help-23

[solemn vault]

Hi again

[solemn vault]

How do i solve this now

<@&286206848099549185>

.close

[safe radish]

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[boreal nexus]

?

[safe radish]

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[compact needle]

theres 2 ways to solve

[urban rivet]

ya

[compact needle]

the simplest is to rearrange one equation so you get one variable on it's own

[urban rivet]

show me

how to rearrange

[compact needle]

7x + 4y = 26

right

[urban rivet]

ya

[compact needle]

you want to re-arrange for one variable

try doing that

rn

try and get x or y on it's own

[boreal nexus]

i think y would be a wiser choice but it doesnt matter much

[compact needle]

get back to your trig equation bro💀

[boreal nexus]

ok bro💀

[urban rivet]

u cant divide

[compact needle]

wydm?

so 7x + 4y = 26

you want x or y on its own

so in this case y looks easier

you move 7x to the other side

4y = 26 - 7x

then divide by 4

[urban rivet]

ya

k

[compact needle]

y = (26 - 7x)/4

now you substitute that for y

in the other equation

5x + 4y = 14

right?

[urban rivet]

ya

[compact needle]

5x + 4(26 - 7x)/4 = 14

now solve that for x

[urban rivet]

4y

[compact needle]

you've solved for y in one equation

now you use that to find out x

[compact needle]

in relation to x

so the final equation you have is 5x + 4(26-7x)/4 = 14

now solve that, and you get x

[safe radish]

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[compact needle]

then you can substitute x in to find y

[urban rivet]

k

ty

[compact needle]

np

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[urban gale]

I don't really get this question.

[urban gale]

are they asking for a in both trig? If so, it would be x

[thin bridge]

no

they're implying that the sum will be a number/constant

the goal is the find that number/constant

[urban gale]

so just solve the problem?

[thin bridge]

yes

[urban gale]

with x as A?

[thin bridge]

no

A is A

x is x

just ignore the A completely

and simplify the LHS

[urban gale]

no like in the formula:
cos(A+B) or sin(A-B)

[thin bridge]

zzz

don't use existing variables to express different things

that will only confuse yourself and others

[urban gale]

you know what i mean?
sum and difference formula

i meant alpha and beta

[thin bridge]

you can apply those if you want

[urban gale]

for cos (x + pi/6):
sqrt(3)/2 * cos x - 1/2 * sin x

and for sin(x - pi/3):
1/2 * sin x - sqrt(3)/2 *cos x

and uhh... they cancel each other out. so its 0

[sharp crane]

Or just use
$$\sin x = - \cos\qty(\frac{\pi}{2} + x)$$

[flat frigate]

nvx

[urban gale]

oh.. but its the right answer

[sharp crane]

To immediately get the answer

[urban gale]

good point

This question confuses me. What does it want and what do i do?

I figured out its subtraction: cos (45-30)

[sharp crane]

yes

so you solved it?

[urban gale]

yes. sqrt(6)+sqrt(2)/ 4

is that correct?

[thin bridge]

missing parentheses

then do a little factorisation and express it in the specified form

[urban gale]

sqrt(3)+1/ 2 sqrt(2)

so what about A and B?

[long wigeon]

Factor put the square root of 2

So sqrt2 (sqrt3 +1)

That’s the numerator

So A is 2

And B is 3

[urban gale]

oh. thanks for the help

[long wigeon]

Np

[urban gale]

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[rigid kernel]

How do I find the value of 27^1/4

[rigid kernel]

I need to know how step by step

[thin token]

you serious?

[solemn vault]

😂

[crisp linden]

You can approximate the value of 27^(1/4) using the Taylor series of (1+x)^alpha

[nova creek]

Optionally you can use a calculator

[safe radish]

@rigid kernel Has your question been resolved?

[forest hemlock]

Hey, since the channels are broken and I don't see supoerman asking questions anymore I'd like to use this channel for help

A slope tangent to the graph of the function at a point (0,-1/3) is -1/18
I need to find A and B

I cannot solve this

Would love if someone will help me out here

[safe radish]

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[forest hemlock]

.

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[forest hemlock]

.

A slope tangent to the graph of the function at a point (0,-1/3) is -1/18
I need to find A and B
I cannot solve this

[forest hemlock]

I didn't

I really struggle with this exercise for 1-2 hours

I'll send you what I got

oh ye

[obtuse jackal]

this is a problem where you can very easily check your answer anyways

[forest hemlock]

let me find it

thanks for your help ramonov but with your assistance i still couldn't understand

[thin bridge]

fk made a typo there

(0,-1/3)
is also a point on the curve

using the derivative gives one equation

[forest hemlock]

A = 9, B = -9

[thin bridge]

using the fact that the point lies on the curve gives another equation

you'll have a system of two equations in two variables

which will give you the values of A and B after solving

[forest hemlock]

i need to see how its being solved because i do not manage to do it on my own

[thin bridge]

after finding the derivative the rest is algebra

[forest hemlock]

this specific exercise i just do not manage

i just solved 20 different] exercises

but not this

[thin bridge]

can you show what you currently have?

