help-27

[pseudo basin]

the third of these should be done carefully

[devout snow]

@finite briar Has your question been resolved?

[finite briar]

well it was tryna avoid that

hmmm

its complicated, yes, But expanding those is much more tiring + we havent been taught abt the pascal triangle thing, so i have to do (a+b)^2, THEN (a+b+2ab)^2 (which is (a+b+c)^2) for (a+b)^4

<@&286206848099549185>..?

[pseudo basin]

oh you have not been taught pascal's triangle

[finite briar]

Like ik how it works, but just bc ik doesnt mean i can use it for my assignment

[devout snow]

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[rose nexus]

y’ = 6- 2y. Determine the line element in the point (0, 4)

[rose nexus]

would the line element be (0, 4 -2) ?

.close

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[next nymph]

can u help me guys

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[exotic lance]

would this be the correct equation A=2.1468+23.146 squared

[exotic lance]

2 x 3.14 x 6 x 8+2 x 3.14 x 6^(2)

there

I got 527.52

is that it?

[devout snow]

@exotic lance Has your question been resolved?

[exotic lance]

Help!

[exotic lance]

.close

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[fallen mulch]

how did we get 4=r^2(2-rad2)

and how can we move on

[torn vessel]

the just simplified this.

BC has length 2, so (BC)^2 = 4

cos(45) = rad(2)/2

[fallen mulch]

how come bc has length 2

[fallen mulch]

why are we applying cosine rule if we know bc

[torn vessel]

sorry, not the arc length of BC, the straight line length BC

[fallen mulch]

ok ok that explains it so arc bc is 2

i mean the straight lune

not arc

[torn vessel]

yes. Then they use cosine rule to find the other sides of the triangle in OBC which both the radius r.

[fallen mulch]

ohh but how did it simfply this into 4=

[torn vessel]

what's (BC)^2?

[fallen mulch]

we use the straight line bc ^2

[torn vessel]

yes, because it's law of cosines which only works for triangles.

[fallen mulch]

true but how did it simplify the unknown radiuses

and what exactly did we find for this

[torn vessel]

^^^ and then solving for r

[fallen mulch]

what is rad4+2rad2

[torn vessel]

rationalized the denominator of 4/(2-rad2)

[fallen mulch]

ohhh true

so to find the area of shaded region we find the area of the circle then we subtract it by the middle unshaded part?

[torn vessel]

yep

[fallen mulch]

so to find the the area of unshaded part we use the cosine rule to find the radius

in which in our case is

rad4+2rad2

after finding the radius how exactly is it going to benefit us

[torn vessel]

either you can use it to find the area of the unshaded region. or do what they did and find the area of the sector defined by AOB

[fallen mulch]

we use to find the area of sector BOC and then sector AOD and then find area of both triangles at the bottom and above that are left out

[torn vessel]

in either case you need the radius

[fallen mulch]

then we add all the 2 traingles and the sectors then subtract by the circle?

[fallen mulch]

is there a way to instantly find the area of unshaded region?

or do we need to find sectors then triangles

and then add all

and subtract by area of circle?

[torn vessel]

instantly? no.

you could probably do a calculus thing, but it's all going to require some work and calculations

[fallen mulch]

yea i cant really yet

but even after finding the area of unshaded part

how are we going to find the area of circle

is the radius same

[torn vessel]

yes

[fallen mulch]

why is the area of shaded region the area of sector minus the area of triangle?

[torn vessel]

sector AOB - triangle AOB = shaded area

[fallen mulch]

ohhh trueee

sorry for the inconvinience

so we find the area of sector by finding the radius through (BC) which is the straight line side and we use the same radius in the exact rule of the area if sector AOB and then we find the area of the traingle which is

how did we get the angle alpha to be 135?

[devout snow]

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[midnight sluice]

im confused about logic notation

[midnight sluice]

