help-19

[lavish frost]

and I set them equal to eachother

[cold sage]

yup

[lavish frost]

but for v'(t) what do I set t to

[cold sage]

youre solving for t

[lavish frost]

oh

wait so then do I plug in tF = 30 and tI = 0?

for the first one

[cold sage]

yeah

[lavish frost]

huh okk

thanks

[cold sage]

nw

[lavish frost]

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[pearl mist]

Nobody helped me last time but I don't understand distributive property

[pearl mist]

Can someone please help me

I don't understand why how I got the wrong answer for 2, 3, 7, and 8

I got the right answer but I always did the opposite operation

I don't understand why the operation changed

<@&286206848099549185>

[fresh ruin]

the way you're doing it, always have a + in between

[pearl mist]

wdym?

[fresh ruin]

positive/negative issues are pretty annoying you basically have to decide on always doing it one way or another

so like for (2-4)(5)
you do 2*5 + -4*5

[pearl mist]

Oh

Okay

[fresh ruin]

like here, you turned one minus into two

[pearl mist]

Oh

But then how would it become a plus

[fresh ruin]

you just have to pick one of those second row circles to be a +

[pearl mist]

But wouldn't that be changing the equation

Since distributive property is just simplifying

[fresh ruin]

it's sort of a matter of where you put the negative in the variables, the equations are
(a+b)c = ac+bc
(a-b)c=ac-bc

OR (a+ [-b]) = ac + [-b]c

[pearl mist]

okay

I think I might be getting it

Wait

So because I'm using a negative number

Like for example -8

if I'm adding a negative number then that means i should subtract not add?

like for this question

[fresh ruin]

the single number won't change anything

like -8 vs. 8

[fresh ruin]

actually this is pretty confusing, instead I'd just always say that it's 10, not -10

idk you can choose which way

[pearl mist]

I kinda get what you're saying

But what I don't understand is what parts I need to change because of the negative and positive numbers

[fresh ruin]

I'd try to make it so the add or subtract in the parentheses carries over

like (5n MINUS 10)(-9) is 5n(-9) MINUS 10(-9)

[pearl mist]

Okay

Thank you

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[lavish frost]

I can't really figure out what it means in the context of the problem

[lavish frost]

I think the integral from 2 10 m'(t) is just m(10) - m(2) which is -35

but does it mean like the amount of meters moved?

[glacial pagoda]

Isnt it coming -70? (-25-45)

[lavish frost]

I thought because of the integral it changes m'(t) to m(t)_

[glacial pagoda]

Yes it does

[lavish frost]

so then that would be 20-55

[glacial pagoda]

Ohh yeahh, I saw wrong values

My bad

[lavish frost]

so then what does it represent

did they move -35 meters ?

[glacial pagoda]

Yes

But in opposite direction

Of the path of their motion

[lavish frost]

I see

so if positive means moving east that would mean they moved 35 meters west

got it

thanks

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[oblique nest]

Not sure where to start, I know how to find a Jacobian but not sure how it can be used here

[odd edge]

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[oblique nest]

<@&286206848099549185>

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[signal perch]

does this mean theres no VA?

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[void crypt]

hello

[void crypt]

Ok so my answer is wrong but I cant tell why

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[rain hedge]

I don't know if this is the right place, but in a MOSFET during pinch off conditions, why does current still flow?

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[rose mist]

can i ask what is that last number on the right side of the paper

[dense parcel]

Its just 1

Math is so much pain

[rose mist]

which part are you stuck at?

[dense parcel]

Factoring 11+sqrt21

How to factor it

In (a+b)^2

Form

A*b=sqrt21/2

Idk further

[clever fjord]

misha

[rose mist]

we assume $\sqrt{11 + \sqrt{21}}$ can be expressed as $\sqrt{x} + \sqrt{y}$

[clever fjord]

misha

[rose mist]

do you understand this part so far

[dense parcel]

Well it can be sqrt3/sqrt2+sqrt7/sqrt2 whole square but theres extra 6

[rose mist]

what do you mean?

[clever fjord]

misha

[rose mist]

take a look at the right side of the equation

[dense parcel]

Yes

[rose mist]

right now, the left side of the equation looks very complex because it contains a sqare root within a square root (also called a nested square root), but if we compare it with the right side, we now have a hint that the left side might also be expressible in a similar form

thus leading me to the assumption of

[clever fjord]

misha

misha

[rose mist]

square both sides

[dense parcel]

Yes

[clever fjord]

misha

[rose mist]

so far so good?

