help-33

[terse gazelle]

in picture 1

samehow they just have one

[brave marsh]

That's what I mean. If the functions are nice enough, the Maclaurin series of the derivative is just the derivative of the Maclaurin series.

So they start with the series of f(x) and differentiate every term in the series.

[terse gazelle]

really?

like differentiate x/sqrt(x^2+1) to the 5th power?

for 4 points?

isnt that a bit of a stretch

[brave marsh]

That's not what I'm saying

In a) you have already computed the series for f(x).

Now, you differentiate every term of that series, instead of f directly.

[terse gazelle]

ahhhh

i see

do you have to find the first term indivisdually?

since like

when u derive the series

you cant get 1

which is the first term of the derived series

[brave marsh]

What is the derivative of $\frac{1}{2}x^2$?

[elfin berry]

Azyrashacorki

[terse gazelle]

x?

[brave marsh]

Okok I see what you mean one sec

There shouldn't be a 1 there then

[terse gazelle]

ok probably a mistake

in the answers

thanks so much for your help

[brave marsh]

Seems so

[terse gazelle]

/close

[brave marsh]

.close

[marsh citrus]

@terse gazelle Has your question been resolved?

[marsh citrus]

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[trim mountain]

if f(x)=cos^2(2x) & h(x)=cos^2(4x), then what is (f ° h)' (pi/4)?

[trim mountain]

so far i got the derivate cos^2(4x)(-sen(2x)*2)+cos^2(2x)(-sen(4x)*4)

but i don't know what I'm supposed to do with "pi/4", so far i try evaluating x as pi/4, but i got -2, than that is no one of my options

[brave marsh]

Those should probably be multiplied, not added.
cos^2(4x)(-sen(2x)2)+cos^2(2x)(-sen(4x)4)

[paper crag]

are you sure that the derivative is right?

[brave marsh]

And the function h should be plugged in the first part as well, so I guess you should just compute h(pi/4) and h'(pi/4) ahead of time.

Otherwise it becomes gnarly when you could just have those values computed before. It will make it more readable and avoid mistakes.

[paper crag]

i mean

[trim mountain]

i redo the derivate it and i was wrong, the derivate should be -5sen(8x)cos^2(2x)+sen(8x)sen^2(2x)

[paper crag]

good

[trim mountain]

this time i got 0 when evaluating x as pi/4, so i think it should be right now

[paper crag]

nice

[trim mountain]

But now i got a different problem

what this is supposed to be

[brave marsh]

I hasn't been compiled, it's LaTeX.

[paper crag]

dude

[brave marsh]

And it's missing most of it.

[trim mountain]

i guess I just mark the question and talk the teacher about it later

[paper crag]

yeah

[trim mountain]

ok in this one I'm lost, not an expert of graphics

wait, no i think i can actually do it

thank you

[paper crag]

k

[trim mountain]

.close

[marsh citrus]

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[crystal valve]

Someone tell me im not losing my mind

[crystal valve]

im not losing my mind right

thats correct

so how df does my teach go from that to this

-5/4 a to - 1/4

??/

[void elm]

I'm a bit confused

is this all your teacher's work?

[crystal valve]

white background

its my teacher

[void elm]

gotcha

[crystal valve]

dark background is me

ME

my teacher

[jagged oracle]

Fr AP gp moment

[void elm]

which part is the confusing part?

[crystal valve]

i got the answer R = 2

he has gotten R = 6

what i did

what he has done

its 5am

im gonna head to bed in 2 seconds but

wtf

oh wait nvm

Yeah i was being dumb af

since a can not be 0 slap 1 on it

😭

.close

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[still temple]

Im getting no solution??

[late geode]

can you show your work

[marsh citrus]

@still temple Has your question been resolved?

[still temple]

Its messy

[late geode]

i don't understand what you're doing

after the first step

[hardy ether]

we cant see the full image

[still temple]

Took log with base y both sides

[late geode]

that was the first step,
what happened after that

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[rough quarry]

How do I solve problems with logarithmic power?

[late geode]

for 75, try applying log_2 to both sides

[rough quarry]

oh yeah, i see

what about 77?

is there some special shtick or same concept?

[glossy flint]

I would apply log_3 to both sides

[rough quarry]

if i do this like 75 then the answer is 81, which is wrong since, its supposed to be 9

[glossy flint]

Can you show your work? Because I get the correct answer

[split fractal]

log_3 isnt the same as ln

[rough quarry]

how come?

isnt ln just log without a base?

[split fractal]

log_e is ln

nope

ln has a base e

[rough quarry]

i see

then what was i supposed to do

[split fractal]

log_3 of 81

cause u need to do both to same side

[rough quarry]

and how does this change things

[split fractal]

oh u also did the left side wrong aswell

[rough quarry]

hm?

[split fractal]

here u forgot the x after the second log_3

[rough quarry]

wouldnt it just be log_3(x)*log_3(x)?