[forest hemlock]

i tried to to f'(x) again

i send

[thin bridge]

iirc before you made any additional attempts at simplification/manipulation,
the expression you initially had seemed fine

you could simplify that further (properly) if you wanted
or just go ahead and apply

tangent to the graph of the function at a point (0,-1/3) is -1/18
directly

that will give you one equation

[forest hemlock]

i have net issues so it will be uploaded in a sec

[thin bridge]

have you done anything after that?

[forest hemlock]

i tried few times but i already erased it

[thin bridge]

why erase

attempt it again and show me

[forest hemlock]

Because it was wrong

Aight i will try

[safe radish]

@forest hemlock Has your question been resolved?

[forest hemlock]

I'm sending

[thin bridge]

your handwriting needs improving as some of those + signs look like - signs

also you really stuffed up your algebra

with multiplication, squaring etc

also choosing not to simplify stuff like 0+A to A for multiple lines is also concerning

[forest hemlock]

so how do i need to solve this correctly

[thin bridge]

fix your handwriting

and work on your basic algebra

try again from

[forest hemlock]

they are both +

[thin bridge]

?

[forest hemlock]

on my page i mean

[thin bridge]

the vertical line is so faint / near invisible that noone can really tell

it shouldn't take me more than a second to stare / look closely to see what sign its supposed to be

anyway describe exactly what you're doing after that step

[forest hemlock]

multiply cross numerator with denominator

[thin bridge]

ok and show the full work for that

for convenience, do you want the - sign on the numerator or denominator for the -1/18?

[forest hemlock]

numerator

[thin bridge]

what's the numerator of the fraction on the left?

[forest hemlock]

-18B+36A

[thin bridge]

no

of the thing in the screenshot

[forest hemlock]

oh

[thin bridge]

or perhaps post your updated work

since it seems that you've since changed a few things

[forest hemlock]

sending

[thin bridge]

,rotate

[thin bridge]

that seems ok now
but still fix that + sign

looks like a $\perp$

[flat frigate]

ℝamonov

[thin bridge]

now use the fact that

(0,-1/3)
is a point on the curve to get another equation

[forest hemlock]

on which equation do i use it

[thin bridge]

use it on what you're given

[forest hemlock]

on this?

or on f'(x)?

[thin bridge]

on f(x)...

[forest hemlock]

I still dont understand

[forest hemlock]

how could i use this on f(x)

[thin bridge]

you don't

now use the fact that
(0,-1/3)
is a point on the curve to get another equation

[forest hemlock]

so which equation do i do
-1/3 = √0+a/0+b?

[thin bridge]

missing parentheses

and messed up capilisation

$\frac{\sqrt{A}}{B} = -\frac13$

[flat frigate]

ℝamonov

[forest hemlock]

how did you come up to that one

[thin bridge]

wdym

[forest hemlock]

how did you come to that equation

[thin bridge]

(0,-1/3)
is a point on the curve

[forest hemlock]

so why we dont include x'es too

if we use it on f

[thin bridge]

wdym

[forest hemlock]

f(x)

this is f(x)

[thin bridge]

f(x) is what you're given

yes

[forest hemlock]

you wrote only √A and B

[thin bridge]

so what does

(0,-1/3)
is a point on the curve

tell you

[forest hemlock]

not sure

[thin bridge]

don't overthink this

how would you determine if a point lies on a curve

algebraically

[forest hemlock]

curve refers to X and Y?

[thin bridge]

curve refers to a graph of a function

tangent to the graph of the function f(x) at a point (0,-1/3) is -1/18
gives you MORE information than just f'(0) = -18

specifically that if you were to graph y=f(x)
(0,-1/3) will be a point on that curve / graph of the function

do NOT overthink this

[forest hemlock]

idk

[thin bridge]

lets say there was a line
y = 3x + 5
how would you determine whether the point (1,3) is on that line?

(algebraically)

you're either overthinking this and/or have a major coordinate geometry knowledge gap to be doing calculus

[forest hemlock]

we see if y matches 3x + 5

[thin bridge]

show exactly what you'd do

what you said doesn't mention the point of interest (1,3) at all

[forest hemlock]

i'll do 3 = 3*1 + 5

it wont match each other which means its not on the graph

[thin bridge]

now if you were given
y = mx + 4
how would you determine the value of m such that the point (3,6)
lies on that line

[forest hemlock]

6 = m3 + 4

[thin bridge]

and apply the exact same idea to your question

[forest hemlock]

so y-0=-1/18(x-(-1/3)?