is

A      A->B
-----------
B
```same as

A->B
A

B

[magic thicket]

never seen a notation that stacks them vertically

[restive river]

it is common

i believe with rules of inference

[midnight sluice]

its on wiki

[magic thicket]

if it's to mean anything then yeah I guess

[midnight sluice]

https://en.wikipedia.org/wiki/Rule_of_inference

[past void]

Yes, that should be the same

[midnight sluice]

'standard form'

[past void]

Because if A implies B and A, then B must be true

[midnight sluice]

Premise#1
Premise#2
...
Premise#n
Conclusion

[magic thicket]

it can be more convenient

[midnight sluice]

but in this notation, does premise#1 have to imply premise#2?

[magic thicket]

he's asking about the notations for proofs in logical inference systems, not whether it holds

[past void]

Oh

[magic thicket]

no

[midnight sluice]

but what happens

when we get to more complex forms like this

is this an inference rule at all?

To help writing cascades of rules forming a derivation, inference rules can also
be aligned on their bottom line. For this, we use the star-version:

actually im only asking because im not sure how to typeset these in latex lol

[magic thicket]

that's why we only ever learned to write them horizontally I guess

looks ugly af here

and extra confusing

[midnight sluice]

hmmm

[devout snow]

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[mild basin]

Is the right matrix considered elementary?

[restive river]

yes simply interchange the rows

[mild basin]

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[iron laurel]

thanks for help

[devout snow]

@iron laurel Has your question been resolved?

[iron laurel]

yo friend i asked no Qs

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[uncut robin]

can someone teach me with solving these?(not all just pick problems from each columns)

[feral agate]

seems pretty self explanatory

which question are you having trouble with exactly?

they've mentioned which question uses which method

well

the first column is called u substitution

[uncut robin]

all of them but i only need someone to teach me one from each columns since i dont want to waste their time

[feral agate]

do you know substitution?

[uncut robin]

wdym?

[feral agate]

https://www.youtube.com/watch?v=sdYdnpYn-1o&pp=ygUNdSBzdWJzdGl0dWlvbg%3D%3D

id suggest looking at this and then proceeding

[uncut robin]

oh okay should i ping you after im done?

[feral agate]

you can open a new channel then

[uncut robin]

oh wait is it the u du?

[feral agate]

yes

[uncut robin]

oh i know that but only the u du part

[feral agate]

the first column is solved precisely like that

the second one also uses substitution but you will also need to know the logarithm integral

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[clever portal]

I need to find the angle B'A'C' but I really have no idea where to start. I looked at the answer a bit, I don't even understood it.

[clever portal]

Could someone please put me on the right path ?

[vagrant steeple]

do you know cyclic quadrilaterals and their properties

[devout snow]

@clever portal Has your question been resolved?

[clever portal]

Yes but I can't figure out how they find the angle.

[devout snow]

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[pseudo quail]

How do I find the shortest distance between a line and a plane

[pseudo quail]

Do I find where they intersect

And then use the formula between a point and a plane

Someone explain the method ples

[restive river]

If a line and a plane intersect, their distance is 0

Like... That's from the definition of intersection itself, no?

[restive river]

The point being the intersection between the line and the plane, it would be in the plane, hence the distance from that point to the plane would be 0

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[mortal ice]

please help

[frosty cradle]

does Dijkstra's algorithm work here

not necessarily

[final scarab]

ahk mb

[dapper widget]

Anything that moves up and right, with no lefts or downs, are the shortest walks, assuming all lines have equal weights. Add the number of shortest walks from below the point and left of the point to get the number of walks. Here's a quick, sloppy, partial solving of it.

I forget the name of the method sadly

[devout snow]

@mortal ice Has your question been resolved?

[dapper widget]

@mortal ice if you need anything explained better, let me know 🙂

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[umbral roost]

Best way to do the blue circled integral?

[umbral roost]

My teacher does it with trig substitution but I’m not fluent with trig identities so is there another way?

[restive river]

Trig sub is likely the easiest and most straightforward thing to do here

[hybrid snow]

You have to do a trig sub

[umbral roost]

No other way comes into mind?

[acoustic leaf]

there really isn't a simpler way here

[umbral roost]

It would be simpler for someone who doesn’t know trig properties

[quaint citrus]

Hold on 1 sec

[restive river]

Hyperbolic integration is an option

But again

Still more complicated