[split fractal]

ye

[rough quarry]

oh

oh

i see

then its like

x*x=81

which is like

x^2=81

then i root

oooohh

[split fractal]

yup

[rough quarry]

oh now it makes sense

thanks bestie ^_^

.solved

[marsh citrus]

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[verbal trail]

basically I have this function in desmos where if I input a degree it will convert it to radian and it will take that radian to cos and sin and plot a dot now what I did next was take that sin and cos coordinate and put it and arctan "expecting" it to result the same as d there but however for unknown reasons because of my limits it is not equal

here is the link for the desmos
https://www.desmos.com/calculator/zzqttfojus

[verbal trail]

im confused

using arrcos only gives semi-circle

[marsh citrus]

@verbal trail Has your question been resolved?

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@verbal trail Has your question been resolved?

[marsh citrus]

@verbal trail Has your question been resolved?

[little cove]

I don't know if it helps, but did you know you can change the angle from radians from/to degrees?

[marsh citrus]

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[hazy dragon]

@spark otter lets continue

[hazy dragon]

so basically you would have done what they were doing

they did round_down(n/2) swappings

to get a traingular matrix

right?

[spark otter]

yeah

[marsh citrus]

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[marsh citrus]

@red prism Has your question been resolved?

[marsh citrus]

@red prism Has your question been resolved?

[marsh citrus]

@red prism Has your question been resolved?

[marsh citrus]

@red prism Has your question been resolved?

[velvet cedar]

Zero for b?

I imagined taking a line integral over the field and it seems like the work cancels out but I'm not sure tbh

Could you send the sketch as well

Right, mostly how I pictured it

[velvet cedar]

Oh ok, it shouldn't be zero I think

The electric field is constant and uniform (I assume) so the p.d. across the wire is $2 l E_0$

[elfin berry]

StrangeQuarkAL

[velvet cedar]

And then workdone is $2 q l E_0$

[elfin berry]

StrangeQuarkAL

[velvet cedar]

I still don't understand why the line integral didn't work tho

Wait, it did work

I just calculated it wrong

It's 1AM here so likely won't be able to read it

[velvet cedar]

$$ \int_{-l}^{l} \begin{bmatrix} q E_0 \ 0 \end{bmatrix} \cdot \begin{bmatrix} x \ \sinh(x/a) \end{bmatrix} dx $$

[elfin berry]

StrangeQuarkAL

[velvet cedar]

Tbh I'm not sure. Super groggy rn

[marsh citrus]

@red prism Has your question been resolved?

[round meadow]

What's the meaning of 'tbh'?

[eager tinsel]

I think it is To BE Honest

[jagged oracle]

Yes

[round meadow]

lol
lmao

my brother just poured a whole pack of ice cream on his head

[velvet cedar]

Crazy

OK I got an idea

Rewrite either pv1 or pv2 in terms of pv2 or pv1 respectively

[marsh citrus]

@red prism Has your question been resolved?

[narrow pecan]

Why is this cubed?

Do you know if your setup is right?

[marsh citrus]

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[lucid hound]

How do I complete part d, i know half of the terminal velocity is mg/2k but cant acheive the result in the Q

[dusky viper]

oh i have done that question before

maths extension 2?

[lucid hound]

yes

[dusky viper]

have u subbed in mg/2k for v?

[lucid hound]

yes

[dusky viper]

then what did u do after?

[lucid hound]

i simplified the equation out, I used the part(c) eq

[dusky viper]

what did u simplify it to?

[lucid hound]

(2m/k)ln(mg/mg)=0

[dusky viper]

?

where did u get = 0?

also ur natural log is wrong

[lucid hound]

ln(mg/mg)=ln1=0 ?

[dusky viper]

it doesn't equal 1

u made an error somewhere

[lucid hound]

whoops its equal to 2m/k[ln(mg/(mg/2))] = 2m/kln2 ??

[dusky viper]

yes

[lucid hound]

now it works out nicely thx for helping me spot my error

[dusky viper]

all good

[lucid hound]

.close

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[toxic pelican]

in defining a lcm of two integers for instance, why is it if a divides m, and b divides m, then e divides m given that e is an existing multiple, then e is the least of the common multiples

[toxic pelican]

an example of why I ask this is: let a=100, b=50 and m=400

a and b divide m

but if e =200

then it also divides m

but is not the least common multiple

[crystal lintel]

can you state the definition you’re using precisely

well i’m going to bed

[toxic pelican]

definition of what?

i'm certain I made it clear enough

[south remnant]

that last part doesn't make sense to me, wouldn't m be that least common multiple ?

[toxic pelican]

i saw this in a book

oh wait

sorry

[toxic pelican]

is there anything else useful you have to say?

[south remnant]

well now that is just the definition, but it depends how your book defines it to begin with

[toxic pelican]

unless

given that there are other existing multiples

you continue dividing

until you get the least

yes?

to be more specific

[south remnant]

i don't understand

[toxic pelican]

ok

because this particular value of e does not satisfy the criterion; "least common multiple"

it has a divisor that is also a multiple of a,b

[south remnant]

well it does, if it is a common multiple and divides every common multiple of a and b, then it is the lcm

[toxic pelican]

when I said "particular value"

[south remnant]

this is just the definition, but I need to know the approach your book takes

[toxic pelican]

I was referring to 200

[south remnant]

oh

the criterion says that it needs to be true for every common multiple m of a and b

e=200 doesn't work because for instance m=100 contradicts the criterion

e is the LCM if, for any m such that a and b divide m, e divides m.