[thin bridge]

whut?

no

[forest hemlock]

idk this is taking too long

lets just drop this

[thin bridge]

you're either overthinking this and/or have a major coordinate geometry knowledge gap to be doing calculus

those simplified examples were intended to remind you of what it means for a point to lie on a curve

[forest hemlock]

i've done all the other calculus exercises and i managed to do okay with them

[thin bridge]

and the info you can get from that

[forest hemlock]

i will try to get help from somebody else and i will see if i will understand that way

i'm just wasting too much time on one exercise and its pointless

[thin bridge]

(0,-1/3)
lies on the curve y= f(x) = sqrt(x+A)/(x+B)

applying the EXACT same idea earlier, what info can you get from that?

and I must insist that you do NOT overthink this

(0,-1/3) lies on the curve

f(0) = ?

[forest hemlock]

0?

[thin bridge]

no

(0,-1/3) lies on the curve

what is the value of y=f(x) when x=0?

[forest hemlock]

-1/3

[thin bridge]

yes

apply that

[forest hemlock]

so -1/3 = √0+A/0+B which means it becomes √A/B?

[thin bridge]

missing parentheses

$\frac{\sqrt{A}}{B} = -\frac13$

[flat frigate]

ℝamonov

[thin bridge]

but yes, that simplifies to the above

[forest hemlock]

and how do i continue from there

[thin bridge]

you now have a system of equations

of two equations and two variables

solve using stuff like substitution

[forest hemlock]

not working out

thanks for trying but i will use someone elses help, it just takes too much time as i said

.close

[safe radish]

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[short kestrel]

what's happening here?

<@&286206848099549185>

[lean chasm]

pls only ping helpers after 15 mins

[lyric ingot]

Were you solving for x and then the answer was 2 - log3/log2

[short kestrel]

yes

how did they get to 2-log3/log2

[lyric ingot]

Make a u - substitution for 2^x

Then solve as if it were a quadratic

u(u + 4 - 16)

u^2 + 4u - 16u = 0

[short kestrel]

wouldn't it be u^2-12=0?

[lyric ingot]

Oh

Right, like terms

4u - 16u = -12u

u^2 - 12u = 0

[short kestrel]

u-12=0

u=12

[lyric ingot]

Uh

[short kestrel]

so 2^x = 12

[lyric ingot]

What?

[short kestrel]

u can't be 0, no value of x can make 2^x = 0

so 2^x has to be 12

[lyric ingot]

Actually you can just say that 2^x = 0, 2^x + 4 - 12 = 0

Since if either one is zero, the entire thing is zero

No u substitution requireed

[short kestrel]

xlog2 = log12

[lyric ingot]

2^x = 0 is undefined

[short kestrel]

now I get the wrong answer here

it's meant to be 0.415, not 3.58

[lyric ingot]

2^x + 4 - 12 = 0
2^x = 8
x = 3

I have no fucking clue how they got 2 - log(3)/log(2)

[short kestrel]

naw it's 4-16

not 4-12

[lyric ingot]

Oh

[short kestrel]

yeah that's what I'm thinking

[lyric ingot]

I made it convienient

Whoops

2^x = 12

x = log_2(12)

log(12)/log(2) if you change bases like they did

I don't get it lol

[short kestrel]

apparently the answer is 4 minus that

[lyric ingot]

Bs

[short kestrel]

yep, wrong answer

god I've spent hours on this

[lyric ingot]

What was it supposed to be?

Log(12)/Log(2)?

[short kestrel]

yep

[lyric ingot]

Nice

[short kestrel]

which does simplify to 2+log3/log2

[lyric ingot]

I got that one right

[short kestrel]

hell yee

[lyric ingot]

How????

Lmfao

[short kestrel]

log_2 12

log12/log2

(log2^2 . 3)/log2

(2log2 + log3)/log2

2 + log3/log2

.close

[safe radish]

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[marsh walrus]

hello

[marsh walrus]

I have, i think, the answer

but im really curious if the $\sqrt{n+m}$ in the end is correct?

[flat frigate]

jan Niku (Shuri for Honorable)

[marsh walrus]

or should it be like $\sqrt{n+2m}$ or some min or max or something?

[flat frigate]

jan Niku (Shuri for Honorable)

[marsh walrus]

min and max definitely not i think, but maybe the 2m?

[safe radish]

@marsh walrus Has your question been resolved?

[marsh walrus]

<@&286206848099549185>

our boys in blue 🙏

.close

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[regal ivy]

so qu ii

do I just find the eqn for the tangent of the max/min of the curve

and find the coordinates?

[safe radish]

@regal ivy Has your question been resolved?

[keen meadow]

answer with ❌ to the bot above

so he doesn't close the channel

[safe radish]

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[keen meadow]

<@&286206848099549185>

[safe radish]

Please don't occupy multiple help channels.

[ancient escarp]

that's not even your question??

[safe radish]

@keen meadow Has your question been resolved?

[keen meadow]

i know, i'm sorry

[safe radish]

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[safe radish]

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[tacit fog]

Anything else u have been told?

Lengths or something

Oh

[sharp tiger]

Tan42 = ae/ab

[tacit fog]

Oh

U have been told ab

4cm?