[umbral roost]

Like I don’t know that 1-sin^2t=cost

And I won’t remember it along with everything else I need to remember

I’m just trying to remember the least possible

[acoustic leaf]

just remember that $\sin^2(\theta)+\cos^2(\theta)=1$

[woven radish]

cloud

[restive river]

You avoiding to learn trig will just hurt you more and more as you progress in your studies

[acoustic leaf]

pretty much everything in trig sub comes out of this identity

[umbral roost]

Oh I know that actually

Now it makes sense

Why is sin^2 = (1-cos2t)/2 then

Or is it just algebra

[acoustic leaf]

that comes out of the double angle formula

[umbral roost]

Can I ask wolf for the formula somehow?

[acoustic leaf]

$\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)$

[woven radish]

cloud

[hidden dragon]

$$cos(a+b)=cosa•cosb-sina•sinb$$ on substituting a and b by theta you will have the above formula

[woven radish]

Calamity

[acoustic leaf]

you get that out of the combination of these two identities

[umbral roost]

Jeez..

So cos2t=1?

Oh no

Never mind

Would you say these 2 formulas are the most useful to learn to pass Calc?

[umbral roost]

In terms of all trig identities

[acoustic leaf]

$\sin^2(\theta)+\cos^2(\theta)=1

\cos^2(\theta) = 1 - \sin^2(\theta)

\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)

\cos(2\theta) = 1 - \sin^2(\theta) -\sin^2(\theta)

\cos(2\theta) - 1 = -2\sin^2(\theta)

\frac{1 -\cos(2\theta)}{2}= \sin^2(\theta)$

[umbral roost]

A lot of these are just the same think]g written differently

[woven radish]

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

[acoustic leaf]

a lot of useful trigonometry in calculus comes out of manipulating known identities like this

[acoustic leaf]

the other useful thing for trig sub comes out of dividing this whole thing by $\cos^2(\theta)$:

$\sin^2(\theta)+\cos^2(\theta)=1

$\frac{\sin^2(\theta)}{\cos^2(\theta)} + \frac{\cos^2(\theta)}{\cos^2(\theta)} = \frac{1}{\cos^2(\theta)}

\tan^2(\theta) + 1 = \sec^2(\theta)$

[woven radish]

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

[devout snow]

@umbral roost Has your question been resolved?

[umbral roost]

Hmm I guess learning the properties is not hard, but getting the intuition needs practice. Anyways thank you @acoustic leaf

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[delicate forge]

guys is there a mistake

I calculated Q_D(p) = -20p + 130

.close

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[craggy plaza]

i understand 0.025 is come from the 95 percent confident interval alpha=0.5 so X(alpha/2)

and the degree of freedom is 16-1= 15

but how do i get part b

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[junior thistle]

hey, so i have a seemingly simple question that i can't seem to find an answer to.
I'm trying to figure out what's the probability of getting a C major chord by hitting 3 random notes on a 88 key piano
I thought i had the right idea but the result seems too high

[devout snow]

@junior thistle Has your question been resolved?

[tired thistle]

@junior thistle

could you describe what a C major chord is?

[junior thistle]

oh yeah sure sorry

[tired thistle]

and is it any C E G

[junior thistle]

yeah :)

[tired thistle]

such that they are consecutive

[junior thistle]

exactly

[tired thistle]

okay how many C's are there

on the 88 key one

11?

[junior thistle]

7 i think

yeah

7

[tired thistle]

okay so then there are only 7 possible C majors right?

one in each octave

[junior thistle]

yes

that's right

[tired thistle]

okay so that almost answers the question

how many ways are there of choosing any 3 random keys

?

[junior thistle]

i was trying to figure out the probability for only one octave to make it easier but i guess that works too

[junior thistle]

that many (?)

[tired thistle]

yep

so what's (no. of ways to get a C major)/ (no. of ways to hit 3 keys)

[junior thistle]

7/(no of way to hit 3 key)

is that correct?

[tired thistle]

yep

so then what's the final answer for the probability

[junior thistle]

is it that?

[tired thistle]

where has the 88^3 come from?

[junior thistle]

mmm

i just confused myself

[tired thistle]

also, there are 2 interpretations to your question, do you mean choosing any 3 unique keys, or can you pick a key more than once

[junior thistle]

3 unique keys obv

[tired thistle]

okay cool

[junior thistle]

since when a key is pressed down, you can't play it anymore

[tired thistle]

so then 88^3 doesn't make sense to play a role

so your answer will be

valid ways/ total ways

= 7/88C3

[junior thistle]

just that?

so if i were to ask the same question but i can only choose one octave to play the chord (still on an 88 key keyboard) what would the answer be?

@tired thistle ?

[tired thistle]

so you can only pick keys from that octave

and you want the probability that those 3 form a C major

correct @junior thistle ?

[junior thistle]

yes

but still on the whole 88 keys

if that makes sense

[tired thistle]

okay fine

but you can ignore the rest of the keys

[junior thistle]

yes

[tired thistle]

so an octave has 11 keys?