[toxic pelican]

.close

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[obsidian bobcat]

I’m unsure why we stop where we stop. Why don’t we fully reduce to rref? How can we just say b3=b1+b2 is the only condition?

[marsh citrus]

@obsidian bobcat Has your question been resolved?

[obsidian bobcat]

<@&286206848099549185>

[wanton willow]

wait im sorry is the final form here
1 1 2
0 1 1
0 0 0
?

[obsidian bobcat]

That’s what the professor stopped at yes

I can see that after that we already have a zero row

And therefore a free variable

Right?

[blazing pendant]

Gaussian Elimination

[obsidian bobcat]

But the first row could be 1 0 2 for example

By reducing the middle 1

I’m just confused why we stopped where it stops

[wanton willow]

because you can solve a system of equations once youve reduced it to row echolon form
going all the way to reduced row echolon form is just a personal preference

[obsidian bobcat]

Hmm to me, it looks like each row has its own conditions because none of them is just flat out x1=this

That sort of was confusing me

[wanton willow]

to be honest i dont know what the b's here are referring to exactly
are you doing gaussian elimination?

[obsidian bobcat]

Yeah, this is just basically algebra but in the context of linear algebra so like matrix algebra

[wanton willow]

i know that

[obsidian bobcat]

The bs are constants

Like Ax=b

[wanton willow]

ohh

[obsidian bobcat]

So A is our coefficient matrix

X are the unknowns

Basically, the final assertion at the bottom right that’s written out is what’s confusing me

[wanton willow]

yeah it wont always just be x1 = c or x2 = b
when youre just reducing it to row echolon form, youll usually end up with a system of equations like
x3 = c
x3 + x2 = b
x3 + x2 + x1 = a
where a b and c are constants

so you plug the value for x3 into the second equation, then the values for x3 and x2 into the third equation, so youve solved for all of the variables

[obsidian bobcat]

I get that but here what will you plug into what concretely?

[wanton willow]

the b3 = b1 + b2?
basically that zero row means that b3 - b1 - b2 is equal to zero
so that means that b3 = b1 + b2

[wanton willow]

sorry can you clarify what you mean exactly?

[obsidian bobcat]

Yeah, get the logic for that actual row, but I’m just confused how we can immediately know that that is the only condition we need to satisfy in order for it to be consistent

Like we still have row 1 being
X1+x2+2x3= b1

And row 2
X2+x3=b1-b2

[wanton willow]

well because if its true then that means that the bottom row imposes no problems, and the middle and top row can just be solved to get a solution
but if its false, then you get a contradiction that 0 doesnt equal 0. so that means it isnt consistent in the first place

[obsidian bobcat]

For me its more the if it true that is a problem. If it’s true, what about the other rows conditions?

[wanton willow]

they dont pose a problem since solutions can be found for them im pretty sure

[obsidian bobcat]

Hmmm ok

[marsh citrus]

@obsidian bobcat Has your question been resolved?

[marsh citrus]

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[brazen lantern]

Help with the geomtery

[brazen lantern]

My teacher told me that I got wrong for the ambiguous case check

I got the angles right, but really don't know what to check

It's law of sine problem

[blazing pendant]

siny/8=sin30°/6

siny=8/12=2/3

y has 2 possible values

[late geode]

your values keep changing

[blazing pendant]

You need to check the 2 values

[late geode]

improper notation at the start
your angle for y in the ambiguous case is also wrong

[brazen lantern]

Yeah that's what I wrote 2 triangles exist

[blazing pendant]

sometimes only 1 exists

[brazen lantern]

Are the equations you mentioned on top the ambiguous case checking equation?

[blazing pendant]

If another , more than 180°

if 180°>y >90°

maybe

[spark fulcrum]

does the set {{null}, null} have two elements or one?

[blazing pendant]

you need to check this

[ruby mulch]

That's out of topic

[late geode]

there are so many errors there

[blazing pendant]

using sin(a+b)=sinacosb+cosasinb

[spark fulcrum]

im in the wrong chat nvm

[brazen lantern]

This is the full values I got

And figured out that the 2 triangles exist

[late geode]

you shouldn't be setting
<C = sin(C)/8 = ...
like that

[brazen lantern]

wdym law of sine

[blazing pendant]

I agree

[brazen lantern]

That's the way to figure angle C out

ain't it?

[blazing pendant]

why C is a value?