[sharp tiger]

Oh

That angle 42 is bd

Not be

[tacit fog]

I see

[sharp tiger]

It's a diagonal of base

De is diagonal of that left wall

[tacit fog]

Oh man this image is like a optical illusion

[sharp tiger]

Inclined at 30

[tacit fog]

To me

[sharp tiger]

Lol yea

Ok so tan42 will be = ad/ab

In traingle adb

[tacit fog]

Ya

Well

I get why they calculated AD

First

[sharp tiger]

Exactly

Cuz they've given us 2 angles

[tacit fog]

Yes

They gave 2 angles

Is it's not bd

Then

[sharp tiger]

B-d-e is not a straight line

[tacit fog]

The answer is just tan(42) * ab

So it's quite obvious

It's bd

Coz ig we have been told the other angle for a reason

Which only pops up

When it's bd

[sharp tiger]

How

[tacit fog]

Uh

BDE is

[sharp tiger]

[tacit fog]

Not straiggt

Line

[crisp linden]

I am failing to see how BDE is a line

[tacit fog]

It bends diagonally in ADH

Umm so did you got how to solve it?

Yup

No

Ur making a mistake here

ABE is not one of the angles of the triangle ABE

But rather of ABD

Well it's pretty obvious

Ig

As u have been told that 30 dergeess for a reason

Which only comes to use

When it's an angle of

Triangle ABD

This is what I generally see

Ya like it generally happens like that

Rip

[safe radish]

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[ocean karma]

hi I have a question

[ocean karma]

If I have a 0,3% chance of winning something, and I win it 4 times in 161, what would that be in %

[plucky elk]

,calc 4/161 * 100

[flat frigate]

Result:

2.4844720496894

[plucky elk]

2.4845%

[ocean karma]

what if there's a 0,3% chance to win it and you win it 4 times in 161, is it still 2,48?

[plucky elk]

Is what still 2.48?

[ocean karma]

%

[plucky elk]

% of what

[ocean karma]

._.

I'll do my best to explain it more clear

let's say I'm bad at rock paper scissors and I have 0,3% chance of winning in rock paper scissors

[plucky elk]

Stop

Your example is already too complicated

Just pick a dice example

[crisp linden]

or coins

[plucky elk]

There's a 1/6 chance or 16.66% percentage of rolling a 1

[ocean karma]

hm ok

[plucky elk]

But you roll a 1 13 times out of 100 rolls

What question are you asking about

13 in 100 as a percentage?

Or 16.66% for rolling a 1?

Those are two different questions. Pick one

[ocean karma]

this was my question

[potent bay]

can you send us the original question? or was this just your own idea?

[ocean karma]

this was just my own idea

[plucky elk]

Then this is the answer

[crisp linden]

The chance of winning it doesn't factor in at all

[ocean karma]

oh I see

[crisp linden]

As we understand your question

[plucky elk]

Your own idea is vague and you have to clarify

[crisp linden]

Just to take an extreme example: I have a 99% chance of winning, but win one time out of 100 tries? What is that in percent?

That is a confusing question.

[potent bay]

if you have a 50% chance that u get head on a coin(normal coin), and you flip it 10 times and you got ALL 10 heads, the % would still be 50% that u get a head on the next flip

the outcomes doesnt change the probabilities given

[potent bay]

if that was your question

[ocean karma]

.close

[safe radish]

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[lean otter]

How does a^1/n work?

[potent bay]

wdym how does it work?

[worthy hemlock]

https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:exp/x2ec2f6f830c9fb89:rational-exp/v/basic-fractional-exponents

See if that helps

[lean otter]

how do you get the exponent 1/n to be n squareroot?

[lean otter]

will check this

[potent bay]

sry im not good with latex but: a^(1/n)* a^(1/n)* .... * a^(1/n)= a^[(1/n)*n]= a^1= a

[lean otter]

Doesn't really explain it

[potent bay]

so what number do we need to get so if we take it n times its product gives 'a'

well here we are at n root 'a'

[lean otter]

Ok I think I get it now. For example 4^1/2 * 4^1/2 = 4 ^2/2 = 4 and Sqr(4)* Sqr(4) = 2*2=4

[worthy hemlock]

http://www.andrusia.com/math/preliminaries/ExponentiationTheorems.pdf

[stray socket]

Are you asking about $a^{\frac{1}{n}}$

[flat frigate]

Umbraleviathan

[lean otter]

yes

[stray socket]

Yeah so basically anything in a denominator becomes the base of an nthroot

1/2 -> square root

1/3 cube root

1/(46838383383) would be the 46838383383th root

[lean otter]

does it change if the exponent is negative?

[stray socket]

Well

-1/2 would be 1/sqrt(x)

-1/3 would be 1/cuberoot(x)

-1/n would be 1/nthroot(x)

[potent bay]

yeah, as your picture shows

[lean otter]

thanks

.close

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[winter pivot]

prove that:

[winter pivot]

$\int_{\infty}^{\infty} \exp(-x^2/2) = \sqrt{2\pi}$

[flat frigate]

EndTimes

[pallid nymph]

whatve you tried?