[dense parcel]

Ye

I get this part

And tried doing it

[rose mist]

now compare the lhs and rhs

[dense parcel]

Sqrtxy=sqrt21/2

[rose mist]

x + y = 11

and

[clever fjord]

misha

[rose mist]

??

ignore that guy

@dense parcel do you understand this part

[rose mist]

yes

and then we just square that again

[clever fjord]

misha

misha

[rose mist]

now if you look closely

[clever fjord]

misha

misha

[rose mist]

now we can finally substitute the values back

[clever fjord]

misha

[rose mist]

x = 21/2

y = 1/2

so a = 21, b = 2, c = 1, d = 2

now we check the gcd conditions

gcd(21,2) = 1

gcd(1,2) = 1

now a + b + c + d = 26

and that is the final answer

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[dense parcel]

I dont get it here

.reopen

[odd edge]

✅

[native tinsel]

It’s sum and product of roots

[dense parcel]

But im supposed to factor it in the form of (a+b)^2

[native tinsel]

wdym by that

[dense parcel]

So it can get out from the surd?

[native tinsel]

You don’t need to get out the surd I think

Looking at your question
We can see that 11+sqrt21 is composed of two parts
One rational and one irrational

[dense parcel]

Yes

[native tinsel]

So we can actually separately solve for x and y

As the two parts are isolated to each other

$11+\sqrt{21} = x+y + 2\sqrt{xy}$

[clever fjord]

Afi

[native tinsel]

Now our object is to solve for x and y

x and y are rational
Giving us 11=x+y and 21=4xy

[dense parcel]

Yes

[native tinsel]

now that we have these
How do we solve for x and y?

[dense parcel]

Square

[native tinsel]

Hmm sure but how does that help us

[dense parcel]

💀

[native tinsel]

One method here is to make use of the sum and product of roots

[dense parcel]

Yes

[native tinsel]

In general, let the equation ax^2 + bx + c = 0 have roots p and q
Then p+q=-b/a and pq=c/a

You know this right?

[dense parcel]

Yes

[native tinsel]

Now we do it backwards

Given x+y=11 and xy=21/4
Can we find a quadratic equation with roots being x and y

[dense parcel]

.close

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[dense parcel]

Ty

[native tinsel]

Np

[full cape]

$ help

[clever fjord]

misha

[rose mist]

the simpler approach would be to recognize that square root of 21 is irrational, and then to compare the terms to form the system of equations

[dense parcel]

@rose mist can you tell me what is E here

[odd edge]

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[mystic saffron]

,rotate

Yo g

,rotate

[mystic saffron]

Part d guys i dont understand

[full cape]

You have to write the value of root 6 in decimals

And then solve it

And the final answer should be in decimals upto 5

[dense parcel]

Lmao allen modules are killing us all

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[nocturne basalt]

If £8,000 is left on deposit for 5 years and growths to £12,000, what was the annual compound interest rate?

[nocturne basalt]

i think this is the formulae for it, but not sure how to approach it

[sand magnet]

Not really

You don't earn the same value each year

[royal ore]

Uh ok

[sand magnet]

You'll get more the fourth year than the first because youll have more to start with year 4

[royal ore]

Ok ok

[solemn umbra]

this formula is quite not clear,
here's the general formula:
V=P(1+r/100)^t, where V=final value, P= initial value, and r=interest rate, and t= time

[solemn umbra]

this is for exponential growth,
and for exponential decay it would be 1-r/100, instead of +

[solemn umbra]

if u insert the values in this formula, and rearrage the whole thing to find r, u should get the answer

[sand magnet]

To understand each year you get your money x 1.somethibg

Let's say interest is 20%

Each year you do money x 1.20

On year 5 you'll have starting money x 1.20 x 1.20 x 1.20 x1.20 x 1.20

Aka starting money x ( 1.20)^5

[nocturne basalt]

yea i see

what i did was doing 12000-8000 divided by 5

then got 10,400

[sand magnet]

In your problem it's not 1.20, you are looking for that value but since starting money and money on year 5 are given you have the equation :
Starting money x (unknown)^5 = year 5 money

[nocturne basalt]

and the answer is 10.4%

so i just lowered the value to a percentage

which is correct

but its not the way to do it

cause i didnt use the formula

[sand magnet]

It's not a mean

For example 20% over 4 years isn't equal to 5% each year

In number 20% over 4 year is starting at 100 and ending at 120

5% per year is 100x1.05 each year aka 100x(1.05)^4

And it's not 120

[nocturne basalt]

yea, how would i do this though?, like rearrange it

[solemn umbra]

oke so the initial equation is V=P(1+r/100)^t right? and we want to know the value of r

so let's divide both side by P first

so we will get V/P = (1+r/100)^t

now we need to move t to the left hand side

[solemn umbra]

how would u do this?