[brazen lantern]

angle C

[blazing pendant]

it should be

sin C

not C

[late geode]

you start with
sin(C)/8 = sin(30)/6
and that leads to C or <C = whatever

[brazen lantern]

yeah basically

[blazing pendant]

not angle C= balabala

[brazen lantern]

Ohhhh mb

So Sin C = 42

Sin A = 108

[blazing pendant]

but sinC=balabala

No

[late geode]

you're also missing () or using the inappropriate sign when calculating <A

[blazing pendant]

sinC<1

[late geode]

180 - (30 +42)
or
180 - 30 - 42

[blazing pendant]

how sin A =108

sin A is smaller than 1

[brazen lantern]

Sin C/8 = Sin 30/6

[late geode]

they didn't say sin(A)= 108

[brazen lantern]

Gives me Sin C=42

[blazing pendant]

nonono

[brazen lantern]

And by using the theorm of triangle, 180 - (42+30)=108

[blazing pendant]

You got a mess

you mix up them

[late geode]

Sin C/8 = Sin 30/6
leads to
sin(C) = 2/3
and that gives
C ~ 42° as one of the solutions

[blazing pendant]

C, is an angle,0° to 180°

sin C is the sine value

[brazen lantern]

Oh wait, since it's ambiguous case, u mean I can't just confirm that the sin c is jsut 42 right?

[late geode]

no

you're again msseeing up the angle with the sine of the angle

sin(C) is not the same as C

sin(C) = 2/3

[blazing pendant]

Angle is angle

sine is sine

[brazen lantern]

Can ya'll write a equations on your paper or smth and upload it

[blazing pendant]

angle 0 degrees tp 180 degrees

[brazen lantern]

It's lwk hard to understand without visual components or

[blazing pendant]

sine 0 to 1

It is not about the solution to the question

[late geode]

$\frac{\sin(\angle C)}{8} = \frac{\sin(30\deg)}{6} \implies \sin(\angle C) = \underbrace{\frac23}_{\text{NOT } 42\deg}$

[elfin berry]

ℝαμΩℕωⅤ

[blazing pendant]

It is about you dont distinguish the concepts

between angle and angle's sine

[late geode]

one of the solutions for <C is approximately 42°
i again must repeat that this is not the same as sin(<C)

[brazen lantern]

So after I plug this in to my calc I got 0.66666666... which is equvialent to 2/3

[blazing pendant]

Also the a can be two values

[late geode]

yes, and that is the value of
sin(<C)

[jagged oracle]

Yo

[blazing pendant]

there are two possible triangles

[brazen lantern]

So what you are saying is <C and sine c is totally different value

[late geode]

yes

one is the angle

[brazen lantern]

The sine(<C) is 2/3

which is another angle right?

[late geode]

and the other is sine being applied to the angle

no

sin(<C) is not an angle

[blazing pendant]

6>8sin30°

[late geode]

sin(<C) is sine being applied to the angle <C

[blazing pendant]

6＜8

so

two possible answers

[late geode]

there are two solutions to
sin(<C) = 2/3
for 0 < C < 180°
taking the sine inverse gives you one of the solutions which you found to be approximately 42°

from the supplementary properties of sine, the other solution would be 180° - 42°

[brazen lantern]

So when i'm checking the ambiguous case check

I shoudl've do 180-42 not 180-72 right?

[late geode]

yes

[blazing pendant]

ASS cannot define a triangle

angle side side

[brazen lantern]

Ok, then let me start it from the beginning

This is the first step

that sin(<C) is equal to 2/3 or 0.66666666...

[blazing pendant]

the second equelity

why you always use incorrect "="

You should turn to another line of your notebook

and changes the formula

[brazen lantern]

Why don't u write it down on ur own notebook and jsut show me?

That will be way easier for me to understand

[blazing pendant]

Nope

that's wrong written habbit

Never write like that

for example

[brazen lantern]

That's why i'm telling YOU to write the equation

So I can differentiate

[blazing pendant]

2x=4
x=2

dont write:2x=4=x=2

[brazen lantern]

Bruh this is just the conecpt

of what it looks like breifly

[blazing pendant]

if you write like that

[brazen lantern]

@late geode Help

[blazing pendant]

You admit 2x=x

[brazen lantern]

.

read it

[blazing pendant]

you should start at another line

[brazen lantern]

That is not what I'm saying tho 💀💀

[late geode]

by using an equal sign like that, you are

$$1+1\foreach~in{2,...,9}{\overset\thonk=~+1}\overset\thonk=10$$
$$2\foreach~in{3,...,10}{\overset\thonk=~}$$

[elfin berry]

ℝαμΩℕωⅤ

[blazing pendant]

Dont write many things at one line and connect them with"="

It is a terrible thing

[brazen lantern]

U can tell ik how to use equal signs from the first attachment

[blazing pendant]

because it is not correct

[brazen lantern]

.

[late geode]

that's why we're telling you to correct the issue

because they're not being used appropriately

[brazen lantern]

Ok i fixed it

[blazing pendant]

Yes

[late geode]

if you really want to write a logical continuation from one equation to another, then use arrows \
$\to, \implies$

[brazen lantern]

Now should i go for Sin (<A) or a

[elfin berry]

ℝαμΩℕωⅤ

[late geode]

now you should to the candidates for <C

[blazing pendant]

if you insist like before, you admit sinC/8=sinC

[brazen lantern]

By 180-42?