[stoic dune]

Haha I have no clue how you'd get this if you are uninformed

There's a tricky double integral way where you convert into polar

[winter pivot]

ok sure let's try that ig

or wait

i guess it would be simpler to convert into complex exponentials?

but that gives us problems

so let's do this as a double integral in x and y. our integrand is 1 dy dx and our bounds are:
0<= y <= exp(-x^2/2)
-inf <= x <= inf
?

then we change it to r dr dt where x=rcost and y=rsint

[stoic dune]

Like I said, it's something you've either seen before, or you haven't and would never get it

Maybe I can do a hint and keep this an interesting puzzle? Lol

[winter pivot]

sure

[stoic dune]

Consider both of these integrals over [0,inf]:
∫ e^(x²) dx and ∫ e^(y²) dy
And find their product. That is, find:
(∫ e^(x²) dx)(∫ e^(y²) dy)

[winter pivot]

can you turn a product of integrals into an integral of products

[safe radish]

@winter pivot Has your question been resolved?

[stoic dune]

@winter pivot
Yes, that's Fubini's theorem

[safe radish]

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[lean otter]

Salutations, what is the point of doing this https://cdn.discordapp.com/attachments/326138772477575180/959883704824627290/unknown.png

[lean otter]

why divide dy with dx then multiply with dx again

[lean chasm]

it's just notation

$\frac{dy^2}{dx^2}$ isn't a "real" fraction

[flat frigate]

illuminator3

[lean chasm]

it also has a typo?

it should be (if my brain is working correctly): $$\frac{d^2y}{dx^2}$$

[flat frigate]

illuminator3

[lean chasm]

which means the second derivative of y wrt x

[lean otter]

yeah but it will result in dy anyways

so might aswell right dy?

[lean otter]

why does dx^2 suddenly become 1

[lean chasm]

$dy \neq \frac{dy}{dx} \cdot dx$

[flat frigate]

illuminator3

[lean chasm]

it didn't

they moved the dx^2 term out

$$a + b \cdot a = (1 + b) \cdot a$$

[flat frigate]

illuminator3

[lean otter]

i see, rules

[lean chasm]

also nvm this

they just squared the first derivative

[lean otter]

this isnt true tho?

[lean chasm]

that inequality is true

[lean otter]

$dy \eq \frac{dy}{dx} \cdot dx$

[flat frigate]

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

[lean chasm]

$\frac{dy}{dx}$ isn't a fraction per-se

[flat frigate]

illuminator3

[lean chasm]

it means the derivative of y wrt x

[lean otter]

yes, but dy = dy/dx * dy

[lean chasm]

in prime notation it'd be $y'$

[flat frigate]

illuminator3

[lean chasm]

no

[lean otter]

?? how come

[lean chasm]

$dy \neq y' \cdot dx$

[flat frigate]

illuminator3

[lean otter]

oh

so its not a fraction? its an operator?

[lean chasm]

yea it's kinda like its own symbol

$\frac{dy}{dx} = y'$

[flat frigate]

illuminator3

[lean chasm]

it's just another way of writing it

a more explicit way

[lean otter]

i see

well, this isnt explained tho

it just went from this

to this

https://cdn.discordapp.com/attachments/903481106819612714/959893094390693978/unknown.png

do you know why @lean chasm ?

[tall bough]

(ax)/x=a

simple

[lean otter]

?

[tall bough]

x cancels out

x/x=1

[lean otter]

i know

but why

is it written like that

whats the point

[plucky elk]

Do an actual problem and you'll see

[safe radish]

@lean otter Has your question been resolved?

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[gleaming spade]

how would i solve this without losing some of the solutions

$\log_x{(2x)}\le\sqrt{\log_x{(2x^3)}}$

[flat frigate]

Pšenica

[gleaming spade]

if you square it you lose some of them

if i try solving it without squaring i get $t=\sqrt{\log_x{2}+3}$ and the equation $(t-2)(t+1)\le0$

[flat frigate]

Pšenica

[gleaming spade]

but one of those solutions is complex so i now have a weird inequality with only one real zero

[safe radish]

@gleaming spade Has your question been resolved?

[drowsy moss]

how do you get that statement with t?

[gleaming spade]

well you do some magic with log properties

and then you see that whats under the sqrt is the same as the outside just 2 more

so you add 2 on both sides and you get that

[drowsy moss]

oooooh ok

yeah, so the one real answer is the only place they intersect

there's also a point of discontinuity

[gleaming spade]

its a bit wonky

but this is supposed to be the answer

$x\in(0,\sqrt[3]{\frac{1}{2}}]\bigcup[2,\infty)$

[flat frigate]

Pšenica

[gleaming spade]

if i graph it in desmos, sure i can see it

but idk how to get cube root 1/2 as a solution

[gleaming spade]

.close

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[lean otter]

im trying to prove the length of a function formula. however i dont understand step 2

[stray socket]

Yeah so they're just rewriting $dy^2$ as $\frac{dy^2}{dx^2} \cdot dx^2$

[flat frigate]

Umbraleviathan

[stray socket]

@lean otter

[lean otter]

but why

[stray socket]

Because you get dy/dx

In there

[lean otter]

is it for the integration?