[solemn umbra]

and also this doesnt work because this way is similar to finding simple interest,
in simple interest, the interest u will get over a certain time only depends on your inital deposit, no matter how long

this one, we are doing is compound interest

[sand magnet]

I just explained it above

[solemn umbra]

yes it is as this bro explained

[nocturne basalt]

yea still trying to wrap my head around it tbh lol

like i kind of get it

[sand magnet]

If you have any questions

[solemn umbra]

no worries, it's quite tough to understand these kind of stuffs, i suggest u to watch a youtube video on it if u still dont understand because some of the videos have cool anology and animation and they make it rlly rlly easy to understand

[nocturne basalt]

when you said the "/100" bit what does that mean

[nocturne basalt]

yeah ill try to find ones, our uni doesnt give us any videos and stuff

[solemn umbra]

ok so normally r is shown as percentage right? like 50%, 20% and so on, it's just in this formula the percentage is turned into r/100, for example 50/100, 20/100

[nocturne basalt]

oh yeah i see

[odd edge]

@nocturne basalt Has your question been resolved?

[nocturne basalt]

cheers guys

[odd edge]

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[sick anvil]

im a bit confused on the derivation of exponents
if someone could explain
derivative of a^x is a^xlna correct?
so derivative of e^x is e^xln e which equals e^x
But if instead of ^x its something else
like 2x
e^2x
I need to use chain rule?
so the derivative of e^2x would be
(e^2x ) * 2?
And the derivative of a^2x would be
(a^2x) 2*lna?

[low locust]

yes

[sick anvil]

ok ty

so the power of the

exponent doesnt change

no -1 or whatever

[low locust]

yes

[sick anvil]

ok thank you

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[frosty plover]

i'm trying to prove its homomorphic but I don't think I'm doing it right

[brittle beacon]

You may want to try write the polynomials f + g and f.g in terms of their coefficients

Oh I need different letters

Anyways, note that your f maps from the ring of polynomials with indeterminate x and real coefficients to the ring of real numbers

So you wanna show that $f(p_1 + p_2) = f(p_1) + f(p_2)$ and that $f(p_1 \cdot p_2) = f(p_1) \cdot f(p_2)$ for polynomials $p_1, p_2\in \bR[x]$

[clever fjord]

@brittle beacon

[brittle beacon]

(and you may need to check another property too-)

[frosty plover]

Yeah. The inverse

and I got that part. But trying to group them is where my proof kinda fall apart

[gritty oar]

In theargument of f, you've written p(x) and p(y) but it should be $p(x) = \sum a_nx^n$ and $q(x) = \sum b_mx^m$

[frosty plover]

ah ok

[clever fjord]

chebyshev's infinite pee norm

[gritty oar]

The idea for + is correct though

what would you get if you evaluate f with what's written in the bottom line? That should be the next step after you fix the details

and as for *

[frosty plover]

so I'd use a,b instead of x,y

[gritty oar]

are you familiar with any tools that help you simplify a product of 2 polynomials

a_n are the coefficients for p(x) and b_m are the coefficients for q(x)

[frosty plover]

and is it something to do with the highest degree?

[gritty oar]

Not quite

[frosty plover]

hmm idk. I've already grouped the like terms no?

[gritty oar]

Well you did do it albeit incorrectly

but you should restart by using two correct elements from R[x]

[frosty plover]

k. I'll try that. I've gtg rn tho. Thx for the help!

[odd edge]

@frosty plover Has your question been resolved?

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[wicked fulcrum]

Hello I know the answer and I have the solution steps written down but I’m still so confused how to get this answer can someone help me thank you

[wicked fulcrum]

For question 3

[zinc glacier]

when you have (a+b)/c, you can split it into a/c + b/c

you can then write the two terms as (coefficient)*(power of x) and then use power rule as normal

[wicked fulcrum]

I do understand that I’m just wondering how come the 63 3/4 stays the same but for 12 why do you have to -1 the 3/4

And why does it turn positive

[odd edge]

@wicked fulcrum Has your question been resolved?

[orchid torrent]

$\int \frac{63x-12}{\sqrt[4]{x}} \dd{x}=\int \left(63x^{3/4}-12x^{-1/4} \right) \dd{x}=\frac{63x^{7/4}}{7/4}-\frac{12x^{3/4}}{3/4}+C$

[clever fjord]

Civil Service Pigeon

[orchid torrent]

idrk what you're asking but this is my best attempt to answer it

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[mystic saffron]

hello

find the number z e C, such that |z| = 1 and | ..... | = 1

[odd edge]

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

[mystic saffron]

well

2

[odd edge]

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

[mystic saffron]

i can use that

i dont have any work

its in mind

all ofi t

so a number z e C ,means that it satisfies z bar = - z

correct?

[dense mirage]

no

[mystic saffron]

no?

[dense mirage]

z bar is the conjugate of z

[mystic saffron]

yes

i know that

[dense mirage]

if z=a+bi, z bar = a - bi

[mystic saffron]

i know that

[dense mirage]

the real part doesnt change

oh

then what

[mystic saffron]

ok then i cant use that

[dense mirage]

z bar =/= -z

[mystic saffron]

lets see

[dense mirage]

try solving like you would a normal fractional equation

turn it into one fraction

[dense mirage]

yurr

[orchid torrent]

just use exponential form

[mystic saffron]

its good right

[dense mirage]

for sure

[mystic saffron]

ok hm

[dense mirage]

you can also just use exponential form yeah

[mystic saffron]

oh on down

its

right

this is true?