[junior thistle]

12

[tired thistle]

okay 12

then there is only one C major in each octave right?

[junior thistle]

yes

but let's say i want to know the probability of someone playing a c major on an exact octave

[tired thistle]

mhm

okay so you want them to pick one specific c majpr

but they're playing on all 88 keys

[junior thistle]

correct

[tired thistle]

okay fine

so how many ways are there to play your one C octave?

[junior thistle]

well

[tired thistle]

and how many ways are there of playing any 3 keys?

[junior thistle]

one

[junior thistle]

88C3

[tired thistle]

and so your final answer will be..

[junior thistle]

yeah ok 1/88C3

[tired thistle]

mhm

[junior thistle]

thank you very much :)

[tired thistle]

np

🙂

[junior thistle]

btw

[tired thistle]

yh

[junior thistle]

i've never seen that notation

88C3

[tired thistle]

which one

oh

C is choose

it's european notation

[junior thistle]

i'm french and i don't really know maths english

[tired thistle]

oka

salut

[junior thistle]

haha salut :)

[junior thistle]

so would you call that a 'choose'

a 'choice' ?

in french it's a combination

[tired thistle]

yes

a choice

how can we choose 3 from 88

par exemple

[junior thistle]

that makes a lot of sense

[tired thistle]

chouette

[junior thistle]

haha do you speak french or do you have google translate open somewhere ;)

[tired thistle]

learnt it in school

but yeah should've said bien

chouette is more like it's great

like describing something

[junior thistle]

ooohh yeah i tend to forget that people learn french in other countries lol

[junior thistle]

yep :)

it also kind of sound old-timey in a way

[tired thistle]

oh didn't know that

I dropped french 2 years back

[junior thistle]

like "neat!" i guess

[tired thistle]

ohh

cool

[junior thistle]

it's an awful language tbh

[tired thistle]

agreed

[tired thistle]

d'accord

[junior thistle]

je suis d'accord *

see

it's awful

it doesn't make any sense

[junior thistle]

.close

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[strange knot]

what is the minimum value of

[silk hill]

Are you able to differentiate it

[strange knot]

yep

and equate that to zero

[silk hill]

Can you show your working

[strange knot]

but i am unable to solve that

which i equate to zero

[silk hill]

Yeah, first remove the common terms like 2 and ln5

[strange knot]

yeah

and then i get

how do i find the values of x that satisfy that equation

[silk hill]

express the exponent as a single trig function

[hollow mantle]

the minimum of the function occurs when sin(2x)=cos(2x), because this is the point where both terms in the sum are equal and at their minimum.

So you need to see when it's true that sin(2x) = cos(2x)

[strange knot]

how are you saying that

[rich summit]

You can use AM-GM to make it a bit more formal

a + b >= 2sqrt(ab), with equality if and only if a = b

[mellow panther]

since the functions are periodic there'll be multiple minima

[strange knot]

i just have to find the value

[strange knot]

why cant that be the place it achieves maxima though

[hollow mantle]

let's go step by step then, what's the extrema over the domain?

[strange knot]

i am sorry i dont know what that means

[hollow mantle]

it's restricted to which interval?

what's the period of f(x)?

[strange knot]

i can only say that when i look at the graph

i dont know how to find it for non standard functions

[mellow panther]

hmm...

,w diff 5^sin2x + 5^cos2x

[hollow mantle]

it's 5pi/8

[woven radish]

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[hollow mantle]

But can you answer any of the questions I did?

[mellow panther]

,w solve (cos2x 5^sin2x - sin2x 5^cos2x) = 0

[woven radish]

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[hollow mantle]

you're missing 1 solution there I think

[mellow panther]

since functions are periodic, even 1 solution should be enough

[hollow mantle]

the largest correspond to the global maximum

and the smallest to the global minimum

f(x) will have the global minimum then at: kπ + 5π/8

remember the period of f(x) is π, because sin(2x) and cos(2x) are both periodic with a period of π

so f(x) repeats its values in π units

so we restrict the domain to [0,π)

we're only considering the values of f(x) when x is between 0 and π (including 0 but not including π).

[strange knot]

yeah

[strange knot]

did you say this when you looked at the derivative or the function

derivative right

because tan would be one and 5 power 0 would also be 1

[hollow mantle]

so the problem is that you don't know exactly how to solve f'(x) = 0?

[strange knot]

yeah

[hollow mantle]

to find critical points?

ok give me a moment

[strange knot]

yep

acutally

i knew even if i solved it

i would get more than one

[hollow mantle]

right

[strange knot]

and a few would be maximas

[hollow mantle]

you must then make a table like i did

and see which one is the smallest one

[strange knot]

but how would i know the value to 5 to a fractional power

i dont know

[hollow mantle]

sry what do you mean, like how do you know how much is 5^(real number)?