[stray socket]

Yeah, kinda

It's to get dy/dx in there

[lean otter]

i see, aight ty

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[stray socket]

No problem

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[next copper]

Using these equations:
y = -4.5(x - 3)² + 46 y = -4.5x² + 27x + 5.5 (same equation, different form)

When y = 30 what is x equal to (I know I can use Desmos to find out but I want to know how to figure it out to)

[cursive spindle]

Quadratic formula would typically work

You can derive it

From completing the square

[next copper]

Isn't Quadratic formula for finding the roots (x intercepts) of a parabola?

[marsh furnace]

Yes, it can then be used to find the value of x

[next copper]

When y = 0

The two occurences in which it is

[next copper]

I'm sorry if this is a stupid question but if by completing the square you mean making it a vertex form equation I did that

Mahdi 👀

[arctic lagoon]

square the both sides and do not change the sign cause any log(x) ϵ R+

you gone delete the sqrt in the second side

DO NOT

use the log rules to make the exponent a coefficient

just multiply by x

2x^2 = 2x^3

x^2 = x^3

[next copper]

Mahdi that was a day ago 😭

[next copper]

<@&286206848099549185> (for my question, not TeXit from yesterday that Mahdis helping)

[safe radish]

@next copper Has your question been resolved?

[next copper]

It has not

[next copper]

This is the question

If there's anyone here

[lost cobalt]

using the first equation you have
y = -4.5(x - 3)² + 46

substitute y=30
so you will have
30= -4.5(x - 3)² + 46

you can solve for x now
30-46 = -4.5(x - 3)²
-16 = -4.5(x - 3)²

-16/-4.5 = (x - 3)²

if you take the square root of both sides you will get
square root (-16/-4.5) = x-3

and from here you can easily find the value of x

[safe radish]

@next copper Has your question been resolved?

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[zenith jay]

I'm stuck on solving for b and c

For b I write it as the square root of the dot product of (2v+3w)

And get sqrt(4+9-12*0.97)

[safe radish]

@zenith jay Has your question been resolved?

[zenith jay]

<@&286206848099549185>

[stoic dune]

Dot product of (2v + 3w) and what?

[zenith jay]

and itself

since | |2v+3w||^2= (2v+3w)*(2v+3w)

[stoic dune]

Ye that works
(2v + 3w)•(2v + 3w)
= 4v² + 9w² + 12v•w

Okay yeah that's exactly what you've got

[zenith jay]

doing tat I get 1.166

But it registers as wrong

[stoic dune]

,calc 4 + 9 + 12(-0.97)

[flat frigate]

Result:

1.36

[zenith jay]

It's sqrt of that

[stoic dune]

Right thank you

And it didn't like it, huh?

[zenith jay]

Yes

I'm very confused

The full value for sqrt 1.36 is 1.16619

It doesn't accept 1.166 though

Would 1.166 be wrong somehow?

[stoic dune]

,calc cos(2.9)

[flat frigate]

Result:

-0.97095816514959

[stoic dune]

,calc sqrt(4 + 9 + 12cos(2.9))

[flat frigate]

Result:

1.1612501962131

[stoic dune]

That's a bit more exact. What about that?

[zenith jay]

IT WORKS

THANK YOU

[stoic dune]

Lol dumb assignment

[zenith jay]

It only works when I put 1.16125

[stoic dune]

Teacher has the digits up too high

[zenith jay]

Yeah... it accepted-0.97 before

well thanks!

I would have wasted much more time lol

[stoic dune]

Np. Good luck with the rest!

[safe radish]

@zenith jay Has your question been resolved?

[plucky elk]

.close

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[storm oxide]

three lines y=x+4 , y=2x-4 , and y=ax-4 are creating a triangle with area of 48, find the value of a

[storm oxide]

i dont know how to start to solve

idk what to do

[safe radish]

@storm oxide Has your question been resolved?

[rotund stratus]

find where those lines intersect

[storm oxide]

i found it

then?

[rotund stratus]

Plot it on a plane

sry forgot can’t do that bc there’s a variable

I think you can use this formula

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@storm oxide Has your question been resolved?

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[lean otter]

Hello

[lean otter]

Shouldn't not this be 6 * sin(6x)?

[crisp linden]

differentiate 1/6 sin(6x)

and see what you get

And if you're still not convinced, take 6sin(6x), and differentiate it, and observe how you do not get cos(6x)

[lean otter]

Got it.

But what I am doing wrong?

I took u = 6x

du = 6

Then I integrated cos(u) * 6

[lean otter]

Shouldn't this be 1/6 * cos(6x)?

[crisp linden]

u = 6x

du = 6dx

what is dx equal to?

[lean otter]

ohh yes

It will be du/6

[crisp linden]

mhm

[crisp linden]

To check whether you integrated correctly

just differentiate the result

If you get the same thing that you integrated, you know you did good.