[dense mirage]

but some courses like keeping it in terms of z and z bar so

[dense mirage]

yeah

[mystic saffron]

good

[dense mirage]

:)

[mystic saffron]

ok i have this

let me think

well down

its just 1

so its

[dense mirage]

mhmm

[mystic saffron]

ok now

hm

do i take

the cases

oh

i cant do that

[dense mirage]

what cases?

[mystic saffron]

forget about it

i forgot i am working in Complex

[dense mirage]

ah

yeah that happens lol

[mystic saffron]

do i just take numbers

like

idk

i doesnt work

[dense mirage]

yeah id convert to a+bi or exponential form

[mystic saffron]

ok sure

correct so far?

oh its not

its b^2

that's more like it

and this has like not really a pretty solution

what do i do

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

<@&286206848099549185>

.close

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[mystic saffron]

another silly linear algebra question

[mystic saffron]

i dont even see my answer

[wanton bison]

how'd you get last row?

i see

[mystic saffron]

yeah

[wanton bison]

technicall you dont need these operations

We can read x_5 = 2 and thus x_4 = 5

[mystic saffron]

my glasses are always dirty

[wanton bison]

x_1 -2x_3 + 2 = 5 -> x_1 = 2x_3 + 3

[wanton bison]

then dont wear them

[mystic saffron]

OH SO I SHOULDNT Rrow reduce

RREF

row reduce echelon

[wanton bison]

you can

i just thought you dont need

[mystic saffron]

oh

[wanton bison]

2x_3 + 3
x_2
x_3
5
2
I get the same as you

[rose mist]

E is just the name of the form or function

(sorry for typing in here just ignore me)

[wanton bison]

it's just the system, does it really depend on x_2

I think I know what they did in b)

[summer hollow]

for the last column, you should pick an arbitrary value for x1 and x3 like x1=5, x3=0 and get one possible solution

[mystic saffron]

Oh

[summer hollow]

(3,0,0..) doesnt work since x4 x5 are fixed

[mystic saffron]

huh

isnt it just the particular solution to the homogenous thingy

[summer hollow]

any solution has to satisfy the equations x1 = 2x3 + 3 etc right, so the initial solution vector u have to find values that u can plug into those equations which work

[mystic saffron]

Yess

Ohh and I can’t even plug anything into x4 x5

[summer hollow]

yeah they stay fixed

[mystic saffron]

So don’t row reduce?

Or how do I do the problem

how do I know what to set it to

Oh 0

3,0,0,5,2

Would be the solution?

[wanton bison]

Now that I see how did you get what you did

If x_4 = 5 and x_5 = 2

[mystic saffron]

Hhuhuuhuhhu

[wanton bison]

you forgot them

[mystic saffron]

I did?

Oops

[wanton bison]

Your vector solution has this form

[mystic saffron]

I haven’t done this in a while

[clever fjord]

𝔸dωn𝓲²s

[mystic saffron]

You take my final

[wanton bison]

which you can rewrite

[clever fjord]

𝔸dωn𝓲²s

[wanton bison]

But in (b) it's somehow 5 0 0 5 2

which I don't get

[mystic saffron]

YEAH

YOU SHOULD TALK TO MY TEACHER

[wanton bison]

and in (c) they only considered x_3 the free variable

[wanton bison]

I kinda hope we have a mistake somewhere

[mystic saffron]

How can I make a mistake in two moves

[wanton bison]

I can't find something else honestly

[summer hollow]

tbh if you plug in the initial solution values they give in b) it doesnt work for the top equation

5 - 2(0) + 2 = 7

[wanton bison]

there you go then farm girl

[mystic saffron]

What should I call you

[wanton bison]

jazz is mean

[mystic saffron]

Always gaming???

[wanton bison]

lmaooo

I game infrequently

[mystic saffron]

LOLOLOLOL

[mystic saffron]

Yeah my professor makes mistakes . At least 1 every class

[summer hollow]

bruh

[mystic saffron]

You saved me

.solved

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[radiant chasm]

Weird question. When integrating, what exactly is your thought process. Is there an order with the way you look at things (eg first you try to algebraicly simplify, then if not that integration by parts etc.)

Please, if possible do keep in mind this is with reference to a high school/college freshman level of student.