[strange knot]

we arent allowed to use calculators

[hollow mantle]

well, imagine you know how to solve f'(x) = 0, and you get your four solutions.

pi/8 is your first one

you have sin(2x) x = pi/8

you know how to put that without using calculator right?

(pi/4)/2

sqrt((1 - cos(pi/4))/2

sqrt(2-sqrt(2))/2

i guess not using calculators might make this a bit harder

but should be doable

cause calculating that sqrt is not trivial either

but you know one thing

5pi/8 is negative

so that will be the smallest

so no need for calculator

right?

[strange knot]

yeah

[hollow mantle]

so you can justify using that argument

[strange knot]

that what i forgot to consider

that trig values would be negative

[hollow mantle]

but the other 3 are possitive

so problem solved

[strange knot]

yep

[hollow mantle]

you only need to solve f'(x)

=0

[strange knot]

hey

the table

i get how you got the last two values

the first two?

[hollow mantle]

those u need calculator, after solving the equation but u can do without being exact

the same, u just need to know they are possitive

[strange knot]

yeah

thanks

[hollow mantle]

no prob have a good day

[strange knot]

you too

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[deep vortex]

Can I give you an alternative way?

[deep vortex]

@strange knot

Similar to https://math.stackexchange.com/questions/2396947/a-question-of-trigonometry-on-how-to-find-minimum-value?noredirect=1, we can use AM-GM: \begin{align*} \frac{5^{\sin(2x)} + 5^{\cos(2x)}}{2} &\geq \sqrt{5^{\sin(2x) + \cos(2x)}} \ &\geq \sqrt{5^{-\sqrt{2}}}\end{align*} by finding the minimum value of $\sin(2x) + \cos(2x)$, thus [5^{\sin(2x)} + 5^{\cos(2x)} \geq 2 \cdot \sqrt{5^{-\sqrt{2}}},] and that is the minimum.

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✅

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[crystal kite]

what is the formula for vertical tangent line?

[supple knot]

find where limit x to a f'(x) is +- infinity

https://mathworld.wolfram.com/VerticalTangent.html

[crystal kite]

thanks

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[livid iris]

SAT No calculator question

got the answer right, it is C, but idk how I got it

March SAT specifically

logically thinking, it would have to do something with finding the radius of the circle

because you can easily find the perimeter then

[dark dawn]

have you tried to draw it?

[livid iris]

its not in standard form is it?

[fleet inlet]

no

complete the square of both

[livid iris]

ohhh I see it

thanks

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[livid iris]

No calculator SAT Math

im pretty sure there is a trigonometry relationship I am missing here

I tried it over again and I have FG/4 = GE/3

[supple knot]

,tex .reflect trig

[woven radish]

riemann

[supple knot]

one of the middle column ones

wait G is unknown

draw a picture

[fleet inlet]

still applies

[livid iris]

it would have to be right triangles cause trig right?

[supple knot]

DFE and DFG will be right angles yes

[livid iris]

oof yeah I missed this relationship

alr thanks

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[opaque solar]

Guys how do I finish this polynomial division

[opaque solar]

Am I even allowed to use 0.5 here?

[supple knot]

,w expand (2x+3) * (3x^2 - 7x + 1/2)

[woven radish]

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[supple knot]

,calc 6(-3/2)^3 - 5(-3/2)^2 - 22(-3/2) - 24

[woven radish]

Result:

-22.5

[supple knot]

you just leave -22.5 / (2x+3) as the remainder

https://mathbitsnotebook.com/Algebra2/Polynomials/POPolyDivide.html

[opaque solar]

So exactly like I did it + that as a remainer

[supple knot]

well almost

[supple knot]

,w expand (2x+3) * [(3x^2 - 7x + 1/2) - 22.5 / (2x+3)]

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[supple knot]

you're off by a tiny bit

sign error here

[devout snow]

@opaque solar Has your question been resolved?

[opaque solar]

Oh - 25.5

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[restive river]

what is this omega and omega^2 ?'

[umbral raft]

omega and omega² are two of the three complex roots for the equation z³-1=0

sort of like iota you could say

[restive river]

oh ok

tysm

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[worn kraken]

anyone want to see if i made a mistake solving this? its showing me the anwser is pi^2

[restive river]

why are the y bounds (1,2)?

[worn kraken]

omg........ i wrote it from a different question

noooooooooooo

thanks hahahah

[restive river]

yw

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[proven sparrow]

Can anyone by any chance help me out with part 5

[proven sparrow]

i had tried it solve one side at a time but it seems it wont go any further than that

[restive river]

are you supposed to prove that they're equal?

[proven sparrow]

oh my bad we had to solve for x

[restive river]

in question 5?

[proven sparrow]

yes

[devout snow]

@proven sparrow Has your question been resolved?

[inland seal]

I would first raise 5 to the power of both sides

Or get each side inside one big logarithm first