[lean otter]

Yes, but one thing which I am not understanding is:

Of course, I am missing something. But I am not able to understand what.

y = 1/6 * sin(6x)

Here shouldn't 1/6 go outside?

[crisp linden]

sure, you can pull a constant out

what's the derivative of sin(6x)?

[lean otter]

cos(6x) * 6

[crisp linden]

Now we need to remember the 1/6 we previously pulled out

1/6 * 6 * cos(6x) = cos(6x)

[lean otter]

Yes, Finally

[crisp linden]

and ey, that's the thing we integrated!

[lean otter]

True 😍

Thank you @crisp linden 😊

.close

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[zenith pivot]

i have no clue where to start

[feral linden]

You didn’t learn sum of geometric sequence?

[zenith pivot]

i have

but its like the reverse right

[feral linden]

?

[zenith pivot]

o btw im trying to find n

[feral linden]

It’s a geometric sequence

[zenith pivot]

yes

[feral linden]

So?

[zenith pivot]

it gave me the sum

[feral linden]

Calculate that sum

[zenith pivot]

the sum is -341?

[feral linden]

I am asking you to calculate that sum using n

You said you knew how to calculate sum of first several terms of a geometric sequence

[zenith pivot]

ok

maybe

i dont

[feral linden]

?

[zenith pivot]

would the first term be 1

[feral linden]

Let’s say you want to calculate T=1+b+…+b^(n-1)
bT-T=(b+b^2+…+b^n)-(1+b+…+b^(n-1))=b^n-1 so when b doesn’t equal 1, T=(b^n-1)/(b-1)

That’s how to calculate sum of geometric sequence

Go ahead Apply it

[zenith pivot]

im confused

[feral linden]

What?

First you said you knew sum of geometric sequence, next thing it appears that you don’t know it. Second I just told you how to find the sum of geometric sequence then you told me you are confused

What exactly do you know or don’t know

[zenith pivot]

i know nothing

the equation

[feral linden]

Then I explained

[zenith pivot]

alright

[feral linden]

If you are not satisfied with the explanation you can read textbook, probably gives you exactly the same explanation but with better writing

[zenith pivot]

ik the equation to find the sum

but

in the problem

[feral linden]

and I told you how to find the sum

[zenith pivot]

its saying to bascailly go backwards

using the sum to find the number of terms

[feral linden]

I say it again, you calculate that sum using n

Express that sum in n

[zenith pivot]

alright

so

[feral linden]

So do it

[zenith pivot]

yes?

[feral linden]

No

It’s -4^n

Not (-4)^n

[zenith pivot]

so

[feral linden]

No

Not -4

4

[zenith pivot]

wait

the r isnt -4

[feral linden]

Told you

[feral linden]

It’s - 4^n

Not (-4)^n

It’s -(4^n)

[zenith pivot]

surely

i got it

[feral linden]

Again no

4 not -4

[zenith pivot]

on the denominator

[feral linden]

I keep telling you it’s-(4^n)

[zenith pivot]

yeah

on the numerator i put that

[feral linden]

What

[feral linden]

Wrong

You didn’t fix it at all

[zenith pivot]

ok

i will try again

[feral linden]

yeah

[zenith pivot]

let sgo

ok

so

i need a1

which is not given in the problem

[feral linden]

-1

[zenith pivot]

how exactly do i find that

[feral linden]

What?

Observation

[zenith pivot]

https://cdn.discordapp.com/attachments/903481106819612714/960715021791408128/unknown.png

[feral linden]

Why can’t you see it

What’s blocking you

[zenith pivot]

its being raised to the i-1

and usually

[feral linden]

So?

[zenith pivot]

the change is always being raised

[feral linden]

God I give up

-(4^n-1)/3=-341

4^n -1=1023

n=5

I am done

[zenith pivot]

wow

did u use log

@feral linden

[zenith pivot]

thanks for helping

.close

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[dreamy finch]

if im trying to figure this out , i would use the remainder formula which is m/(n+1)! multiplied by (x-c)^n+1, right

[dreamy finch]

then i could sub in values

x=.4

c=0 (maclaurin series)

and then m is d/dx sin(x) at x=.4

can you tell me if im doing this correctly so far please

oh wait, it would be d/dx at x=0, right?

[tame charm]

Wdym m

Let’s just start with: what is the maclaurin polynomial for sin(x)

[dreamy finch]

is the maximum value

[tame charm]

?

[zinc berry]

They are saying their variable m is the maximum value

As in the remainder theorem normally has you maximize the absolute value of the k+1th derivative between the center and x. But the phrasing and reasoning is a bit sloppy, especially the part where m is assumed to simply the first derivative of sin(x) at x = 0.4.

[dreamy finch]

wait let me try another one

so m is d/dx of ln(x) at x=2, right?

then 2.5 is the variable x in the formula (has nothing to do with the x variable in the derivative above)

and then c is 2

so i would be evaluating .5/(n+1)! times (.5)^n+1, right?

and then plug in values for n to see when that solution is equal to or less than .001

is that correct?