[vapid temple]

it's mostly pattern recognition from doing lots of problems

[radiant chasm]

yeah I do get that, but with someone who doesnt have that experience, what 'alogrithm' would you reccomend.

sorry if its a weird question

I have a final math exam in 4 hours and I have just gone through the concepts 😭

[vapid temple]

i would look at if u-sub would work first

then integration by parts if it's not an option and you see that two differents types of functions are being multiplied

logarithmic, inverse trig, algebraic, trig, exponential

etc.

then if i see a square root in the denominator i'll think trigonometric substitution

[radiant chasm]

Ok thats very helpful

thank you very much

I just need a structured way of thinking about itnegrals

[vapid temple]

if nothing seems to work, partial fractions or simplify things (complete the square, multiply by conjugate)

[radiant chasm]

since I havent praticed

[vapid temple]

yeah it's not gonna be an easy task

[radiant chasm]

I did get a 5 in calc BC

[vapid temple]

integration really is about practice

[radiant chasm]

last year

but I havent touched integration

since

aanyways tysm

[vapid temple]

i would just do like 2-3 integrals of each type

if yo u have the time

[radiant chasm]

unfortunately I dont

[vapid temple]

just to be somewhat familiar

[radiant chasm]

I have to do differential equations

and mclaurin series too

😭

[vapid temple]

oof

calc2 really is a pain

[radiant chasm]

yeah its a tough time

I dont have calc 2 though

[vapid temple]

we'll get through it

ah

[radiant chasm]

I take the IB math aa hl course

[vapid temple]

idk what that is lol

[radiant chasm]

ong

[radiant chasm]

if you ever have children

[vapid temple]

you do series and integrals in calc2 tho

[radiant chasm]

DO NOT MAKE THEM DO IB

[radiant chasm]

yeah

anyways thank you

I have wasted too muhc time

[vapid temple]

no problem!

[radiant chasm]

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[still forge]

how did they make a conclusion of 500?

[thin violet]

'x hundred units' is expressed in units of hundreds

[odd edge]

@still forge Has your question been resolved?

[still forge]

thx

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[amber fable]

I was hoping to get some insights on this problem. Ty.

[odd edge]

@amber fable Has your question been resolved?

[odd edge]

@amber fable Has your question been resolved?

[odd edge]

@amber fable Has your question been resolved?

[amber fable]

...

[versed siren]

Nah I'd win

What's the prob

[odd edge]

@amber fable Has your question been resolved?

[hybrid gyro]

Nah

I'd win

[amber fable]

Yess. Sorry for the late response.

[amber fable]

The problem is given as above. We are given a n sided polygon and we have to find the number of r sided polygons formed such that no side is coincident with that of the original polygon.

[amber fable]

...

[mystic saffron]

you aren't winning

[amber fable]

I will win. For sure.

Well, I think its time to end this. Thank you to all who took a attempt at the problem. Next time.

.close

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[odd edge]

@mystic saffron Has your question been resolved?

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[bitter cliff]

I got 0, pi and pi/2, 3pi/2

but I guess those last two aren't answers

I'm not sure how though

[tender bloom]

0 is the trivial solution

sin^2 pi is = 0

[bitter cliff]

someone in class told me something about pi/2 and 3pi/2 don't work with tangent so that is why they aren't answers. is this a real thing?

[gray dome]

idt the pi/2 soln works

yeah

because tan = sin/cos

sin(pi/2) = 1

cos(pi/2) = 0

and u divide by 0

so its undefined

If you look at the graph tan(pi/2) is undefined

it has an asymptote

@bitter cliff

[bitter cliff]

so do I have to apply this rule only if tanget is in the problem

[gray dome]

what rule ru referring to?

its just u cant divide by 0

big no no in math

[tender bloom]

tan x sin^2 x - tan x = 0

[bitter cliff]

well not rule, but whenever tan is in the problem I have to think about it

[gray dome]

yes

[bitter cliff]

ok that makes sense

[gray dome]

what is the range of values for x

[tender bloom]

you will get tanx (sin^2x-1) = 0

[bitter cliff]

I don't know aren't those the answer or are you talking about having a +2npi

[bitter cliff]

that's what I did

[gray dome]

well u need to have a general form

however...

u can also get the eqn

gimme 1 sec

$\frac{-1}{2sin(2x)}$

[clever fjord]

UsingApp

[bitter cliff]

what is the +2npi for then doesn't that mean the full range

[tender bloom]

pause, what

after you figure out that you can reduce out tan x for all x diff pi, you get sin^2 x - 1 = 0

just solve that

[bitter cliff]

yeah but I think that gives you pi/2 and 3pi/2 but those I guess aren't answers because they don't work with tan

I have to go eah dinner thanks for the help

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[normal snow]

i asked this in another channel but i went afk and it closed

[normal snow]

is the reflection in x or y

[mystic saffron]

what do you mean by reflection

[normal snow]

transformation

[mystic saffron]

?

[normal snow]

uh?

does it have another name?

[elfin zodiac]

Reflected over y-axis

And translated to the left 2 units

[normal snow]

ah thanks

.close

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[plush panther]

$\frac{d^2y}{dt^2}-\frac{4}{t}\frac{dy}{dt}+\frac{6}{t^2}y=t, y(1)=0, \frac{dy}{dt}(1)=1$

[clever fjord]

leo~

[plush panther]

what method would you use to solve this diff eq?