You can write x - 3 as log base 5 of 5^(x - 3)

And then on the left side, you have three logarithms of base 5, which you can combine together

[proven sparrow]

k i'll try it that way

thanks

hehe done thank you

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[agile narwhal]

Let P be the shortest path in a weighted graph G. If we create a graph G_1 from G by doubling the weight of each edge, then P will still be the shortest path in G_1.

[agile narwhal]

does it hold?

[topaz axle]

of course

[agile narwhal]

i dont think so, it is possible to create counter example where some edges are less than 1

path_1 = 1 & path_2 = 1/2 + 1/2 + 1/2, path_1 is shorter

doubled: path_1 = 2 & path_2 = 1/4+ 1/4 + 1/4, path_2 is shorter

[topaz axle]

you halved the wieghts in this case

[agile narwhal]

oh, im dumb i forgot to multiply numerator 🙂

yea, then it holds i think

thx

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[cold current]

hi

[arctic silo]

HI

[cold current]

dude i figured out while writing my question mb lol

sorrys

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[bronze tapir]

Can anyone help me differentiate b / (2a) ?

[bronze tapir]

I have no idea why I'm struggling with this

[dreamy shore]

they both constant?

[bronze tapir]

technically. but I'm looking for max so treat them like variables

[dreamy shore]

Apply quotient rule

[wooden veldt]

What are you differentiating with respect to?

[bronze tapir]

I tried doing 0.5b * a^-1

which turns into -0.5b *a^-2

But it's incorrect

[bronze tapir]

wym?

oh a

[dreamy shore]

If you're diff w.r.t a, then treat b/2 like a constant and take it out of the differentiation sign

then you're left with d/dx( 1/a )

[bronze tapir]

I'm confused

the full expression is d/dx(4a/b + b/2a -1)

[dreamy shore]

you split the differentiation right

on each term

[bronze tapir]

I think the lack of x's are confusing me

even though it's just a name

[dreamy shore]

Maybe just rename all a's to x's lol

[bronze tapir]

I got -b/(2a^2)

that part I mean

which is incorrect

🫠🫠

[dreamy shore]

d/da (b/2a)

= b/2 d/da (1/a)

since b/2 is constant

now d/dx (1/a) = -1/a²

[bronze tapir]

sorry the 2a are together

b / (2a)

I actually got -b/(2a^2)

[dreamy shore]

that's the same thing

[dreamy shore]

you still get -b/2a² this way

cuz 2 is a constant you can take it out of the differentiation symbol

[bronze tapir]

I give up

lol

[dreamy shore]

lol you shouldn't

[bronze tapir]

Can you explain step by step how to do b / (2x) ? isn't it [f(x) / g(x)]'?

[dreamy shore]

see when you have something constant inside a differentiation sign
like d/dx (18x²)

you can take the 18 out of that sign, as it's a constant
Then you get 18 d/dx (x²)

you get it?

[bronze tapir]

no lol

isn't it just 36x?

[dreamy shore]

it is

but what I'm saying is some constant can be taken out of the d/dx

[bronze tapir]

oh okay

[dreamy shore]

Suppose d/dx (c/x)
c is a constant, hence can be taken out of the diff sign
and you'll get
c × d/dx (1/x)

you understand this much?

[bronze tapir]

yeah

but d/dx (1/2x) is 1 / (-2x) no?

[dreamy shore]

no

it's -1/2x²

d/dx (x^-1)
= -1 × x^-2

[bronze tapir]

couldn't I subtract it by -2?

[dreamy shore]

wait what?

[bronze tapir]

-2 / 4a^2

1 / -2a^2?

[dreamy shore]

the answer is correct but I don't think you understand it correctly though

you know why d/dx (x^-1) is -1/x²

[bronze tapir]

yes

sorry it's just very late where I am and my brain isn't functioning lol

[dreamy shore]

lol no problem

[dreamy shore]

Alright, so d/da (b/2a) is -b/2a² you should understand it too?

[bronze tapir]

anyway

the dx would still give me that

yeah

and it's incorrect apparently

it needs to be +(1 /(2a) )

[dreamy shore]

it needs to be +b/2a² ?

[bronze tapir]

no

1 / (2a)

I have no idea why

[dreamy shore]

where tf is 1/2a coming from?

and where is the b gone?

[bronze tapir]

that's the derivative

I dunno that' s why I'm struggling

[dreamy shore]

man you confused me more than you confused youself

yourself*

[bronze tapir]

that's someone else's answer

[dreamy shore]

what the hail, he's differentiating w.r.t b not a

[bronze tapir]

that's w.r.t?

[dreamy shore]

d/dx stands for differentiating with respect to x

so he's differentiating w.r.t b

[bronze tapir]

oh

lmao

it can be either way right?

[dreamy shore]

no, differentiating with respect to b is a lot different than with respect to a

[bronze tapir]

I'm lost

[dreamy shore]

I found you

[dreamy shore]

Tell me the particular step which you don't understand in this answer

or tell me that you don't understand differentiation at all lmao

[bronze tapir]

never mind I got it

sorry

thanks for the help

[dreamy shore]

I didn't help at all lmao

[bronze tapir]

you did

My brain takes time to process the simplest explanations lol

.close

[dreamy shore]

telling you the variable of differentiation was the help xD

[devout snow]

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[hidden dragon]

Want to know if |y|=ln(x+1) is a parabola of equation y^2=4ax because it's graph looks similar

[hidden dragon]