[tame charm]

Why can’t you just find the maclaurin series and plug in .4 repeatedly until you get a good enough error

[dreamy finch]

cuz im trying to learn with the formula

and also its saying it is a taylor polynomial

[zinc berry]

Is that a question, I don't understand what you are trying to get at

[dreamy finch]

all my messages below this image is my question

or are my questions i should say

i just want to know if i got the variables right

[grizzled stirrup]

Do you mean slope by m?

[dreamy finch]

i dont think so:

[tame charm]

This is what you’re using?

What is m exactly I didn’t really get it sry

[dreamy finch]

m is just the max value between a and x (or c and a if you're using the variables i mentioned)

[grizzled stirrup]

The question just wants you to calculate ln(2.5) with an accuracy of 3 decimal points

[dreamy finch]

yes

[grizzled stirrup]

I think its the maximum value of the 2nd derivative of lnx

Yep

[dreamy finch]

dayum

ive always been doing hte first derivative

i did 3 practice problems and got them all right (by taking the first derivative) and i thought i understood it until this problem. Was just because i was getting lucky?

thank you so much anyways

[grizzled stirrup]

Mvm it is not the second derivative. It is n+1th derivative where n is the order of the Taylor polynomial ur calculating the error for@dreamy finch

[safe radish]

@dreamy finch Has your question been resolved?

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[winter pivot]

Yeah try finding the common denom

[dapper forum]

Do you have to use the limit h goes to zero to find derivative?

I am looking

[flat frigate]

Cringelord

[winter pivot]

Factor out 1/h

1/h (first fraction - second fraction)

I didnt feel like writing it out

Exactly the same

1/h (first fraction - second fraction)

Is equal to
(First fraction - second fraction)/h

[flat frigate]

Cringelord

[winter pivot]

No

Get rid of the extra h

Multiplication distributes

Your numerator is correct

You don't need the denominator

so now we have $\frac{1}{h}(\frac{t^2}{t^2(t+h)^2})-\frac{1}{h}(\frac{(t+h)^2}{t^2(t+h)^2})$

[flat frigate]

EndTimes

[winter pivot]

Multiply

[dapper forum]

I am glad you're getting help. You can get lost in calculation. lol

[flat frigate]

Cringelord

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[broken ether]

hi

[broken ether]

i need assistance with determining Monotonie i think it’s called? idk what it is in english

f(x) = x^2 - 2 x

is my question

the question reads “Investigate the function f with regard to its monotonic behavior using the monotonicity theorem.”

[winter pivot]

Then we have:
[-(2ht+h^2)] / [ht^2(t+h)^2] =
[-(2t+h)] / [t^2(t+h)^2] =
[-(2t+h)] / [(t^2+ht)^2] =
-2t/t^4 = -2/t^3

@compact isle

[broken ether]

uhhhh

[winter pivot]

A monotonically increasing function is one whose derivative is always greater than or equal to 0

Monotonically increasing means never decreasing

[broken ether]

so how would i solve

[winter pivot]

Find the derivative

[broken ether]

on my paper i did a derivation so it went from
f(x) = x^2 - 2 x
f’(x) = 2x - 2

but idk how to continue

i am super horrible at math

[winter pivot]

Find when it is greater than 0

[sharp crane]

find on which interval(s) f' >= 0 and on which interval(s) f' <= 0

[broken ether]

i am very lost 😅

i mean on

both

2x - 2

[winter pivot]

Do you know what the derivative means?

[broken ether]

they are both above

0

[broken ether]

no

[sharp crane]

when is (2x - 2) > 0?

[winter pivot]

The derivative is the rate of change of a function. It is the slope of the tangent line at a point

[broken ether]

isn’t this equal to zee

zero

[sharp crane]

not always. it's the equation of a line, so there will be exactly one value of x where 2x - 2 is zero

otherwise it's either strictly positive or negative

[broken ether]

so how do i determine that?

[winter pivot]

2x - 2 = 0

Solve

[broken ether]

i would not know where to go further from there

[winter pivot]

Try adding 2 to both sides

[broken ether]

sorry i don’t understand

which sides? like to 2x and to 0?

[winter pivot]

Yes

[broken ether]

ok so 4x - 2 = 2

[winter pivot]

Not quite

We can only add things that have the same name

x doesn't have the same units as 2

So 2x+2 is not 4x

[broken ether]

how else can you add 2?

[winter pivot]

To the constant term

The term that doesnt have an x next to it

[broken ether]

oh so 2x - 4 =2?

[winter pivot]

Not quite

We have the equation:
2x - 2 = 0

[broken ether]

yes

[winter pivot]

When we add 2 to both sides we get:
2x - 2 + 2 = 0 + 2

[broken ether]

is it ok to write it as 2 + 2x + 2 = 0 + 2

[winter pivot]

No

Why would you do that?

[broken ether]

ohhh

nevermind

i can see now