[odd edge]

@plush panther Has your question been resolved?

[plush panther]

i hope i can do this without a laplace transform?

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[hexed sparrow]

i used the parametric formula for the arch legnth and i ended up with s(t)=1/2 t^2 +t

i dont really know how to sub part b into this bc my eqn has one parameter and the coords they are giving me have 2

could someone please check my number 1 and show me how to do 2.

please@ me

<@&286206848099549185>

[odd edge]

@hexed sparrow Has your question been resolved?

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@hexed sparrow Has your question been resolved?

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@hexed sparrow Has your question been resolved?

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@hexed sparrow Has your question been resolved?

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[pine drift]

Hey sorry I have a new question lol um is this not correct?

[pine drift]

8 has an invisible exponent of 1 right

And i can’t bring anything out of the square root bc 8 doesn’t have 4 2s. It only has 3 2s

[plain lagoon]

don't forget that 8 is also a number raised to the power of another number

[pine drift]

Wdym

That number is 1 right?

[plain lagoon]

yes, that is 8's exponent

[pine drift]

Ok!

[plain lagoon]

But I don' tthink the second option is correct

_ _:)

[pine drift]

Okay

Im not sure what to do

[plain lagoon]

okay, so.. a hint

[clever fjord]

Fω 2.718ℓ

[pine drift]

Yeah 8^1

Idk what to do next haha

[snow ivy]

factorize it more

[plain lagoon]

^^^

[pine drift]

2 2 2

[plain lagoon]

you don't stop at 8^1

[plain lagoon]

yesss

[pine drift]

I can’t bring anything out tho

The root thing is 4

And I only have 3 twos

[plain lagoon]

yeah you don't

[plain lagoon]

so 8 = ?

other than 8^1

[pine drift]

2 * 2 * 2

[plain lagoon]

another way to write that?

[pine drift]

?

[snow ivy]

which is equal to?

[pine drift]

Um 8?

[snow ivy]

write in term of exponent

[plain lagoon]

^^^

[pine drift]

Idk guys

They didn’t teach us

[plain lagoon]

okay maybe lets start from the beginning

okay so... what is 3^2 ?

[pine drift]

3 times 3 = 9

Oh

Oh

[plain lagoon]

_ _:D

[pine drift]

2^3

Right

[plain lagoon]

yeah

[pine drift]

Okkk

Now I have this

Right

[plain lagoon]

do the same thing you did here

[pine drift]

Ok! Answer is D?

[snow ivy]

yes

keep doing maths

and free this channel if your doubt has been resolved

[pine drift]

.close

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[soft vessel]

hey how do i find the centre and radius for this circle?

[late flint]

complete the square maybe

[soft vessel]

yeah thats what its saying to do im just not sure how

like what is a b and c

assuming -16 is c

[snow ivy]

you need to cram the general equation of circle

[soft vessel]

like rearrange

cram?

[snow ivy]

keep in mind

[onyx cloak]

general equation of circle is x^2 + y^2 + 2gx + 2fy + c = 0

compare your equation to this

[late flint]

this does not directly lead to an equation for y in the form (y-...)^2, so you add and subtract a term of (b / (2*a) )^2 (completing the square)

[onyx cloak]

and find center and radius

[soft vessel]

oh i didnt know that i thought it was (x - h)^2 + (y - k)^2 = r^2

[onyx cloak]

i mean that is too

expand and you'll reach the equation i wrote above

[snow ivy]

this is standard equation of circle

[late flint]

your b is -6, and a is 1 due to y^2 - 6y

so to complete the square you add and subtract 9

y^2 - 6y + 9 - 9 = (y-3)^2 - 9

[soft vessel]

how do i find it in that equation

im confused two things are being said

[snow ivy]

cram

[late flint]

leaving you with $x^2 + (y-3)^2 - 9 - 16 = 0$

[clever fjord]

Triaxyz

[onyx cloak]

go on with what you know

[soft vessel]

what does that mean

[onyx cloak]

(x - h)^2 + (y - k)^2 = r^2

this one

[soft vessel]

okay

[onyx cloak]

transform your eqution into this

[soft vessel]

ah okay im getting it

[snow ivy]

i ll text when your doubt has been solved

else you ll get confused

[soft vessel]

is it y - 3 or y + 3

i got x^2 + (y + 3)^2 -27 = 0

[soft vessel]

^

[late flint]

it's (y-3)^2

[soft vessel]

how?

ok wait i see

[late flint]

by adding and subtracting 9, you can group y^2 - 6 y + 9 to (y-3)^2

[soft vessel]

i was mistakig -6y for + 6y

[late flint]

also -9-16 is -25 not -27

resulting in a really nice radius and center

[odd edge]

@soft vessel Has your question been resolved?