[late geode]

arcsin(2/3),
and 180° - arcsin(2/3) for the potential ambiguous case

[brazen lantern]

So 180-42 gives me 108

sry

138

[late geode]

yes

[late geode]

and then try determining <A
when <C = 138
and see if you get a valid value

[blazing pendant]

NEVER use "=" the way you used before

it is terribly wrong

[brazen lantern]

So when <C = 138, I get <A for 12

[late geode]

yes

[brazen lantern]

<A value is for the other potential angle of <C right? the 138 degree

[blazing pendant]

6 is between 8 and 8sin30°

so there are 2 triangles

if it is shorter than 8sin30°

No answer

[late geode]

each <C has a respective <A

when <C = 138, <A = 12
when <C = 42, <A will be different

[brazen lantern]

When <C is equal to 42, <A should be 108

Ohhh i see the problem

So just to clarify

[blazing pendant]

if AC is longer than 8, one answer

[brazen lantern]

Whenever I'm checking for the ambiguous case, I should seek for the angle C right

[blazing pendant]

You can draw a vertical line from A to BC

[late geode]

use whichever angle you're trying to solve for

[blazing pendant]

Name it AD

[brazen lantern]

Wait then I also have to calculate for a right since we got 2 different triangle

[blazing pendant]

and compare AC and AD

[brazen lantern]

So it should be 2 different value

[blazing pendant]

Value of sine cannot define an angle 0to180°

you need to check the total sum of 3 angles

you need to check 2 possible triangles

[brazen lantern]

.close

[marsh citrus]

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[marsh citrus]

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[still temple]

help with derivatives

[hearty thicket]

ok hi

[still temple]

there

hiiii

okok

lemme do the problem

wait does it matter

which order

in the quotient rule

[hearty thicket]

so $\frac{x^2+1}{x-1}$

[elfin berry]

Dinciii

[hearty thicket]

right?

[still temple]

yes

[hearty thicket]

you wanna go through it or wanna try alone first?

[still temple]

can I try it first

[hearty thicket]

okok

[still temple]

I think I can do it

[hearty thicket]

tell me when you're done

[still temple]

-1?

[hearty thicket]

?

[still temple]

uh oh

I got -1

after simplifying

ummmmm

I took the derivatives

and used the quotient rule

[hearty thicket]

so lemme remind you, quotient rule: $(\frac{a}{b})' = \frac{a'b - ab'}{b^2}$

[elfin berry]

Dinciii

[still temple]

right

[hearty thicket]

in your case a is what

[still temple]

x^2 + 1

[hearty thicket]

so if you differentiate x^2+1 u get what

[still temple]

2x?

[hearty thicket]

yep

what abt x-1

[still temple]

1

[hearty thicket]

so if u sub those back in

[still temple]

(x - 1)(2x) - (x^2 +1)(1)/(x - 1)^2?

[hearty thicket]

yeah thatss it

now just simplify

[still temple]

ahhhhh okay

1???

[hearty thicket]

does it simplify to 1?

[still temple]

yes for me

wait

yeah i got 1

[hearty thicket]

alright then that's it

[still temple]

YAYY

can we do another one

[hearty thicket]

sure

can I give u an example

[still temple]

oh I was about to you give one from my previous test

can we do mine if you don’t mind

[hearty thicket]

okok

[still temple]

f(x) = 3x times square root of x

[hearty thicket]

$3xsqrt(x)$

[elfin berry]

Dinciii

[still temple]

yes

just the x under the radical

[hearty thicket]

do yk the definition of a derivative of sqrt(x)?

[still temple]

power rule?

[hearty thicket]

yeah

[still temple]

ahh yeah

so square root of x

would also be written as x^ 1/2?

is that the only law I use????

[hearty thicket]

yes

[still temple]

oh

I got 3/2 x^ -1/2 is that right

[hearty thicket]

3/2?

[still temple]

yup

[hearty thicket]

try again

[still temple]

hmmmm

is the derivative of square root of x

x^ -1/2?

[hearty thicket]

power rule is $(x^n)' = nx^{n-1}$

[still temple]

right

[hearty thicket]

yeah so

x^1/2

you bring the 1/2 down

and -1 in the power

[elfin berry]

Dinciii

[hearty thicket]

so we get

[still temple]

1/2x ^ 1/2 - 1?

[hearty thicket]

$\frac{1}{2}x^{-\frac{1}{2}}$

[elfin berry]

Dinciii

[still temple]

yes

[hearty thicket]

and we can rewrite

$x^{-\frac{1}{2}} = \frac{1}{x^{\frac{1}{2}}}$

[still temple]

ohhh right

[elfin berry]

Dinciii

[hearty thicket]

yeah

so now we have just

$\frac{1}{2\sqrt(x)}$

[elfin berry]

Dinciii

[still temple]

OHH

wait is that the answer

[hearty thicket]

thats the derivative of $\sqrt{x}$

[elfin berry]

Dinciii

[still temple]

oh nvm

Right

wait so

ohhhh now 3x

[hearty thicket]

but we had $3x\sqrt{x}$

[elfin berry]

Dinciii

[hearty thicket]

right

[still temple]

yes

[hearty thicket]

remember the product rule: $(ab)' = a'b + ab'$

[elfin berry]

Dinciii

[still temple]

yup

I have to use that too?

