig for nothing

[sharp flame]

radians are the SI unit of angles

[pale cargo]

i aint switching to radians for no one

[sharp flame]

degrees are useful in geometry and physics though

[pale cargo]

im not the fart, im the shit

[pale cargo]

why do u think i didnt take physics

[rugged void]

lmao

radians r useful though

calculus heavily depends on radians

[pale cargo]

true

wait how tf did radians come in calculus

[rugged void]

2pi theta, where theta is in rad, is arc length

and stuff

like that

and the sine limit thing

!done

[full forum]

If you are done with this channel, please mark your problem as solved by typing .close

[rugged void]

@pale cargo

^

[pale cargo]

oh yes

.closs

.closw

.close

[full forum]

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[desert moat]

How to solve this

[full forum]

Please don't occupy multiple help channels.

[sleek spoke]

You take the formula V=IR

@desert moat

[desert moat]

Can you solve

I really need your help

I have my university exam in almost half an hr

Like its an online quiz

I dont know a thing

[sleek spoke]

You need to understand the concept first

I is Ampere

V is voltage

R ΩΜ

[versed crystal]

ohms

[sleek spoke]

yeah

W

[versed crystal]

is that thing parallel

[sleek spoke]

Ω

I cant really explain it in english

I have learned it in another laguage

[desert moat]

Its okay i just need the answer

[versed crystal]

??

doesnt it already give you the variables/values for it?

so just input it to V=IR?

[sleek spoke]

yea

[versed crystal]

R4= 4 Ohms ( i cant type omega ) I=5.7A

multiplication?

ok it seems like you dont know the current 💀

[desert moat]

Yesssss

😭😭😭😭😭

The test wouldnt be that basic

[warm kite]

uk kirchoff volt laws?

[versed crystal]

tsk, ive had my problems with parallel circuits

some guys that knows what he's saying ( i don't I only did the basic stuff )

im out

[desert moat]

Yes i know the law but idk how to solve

Can you guys help me in the quiz

Please guysss

Seee

Please in fucked up

[tender sonnet]

Not sure what you need help with

I=V/R

[desert moat]

How to solve this you know

Like how to solve how to calculate

[tender sonnet]

Ohms law

[desert moat]

Okayyy can you solve please how the voltage is 9.79?

How to apply

Are you good at physics?

[tender sonnet]

One sec

Trying to multitask my classes rn lol

[desert moat]

Oh ok but can you help me with the quizz please i dont wanna fail

[tender sonnet]

Also this looks like a series

So you V = V_1 + V_2 … V_n

[desert moat]

Broo listen

[sleek spoke]

send the quiz first lol

[desert moat]

Ill send youu

The question

Will you guys will solve it please please

[sleek spoke]

I will try