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[rain palm]

given that points (-4,2) , (3,4) and (p,7) lie on the same straight line,
find the value of p

[onyx cloak]

points on straight line are collinear

so u can use determinant

or use slope of straight line

since all points on are a line

slopes written using them should be same

[rain palm]

wdym

their gradient is the same?

[elfin zodiac]

Do you know what is the slope?

[rain palm]

idk

[elfin zodiac]

Are you studying in english?

[onyx cloak]

yes (the magnitude)

[rain palm]

yes

[onyx cloak]

maybe they're from UK

[elfin zodiac]

Didnt know in uk slope is not used

[rain palm]

so

how do we proceed?

find out the gradient?

[onyx cloak]

u have 3 points

find gradient using 2 at a time

then equate

since they're in straight line

so slope must be same

[elfin zodiac]

Use the 2 first to find it

Then use the result to find p

[rain palm]

m = 2/7

what next? do we substitue or something?

[onyx cloak]

write gradient using (p,7) and (3,4)

and that's equal to 2/7

[rain palm]

so
m = 4 - 7
-----
3 - p
= 2/7
is this what you mean

[onyx cloak]

yes

[elfin zodiac]

You have to do exactly what you did to find the gradient for the first two points, but now you already have the gradient (2/7) and you have to find p using the second and third point

[rain palm]

i don't know what to do
I've tried changing places for y2 - y1/x2 - x1
and got -3/3-p or 3/p-3

[elfin zodiac]

That is equal to 2/7

Solve the equation

[rain palm]

sorry but how exaclty 😅

[elfin zodiac]

What do you mean

3/(p-3) = 2/7

You don’t know how to solve equations?

[rain palm]

i know

but what to multiply to cancel out the denominators?

[elfin zodiac]

If you had 1/a what would you multiply by?

[rain palm]

by a

[elfin zodiac]

So what do you have now

[rain palm]

1

[elfin zodiac]

?

Sorry bad question

What do you have in your exercise

[rain palm]

so multiply by 21-7p?

[elfin zodiac]

Where do you have that?

3/(p-3)

What is in the denominator

[rain palm]

by getting the lcm of 3/p-3 = 2/7

[elfin zodiac]

You can do all in once if u want, but i prefer u do first one side and then the other

But if u can do both

Do both

Ok so what do u get

[rain palm]

21 = 2p - 6

[elfin zodiac]

Good, so find p now

[rain palm]

27/2

thank you for your time and patience

have a nice day

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[dull junco]

Isnt there supposed to be the value of angles written on the pie chart?

[odd edge]

@dull junco Has your question been resolved?

[dull junco]

Helpppp

[odd edge]

@dull junco Has your question been resolved?

[dull junco]

Oh

.close

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[cobalt viper]

How can I do this?

[glass wedge]

which part specifically is bugging you, because they've written exactly the steps you should take

[lusty nova]

In order to show f'(x)>0 you want to show that f(x+h)-f(x)/h is always positive

[cobalt viper]

okay I'll try that

[glass wedge]

Or just say the derivative of x³ is 3x² and x² is a strictly non negative function

[lusty nova]

yea this is easier, but only applies to f(x) =x^3

[glass wedge]

You're doing the same thing except finding the derivative using the definition

[lusty nova]

if you are asked to prove the general case use this else dont bother

[cobalt viper]

what is b?

[glass wedge]

Lol

[lusty nova]

b > 0

[cobalt viper]

yeah but what is that representing in terms of this question

[lusty nova]

b represent any number greater than zero. You interest is not what b is but to show that f(b)>f(0) when b>0

[cobalt viper]

what does strictly increasing/decreasing mean? sorry this is my first day doing calc

[lusty nova]

strictly increasibg means f(x) > f(y) when ever x>y

for example 2^3 > 1^3 because 2 > 1

[upbeat anchor]

If the proceeding y-value is greater than the previous y-value than it is strictly increasing. Vice versa for strictly decreasing

[upbeat anchor]

don't explain it like this, it's confusing for beginners

[lusty nova]

right

[cobalt viper]

oh ok thanks

so

I'd just derive it

and

bro im lost asf 💀

[lusty nova]

f'(x)=3x^2

[cobalt viper]

how can I just show its a strictly increasing function like its obvious

a cubic graph is incrceasing

the derivitive 3x^2 is also above the x axis

but how can I word that

[lusty nova]

take x > y, then cube both sides

[cobalt viper]

this is the answer

can you help me process this information

[lusty nova]

sure

you want to show x^2 > y^2 using the fact that x> y

[cobalt viper]

okay

and how can I do that

[lusty nova]

if b>0,
f'(0)=0
therefore
f'(b)=3b^2 > 0 = f'(0)

this shows f'(b)> f'(0)

here i used the fact that b>0

[cobalt viper]

ohh

ur right

what about for this?