[hearty thicket]

ofc

[still temple]

I thought you just find the derivative of 3 and multiply it together?

[hearty thicket]

so now we have $(3x\sqrt{x})' = (3x)'\sqrt{x} + 3x(\sqrt{x})'$

[elfin berry]

Dinciii

[hearty thicket]

right?

[still temple]

right

[hearty thicket]

now that is $3\sqrt{x} + 3x\frac{1}{2\sqrt{x}}$

[elfin berry]

Dinciii

[still temple]

right sorry

[hearty thicket]

but

[still temple]

yea

[hearty thicket]

nvm you can do this way easier im a dummy

[still temple]

do I have to rationalize it

nahhh you’re good

[hearty thicket]

so $3x\sqrt{x} = 3x^{\frac{3}{2}}$

[elfin berry]

Dinciii

[still temple]

wait what

[hearty thicket]

and we just use power rule 😄

im a dummy

[still temple]

hold up

[hearty thicket]

well

$a^b * a^c = a^{b+c}$

[elfin berry]

Dinciii

[hearty thicket]

right

[still temple]

yes?

right

[hearty thicket]

now we just use

power rule

so $(3x^{\frac{3}{2}})'$

[elfin berry]

Dinciii

[still temple]

right

[hearty thicket]

just differentiate that

so you bring the 3/2 down

[still temple]

3/2x ^ 1/2 times 3?

wait no 9/2 x^ 1/2

[hearty thicket]

$3\frac{3}{2}x^{\frac{1}{2}}$

[elfin berry]

Dinciii

[hearty thicket]

yes

[still temple]

yes

[hearty thicket]

thats it

[still temple]

can you simplify the 3 and 3/2?

[hearty thicket]

yeah thats your 9/2

[still temple]

ahhh okay

yay

okay wait

so how we did it

first we used the product rule?

for the 3x times the square root of x?

[hearty thicket]

well u can also do it like that

but easier isjust

[still temple]

and then we used the power rule for that?

[hearty thicket]

to use the properties of powers

[still temple]

oh

[hearty thicket]

we just did 2 ways of doing it

[still temple]

the power rule?

[hearty thicket]

yes so

[hearty thicket]

1st way

[hearty thicket]

2nd way

[still temple]

OHHHH

gotcha

thank youuuuuuu

I have about 8 mins I’m gonna review some stuff

.close

[marsh citrus]

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[limpid bison]

I got to this step

I mean here

and the method I thought I could use

would be to square y

and reaarange it in terms of sint

idk why

that doesn't give me the right answer

it is a valid method isn't it?

fak nvm

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[limpid bison]

.reopen

[marsh citrus]

✅

[limpid bison]

yeah no

Im totes lost

[velvet cedar]

do you mean x?

[limpid bison]

both terms

[velvet cedar]

could you elaborate on what you did?

[velvet cedar]

damn, that's gotta hurt

[limpid bison]

wdym haha

[velvet cedar]

$$ y = a + b $$
$$y^2 =/= a^2 + b^2 $$

[elfin berry]

StrangeQuarkAL

[limpid bison]

oh

[velvet cedar]

not equal

[limpid bison]

shit right

wat

😭

But i see so many times

'square root both sides'

'square both sides'

oh

it would be

(a+b)^2 innit

ah ok thank you

.close

[marsh citrus]

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[velvet cedar]

no problem

[marsh citrus]

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[toxic pelican]

how do I type a matrix in the online office-365 (website) version of Microsoft word?

[proud ice]

\matrix()

[toxic pelican]

website

[proud ice]

Then & to create new columns and @ to create new rows

[toxic pelican]

oh?

so for instance

would I type

\matrix(&1,2,3@4,5,6)?

[proud ice]

\matrix(1&2&3@4&5&6)

[toxic pelican]

just to be sure

you do know what I mean by website version right

[proud ice]

I imagine you mean the version on the website

I'm giving you offline instructions, but if it's different for website, then I cannot help you

[toxic pelican]

it's different

[toxic pelican]

thank you anyway

[marsh citrus]

@toxic pelican Has your question been resolved?

[toxic pelican]

.close

[marsh citrus]

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[eager dagger]

How to do this??
We only got the area formula for a regular polygon A=1/2ap

[eager dagger]

I did this

[tight furnace]

Do you know what the other sides are

[eager dagger]

No

[tight furnace]

Have you learned the side lengths of a 30-60-90 triangle

[eager dagger]

Yes

When I tried to find the long side it was wrong

Or idk

It’s Times Square root 3 right?

[tight furnace]

For the longer leg yes

For the hypotenuse it's double

[eager dagger]

Yeah

Okay I got 16 square root 3 for the area

Thank you

For your help

[tight furnace]

Np

[marsh citrus]

@eager dagger Has your question been resolved?