,w graph |y|=ln(x+1)

[woven radish]

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[hidden dragon]

,w graph y^2=x

[woven radish]

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[devout snow]

@hidden dragon Has your question been resolved?

[fleet inlet]

no

,w dln(x)/dx

[woven radish]

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[hidden dragon]

Ohh tysm

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[jaunty wasp]

how do i draw this vector difference?

[zenith jacinth]

opposite direction

[mighty knoll]

a - b = a + (-b)

[jaunty wasp]

so is it basically a - b + c

[fleet inlet]

yes

[jaunty wasp]

would i start with drawing - b and c vectors first because it was originally in brackets?

[devout snow]

@jaunty wasp Has your question been resolved?

[fleet inlet]

addition is commutative

it does not matter which order you draw them in

[devout snow]

@jaunty wasp Has your question been resolved?

[jaunty wasp]

ok

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[restive river]

Hi , could someone teach me how to do this question? I don’t know where to start!

[devout snow]

@restive river Has your question been resolved?

[restive river]

<@&286206848099549185>

[restive river]

Anyone?

Please?

[indigo moth]

are they asking you the value of x to make both equations the same?

[supple pasture]

In other words, the question is asking for a value of $x$ such that $-20x^2-50x+200=-5x^2-120x+175$

[woven radish]

Math Is Fun

[supple pasture]

That is, $-15x^2+70x+25=0$

[woven radish]

Math Is Fun

[supple pasture]

are you able to do the rest yourself?

[devout snow]

@restive river Has your question been resolved?

[restive river]

do you still need help

[supple pasture]

@restive river

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[restive river]

.reopen

[restive river]

Yea I do

[restive river]

I’m sorry I didn’t respond

@restive river @supple pasture are you guys still available to help?

I’ve done this much

[restive river]

do you know how to use the quadraticequation?

Yes

But can’t I just factor?

sure

Seems about right

Find y vals now?

you already know the y value

its 0

How

you found the zeros of the function

which were x=5 and x=-1/3

But now don’t I plug in the x values to get a y val so I can have a point?

you want to know the y value of (-1/3,y) ?

Yes

Don’t I need to..?

you alredy know that

you solved the equation for the zeros of the function

the zeros are the x values that return zero

Don’t I have to go do (-5(5)^2-120(5)+175

=-550

So y =-550

@restive river

They are asking for the break even price

they have costs

and they have the money they make from selling

to break even you must make more money then it costs

Ok, what do I do after finding my x values

@restive river

We are here

stop pinging me

[restive river]

Sorry

But what do I do now

[supple knot]

Read the question again

Which of your two answers make more sense

[restive river]

5

5 makes more sense

BUT, the answr in the back is 50. So that’s why I’m confused

@supple knot

[untold pivot]

x is the price per $10 for each part

x=5

What does this mean

[restive river]

5$ is the price per 10$ per each part

That wording isn’t clicking in my head

Obviously they must be multiplied

But the language isn’t making sense for me I don’t know why

5 is the price per 10 per each part

[untold pivot]

No I agree wording is strange

[restive river]

So how should I know to multiply them?

[untold pivot]

Well clearly x is the price per $10 for each part, there are only a couple of ways to understand this. Multiplication just makes the most sense here

If it's just 5, they would just use the wording x is the price per part

[untold pivot]

You can try to reconcile this by thinking of it as x is the price per no. of $10 notes for each part

[restive river]

Alright, so I just need to rmember when they say something like that. Multiply them

Is there a way to check if my answer is correct?

[untold pivot]

Not really, no but like I said the question phrasing is poor, it just so happens that multiplication is the best interpretation since it's "per $10"

[restive river]

Okay, @untold pivot can you help me with the next question. I’ve tried multiple times and I keep getting it wrong

[untold pivot]

I can take a look

[restive river]

It’s this let me show you what I’ve been doing

[untold pivot]

Step 2 to step 3

What did you do to the 6

[restive river]

In the bottom portion?

[untold pivot]

Yes

[restive river]

I’ll show you

[untold pivot]

No I mean look at your working in the picture you sent

[restive river]

I added it in like that

[untold pivot]

But there's a - sign outside

[restive river]

I applied the negative after I added the 6 in

So I got this

[untold pivot]

(-(x-2)(x+2))+6
(-(x^2+4x+4))+6

You cannot add the 6 into the brackets

[restive river]

Oh…

[untold pivot]

Because the - does not apply to the 6

[restive river]

Oh

So apply the negative first ADD THE 6 AFTER

[untold pivot]

Sometimes if your stuck, it's good to just plug in values into your calculator to test which line exactly you went wrong, it looks like you erased and redid the question alot of times