[lusty nova]

use the fact that f(x)=x^3 is increasing

when x>y , x^3 > y^3
this implies
-x^3 < - y^3(because minus changes greater than to less than)

[cobalt viper]

oh aight

appreciate it

.close

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[crystal lodge]

have the following question on a practice exam, would appreciate a sanity check/any better proofs on the following

[crystal lodge]

If i have sets A and L then $|A^L|$ represents how many different possible functions from L to A there are. Every element in $L$ could be mapped to any of the $|A|$ elements, so there are $\overbrace{|A| \times |A| \times... \times |A|}^{|L| \text{times}}$ possible functions, or $|A^L| = |A|^{|L|}$ functions. Then, as $|A| = |B|$ and $|L| = |M|$, and $ |A^L| = |B^M| \leftrightarrow |A|^{|L|} = |B|^{|M|}$, this is always true.

[clever fjord]

ChiliLion

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@crystal lodge Has your question been resolved?

[crystal lodge]

<@&286206848099549185>

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[steel citrus]

A rectangular cardboard with dimensions of 9.0 cm x 18.0 cm is folded from the ends into an open box according to the pictures below. Create a function that models the volume of the bag, whose variable is the length x. Also specify the function definition condition.

[steel citrus]

I don't know what to do...... i did 2x+2y=18 ==> x+y=9 ==>y=-x+9 ==> v(x)=x(-x+9) ... the answer should be -9x^+81x but idk what to do

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@steel citrus Has your question been resolved?

[orchid torrent]

You forgot this dimension

[steel citrus]

so i need to do 9*x(-x+9)?

i feel dumb

thanks lol

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[cold swift]

do you know the double angle formula?

[thick pine]

yupp!

but thats sin3x

and cos 3x :((

[cold swift]

so we essentially we have to invert the double angle formula

[thick pine]

huhuh

[clever fjord]

y0shi

[cold swift]

see how the last part there resembles the double angle

[thick pine]

oop ya

wait but then if i use double angle

wont it be 1/2*3?

im confufu

[clever fjord]

y0shi

[cold swift]

since we can replace that x in the double angle formula with 3x

so that we would get this

[thick pine]

you're a genius

15 * sin6x

:0:0:0:0

[cold swift]

yep that’s it

[thick pine]

ok now finding period

is there a formula for that? or do i have to graph and count?

[cold swift]

usually for a sine or a cosine graph in the form: $sin(bx), cos(bx)$, the period would just be $\frac{2\pi}{b}$

[clever fjord]

y0shi

[cold swift]

graphing would do just fine as well

[thick pine]

and b is..?

[cold swift]

sin(bx) or cos(bx)

[thick pine]

where can i learn like how to calculate all the things in a trigo func

[cold swift]

the coefficient

[thick pine]

oo

but like wb the other letters?

[cold swift]

you can probably fine this on organic chem

[thick pine]

where can i find like a key

[thick pine]

HUH

[cold swift]

organic chem tutor

[thick pine]

is this all i need to know?

anything else?

[cold swift]

yeah basically

and also the thing i said above

with the period

but make sure that the B

[clever fjord]

y0shi

[cold swift]

you would have to factor it out

[thick pine]

oh okay

i'll make notes

so

f(x) = a (trig) (b [ x-c ] ) + d
a = amplitude
trig = function
b =
c =
d = vertical shift

what is phase shift?

[cold swift]

c is the horizontal shift

and b is either called the vertical compression or horizontal compression i don’t remember

you can search that up if you like

but just know that we can use b to get the period

[thick pine]

NICE got it

wait what is b in the function then

[cold swift]

it would be 6

since there is no horizontal shift

[thick pine]

oh shoot obv

[cold swift]

it’ll just be the coefficient

[thick pine]

ok so just 2pi/6, wb part c?

15 sin 6x

[cold swift]

so we have to consider $15sin(6x)=0$

[clever fjord]

y0shi

[cold swift]

yep

we can divide 15 on both sides to just have

[thick pine]

okay soo

sin 6x

[cold swift]

$sin(6x)=0$

[clever fjord]

y0shi

[cold swift]

yep

and now we have to consider

what input would make the sine function 0?

[thick pine]

wait what then

[cold swift]

recall the unit circle

or the sine function

[thick pine]

y messages have delayy

UM, cos 90-x? haha i dont know

[cold swift]

for what values of x does the sine function give us 0?

[thick pine]

oh oops

[thick pine]

sin 0 gives 0-?

does that work

[cold swift]

yep that’s one of them

[thick pine]

one of them>'

[cold swift]

there’s one more tho

[thick pine]

2pi?

[cold swift]

considering everything from 0 to 2pi

[thick pine]

360/6?

[cold swift]

mm yep

one more

[thick pine]

YAYYY

what

4pi-?

[cold swift]

so x=0, 2pi and

that’s not wrong but we want to consider for x values between 0 and 2pi

[thick pine]

oh right totally forgot

i thought thats it what else?

iwe cant do pi can we?

[cold swift]

pi is correct