[marsh citrus]

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[still temple]

is just a way of saying the the limit goes to infinity in both directions

yeah, but it says limmit to a, so may be something missing

[old sparrow]

basically he meant to say that limit in a+ it's +infinity
and limit in a- is -infinity

no

got me confused now ;-;

[brave marsh]

It looks like it`s just an explanation for the general case.

Like if the derivative tends to either infinity or -infinity at some point a, then the tangent is vertical.

And x^(1/3) is drawn as an example

Yes essentially. Either f'(x) -> infinity or f'(x) -> - infinity.

[old sparrow]

what an embarrassment ._.

about my answer xd

how dare i say it's even negative... x)

[marsh citrus]

@untold brook Has your question been resolved?

[brave marsh]

Like understand why an infinite derivative hints at a vertical tangent?

Well if you think about it in terms of the slope of a tangent line, then you might recall that high slope are very steep.

So if you imagine this happening to infinity, it would look like a vertical line

[marsh citrus]

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✅

[brave marsh]

Yes essentially.

[marsh citrus]

@untold brook Has your question been resolved?

[marsh citrus]

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[wheat gull]

hello

[wheat gull]

where did i go wrong here?

i followed previous examples in my notes exactly but i got the wrong answer

i can’t spot my mistake

[eager tinsel]

you just missed the negative sign of first term in the last step

[wheat gull]

oh i see that

but the answer sheet says its completely different

[eager tinsel]

like what?

[eager tinsel]

If you want exact answer then you may try to convert cosine to sine in your answer

[wheat gull]

how do i do that?

using trig identity?

[eager tinsel]

There are dozens of forms by trig manupulations, so I think any answer should be considred correct

[wheat gull]

oh yeah i didn’t think of that

[eager tinsel]

yeah
or you may try by substituting sinx as u
instead of cosx

[wheat gull]

ohhh that makes sense

thank you

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[fast stream]

How do I find the parallel line of ax+by=c?

[proud basin]

which question

[fast stream]

b

It’s just I have to convert the linear equation into the parallel line of it

Where it says 3x+2y=2

@proud basin

[raw quest]

ok do u know what it means if the line is parallel to the other

[fast stream]

It’s like same slope but different y-intercept

It’s just idk about how to find the y-int for that parallel line

[proud kiln]

isn't y intercept just the value of y for which x=0?

you don't need to do it through y intercept

[fast stream]

Ik y intercept is (0, y)

[proud kiln]

is it question b?

do you have a specific point that needs to lie on the line or something?

because i don't see the second condition

(unless you need to find just some parallel line)

[fast stream]

It’s like I just need to know what the parallel line of 3x+2y=2 is

[proud kiln]

infinitely many

[fast stream]

So it can be any y-int?

[proud kiln]

3x+2y = z for z in R\{2}

[proud kiln]

yes

[fast stream]

Oh ic thanks

[marsh citrus]

@fast stream Has your question been resolved?

[marsh citrus]

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[earnest sundial]

Are all scalar products of (real) vectors always commutative or is it just the standard scalar product that happens to be?

[earnest sundial]

ie. does the commutative property hold in general for some operation <a, b> of form

[elfin berry]

Bob Goldham

Bob Goldham

[earnest sundial]

I would assume that it should, since this just expands to a sum of products of real numbers and swapping a, b should only really affect the order of factors within products, so this should follow from the commutative property of multiplication in the real numbers?
But I'm not sure that's right

[devout mauve]

no, you need that A is symmetric

convince yourself in the 2x2 case

[marsh citrus]

@earnest sundial Has your question been resolved?

[marsh citrus]

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[jolly ginkgo]

evaluate the integral of x3/sqrt(a4+x4)dx

[jolly ginkgo]

nvm

i thought tansec was a problem turns out is derivative of sec

.close

[marsh citrus]

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[jolly ginkgo]

o shoot

.reopen

[marsh citrus]

✅

[jolly ginkgo]

ok so i had x2=a2tanu

so x4=a4tan^2u

2x=a2.du/dx.sec^2u

and so the original simplify to integral of a^2.secu.tanu/2 du

[vital oracle]

bruh you edited the power here to be 3 instead of 5

[jolly ginkgo]

3

mb

[vital oracle]

do you realize what that means

[jolly ginkgo]

what

[vital oracle]

seeing that theres an x4 in the denominator

and an x3 in the numerator

[jolly ginkgo]

that means?

[vital oracle]

we dont give out answers on the server

Im not going to spoil what you can do with that x4 that the x3 now allows

[jolly ginkgo]

no tell me

[vital oracle]

but its really simple, once you see it youll know it

[jolly ginkgo]

ive never heard of anything like that before

[vital oracle]

no cheating

[jolly ginkgo]

its not a test

[vital oracle]

bro is not going to bother figuring it out

[vital oracle]

yes you have

I didnt say its a "cool new trick"

if its simple, that means youve already done it before, its a standard trick and this time youre having a hard time seeing whats otherwise really simple

theres a simple trick to do with the a4+x4

[jolly ginkgo]

all i see from that is a trig sub

[vital oracle]

and the numerator just being x3

you gotta think here: in what calculus-related scenario have x4 and x3 related together?