[untold pivot]

Yes

[restive river]

My calculator doesn’t do variables

[untold pivot]

You just use your own values

For e.g. -(1+2)(1+2)+6
Compared to -(1^2+4(1)+4+6)

I replaced x with 1

Plug both equations into calculator and realise that they are different

Then at the very least you know exactly which step is the error

[restive river]

Okay

@untold pivot do you happen to know how to do this question?

[untold pivot]

Just write out the formula for area and perimeter for both objects

Let s be the side length of square and r be radius of circle

Then 4s=2(pi)r according to the question

What is the formula for area of circle, what is the formula for area of square

Try solving it from there

[devout snow]

@restive river Has your question been resolved?

[devout snow]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[restive river]

.reopen

[devout snow]

✅

[restive river]

Area of circle: pier^2

Area of square: side^2

@untold pivot

But I’m still very lost

[final drum]

how may i help u

[restive river]

I need help with this question

[final drum]

alr

[glacial valve]

the diameter(or 2r) * π is perimeter of circle and perimeter of square is just 4*side length

[final drum]

okay

so what i do normally

[restive river]

I should also let you know I don’t know what pi is or means

But I think in this case I don’t need to know it

[glacial valve]

π

oh

[final drum]

u have to

[restive river]

Can’t I just treat it like a variable?

[final drum]

for cicle problems

actually

[weak cove]

$\pi =3.14159265358979323........$

[final drum]

yes sorta

[woven radish]

AustinU

[weak cove]

treat it as a constant

as if it was = 3

[final drum]

let me get a piece of paper

[weak cove]

but just a bit different

[final drum]

okay

i got the solution

@restive river

[restive river]

Yes

[devout snow]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[restive river]

Pls teach me

[final drum]

how do u find the area of a square

mb

[restive river]

Uhhh square the value of a side?

[final drum]

yes

[supple pasture]

Is help still needed here?

[final drum]

no

i got it mate

[supple pasture]

ok

[final drum]

:)

okay let x be the side length

[restive river]

@final drum can we restart from the very beginning.

I’m not understanding Anything that is going on

[final drum]

yes

okay

lets let the variable d be the diameter

okay

[restive river]

Ok so we need to find the ratio of the area of the circle to the area of the square

[restive river]

Okay.

[final drum]

yep

now

with that

[restive river]

But I don’t understand where diamter came from

[final drum]

how do we find the area of a circle

[final drum]

the question

just treat it as a variable

[restive river]

Isn’t it perimeter?

[final drum]

mb

perimeter

im high

yes let p the perimtere

how do you find the perimeter of a square

[weak cove]

@restive river this is just a little sidebar about what "pi" is, since you said you didn't know earlier. Please feel free to ignore until after your problem. $\pi$ represents the relationship between the distance across a circle (the diameter), and the distance around a circle (the circumference), $C = D \cdot \pi$ or what you might know as $C = 2 \cdot r \cdot \pi$ since $2r=d$ It also turns out that $\pi$ is very close to 3, it is like I said earlier $\pi = 3.14159265...$

[woven radish]

AustinU

[restive river]

I don’t know

[final drum]

waht does perimeter mean

[restive river]

The length of the outside?

[final drum]

yes

a square

has 4 same length

[restive river]

Yea

So I guess 4 times a side?

[final drum]

there fore perimeter is 4 times the side length

yes

[restive river]

Okay

So p=4x?

[final drum]

what is the area of a square

yes

:D

[restive river]

X^2?

[final drum]

yes

now we sub p back in

[restive river]

🙂

[restive river]

In place of..?

[final drum]

actually

forget about x

[restive river]

Okay

[final drum]

a side length of a square

is p/4

we can agree right

[restive river]

What

[final drum]

p is perimeter

[restive river]

A side of a square is a 4th of the perimeter?

[final drum]

yes

is it not

[restive river]

Yes it is

[final drum]

now

given that

how do we find the area

(p/4)^2

right?

[restive river]

P/4^2

Yea

[final drum]

p^2/16

we got that

now its time to do the circle

[restive river]

Okay

[final drum]

how do we get the perimeter of a circle

[restive river]

Idk

[final drum]

what grade r u

[restive river]

11

[final drum]

like hs

[restive river]

11

[final drum]

highschool?

[restive river]

Yea

[final drum]

so area of circle is pi*(radius^2)

[restive river]

And what is radios?

[final drum]

and the perimeter is 2pi*radius

[final drum]

you know

a circle

given a center point

[restive river]

Uh

[restive river]

Ok

Oh

Ok

[final drum]

the black line is the radius

[restive river]

Got it

Got it got it

[final drum]

we need the radius to solve for area

[restive river]

O

[final drum]

and to find the radius

2pi*radius = perimeter

so how do we isolate radius