[jolly ginkgo]

oh fuck me

[vital oracle]

Im not seeing you connecting what could plausibly work here

so much for a "never heard of anything like that"

[jolly ginkgo]

thats really embarassing

thank you

[vital oracle]

and next time, type ^s to notate powers instead of placing the numbers and letters next to each other like multiplication

[jolly ginkgo]

multiplication does on the front and power go after

i thought that was how it is?

[vital oracle]

mybrotherinchrist

doyouthinksavingspacewillsavetimeonreading?

itsshorterbutthatsnothowthatworks

[jolly ginkgo]

aight

[vital oracle]

no

[jolly ginkgo]

thank you

[vital oracle]

anything strung together is always multiplication

[jolly ginkgo]

in typing

[vital oracle]

a4 is a common formatting error you get when you improperly copy-pasting math

so if you read it as a^4, thats expected, but heavily not recommended

[jolly ginkgo]

really

sorry idk any of that stuff

it was always 3x2 instead of 3x^2 for me

[vital oracle]

then now's your chance to know how youre supposed to properly type math with a keyboard

[jolly ginkgo]

ok ill change

[vital oracle]

you say that like you were right from the get go

"ok ill change"?? 🥶

[jolly ginkgo]

what else am i supposed to do

[vital oracle]

not say that

[jolly ginkgo]

okay

[vital oracle]

maybe say "ok that makes sense"

or "ok it checks out that theres a symbol for ^"

[hearty thicket]

lmao take it easy

[vital oracle]

gotta put some manners into this guy is all, no hard feelings

[hearty thicket]

alr bro you do you

[vital oracle]

realistically the only real scenario you can do this is if youre working with polynomials only, so theres no room for error like here

[jolly ginkgo]

.closoe

.close

[marsh citrus]

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[upbeat haven]

If x belongs to A and x belongs to B then A is subset of B

e.g. let A = (x, y, z) and B = {x,y}

How

[upbeat haven]

<@&286206848099549185>

[nimble prairie]

?

A isn’t even a set here

And if an element is simultaneously in both sets then that does not imply one or the other is a subset of the other..

[marsh citrus]

@upbeat haven Has your question been resolved?

[marsh citrus]

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[marsh citrus]

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[candid thunder]

Multiple choice question

[eager tinsel]

!status

[marsh citrus]

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

[quaint elm]

are there any of these that you can eliminate right away?

[candid thunder]

Idk where to begin

[eager tinsel]

Okay, see
Let's work through the options

[candid thunder]

Ok

[jagged oracle]

The sun flour

What

Find root from second

[eager tinsel]

Don't you think working with option D is a good idea?

[jagged oracle]

Hint:find roots of second equation

[candid thunder]

Yup ig so

[jagged oracle]

Now iam not gonna tell more

[candid thunder]

How to find the common root

[jagged oracle]

See first find det of second equation

[eager tinsel]

you know about Quadratic formula right?

[jagged oracle]

,w calculate 4^2-(4x1x5)

[elfin berry]

Results provided by WolframAlpha

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Requested by oppailol.

[jagged oracle]

So det. Of second equation is -ve

This means it has ______ roots

Fill in the blank sun flour

[candid thunder]

Yes

[jagged oracle]

@candid thunder

[candid thunder]

Imaginary

[jagged oracle]

Yes

[quaint elm]

oopailol I thought you were not going to say more

[jagged oracle]

And imaginary roots occur in _______

[candid thunder]

I calculated the roots are -2 +/- i

[candid thunder]

Lol

[jagged oracle]

No that's good but imaginary roots occur in

[candid thunder]

Idk

[jagged oracle]

Pair

[candid thunder]

Ohh ya

[jagged oracle]

If coefficients are real

[candid thunder]

I remember

[jagged oracle]

And here they are real

1 4 5

So this means

That if the first equation has one root common with 2nd

This should mean what tell me?

[candid thunder]

It will be the same as the second one

[jagged oracle]

Fr

So both equations have same roots

Fr now do it

Do u need anything now

[candid thunder]

How to check option C??

[jagged oracle]

Why u need option C

Aah

Listen

[candid thunder]

Cause it's mcq

[jagged oracle]

Both equations have same roots

[candid thunder]

Both option c and d are correct options

[jagged oracle]

Does that not mean that both the equations can be same

[candid thunder]

Ohh

Ya

[jagged oracle]

Here you got wrong

@candid thunder if two equations has same roots

[candid thunder]

They will be on same ratio

[jagged oracle]

Then the ratio of their coefficients is same

[eager tinsel]

Since both roots are common, the ratio of coefficients of variables will be same

[jagged oracle]

Yes

[candid thunder]

The coefficients will be in ratio

[jagged oracle]

So as it is 1 4 5

We can say a/1=b/4=c/5=k

Option C is done

[candid thunder]

Thanks a lot @jagged oracle @eager tinsel

[jagged oracle]

Fr

@eager tinsel

Say welcome bro

[eager tinsel]

?

[jagged oracle]

Welcome np

[eager tinsel]

Welcome

[jagged oracle]

To her

[candid thunder]

Ty