help-17

[oak magnet]

Well yes

[snow schooner]

So Mr jay was wrong on y ou tube

[atomic jasper]

you find a=4

[snow schooner]

Yes

[atomic jasper]

now, lets replace that = with a > again

and we get a>4

[oak magnet]

It will be the same process for inequality but with specific rule

[snow schooner]

So I was right

?

[vocal sleet]

@snow schooner Has your question been resolved?

[vocal sleet]

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[ember bronze]

A pyramid with a square base has its vertex in the plane with the equation 𝑧 = 3 Three of the vertices of the base are points in the 𝑂 𝑋 𝑌 OXY plane:A=(1,0,0), B=(1,1,0), and C=(0,1,0).

[ember bronze]

c) If the vertex of the pyramid is the point 𝑉 = ( 𝑎 , 𝑏 , 3 ) , what is the equation of the line that contains the height over the base?

.close

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[agile mica]

Im having trouble understanding how the radical simplifies to what it is below

[vocal sleet]

@agile mica Has your question been resolved?

[agile mica]

Im getting a 2 in the numerator instead of a 1

nvm i think i just made some dumb mistake

.close

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[prisma badger]

can someone walk me through this

[halcyon imp]

Are you supposed to prove LH equals RH ?

[prisma badger]

no solve for x

[halcyon imp]

Oh gotcha, first question. Are you working with only real values?

or also complex

[prisma badger]

wdym?

i dont know what that is so i assume real

[hazy spindle]

You can literally replace the number one on the right side with log7(7), then just have the arguments from left and right be equal

[prisma badger]

oh

but whats the other way

[halcyon imp]

Ok good. First off, the restriction should really be x > 2 Because the log can't take on any negative values as well, not just 0

[hazy spindle]

Have (x-2)(x+4) equal to 7

That is the only time when log7(y)=1

[prisma badger]

mk

[halcyon imp]

Not sure what that means

[prisma badger]

ok

[hazy spindle]

This is the solution btw

[hazy spindle]

this is the explanaition

[prisma badger]

yeah but is there another way

[hazy spindle]

I provided you with 2 ways 💀

[prisma badger]

because what if it dosnt equal 1

[halcyon imp]

You could also raise both sides as exponents with base 7, then the log should cancel out and you get the same result

[hazy spindle]

As long as you know how logarithms work, you should be fine. If it equals 2, then with, log7(y), the y would have to be equal to 49

[prisma badger]

now what if its like 58

[hazy spindle]

That would be quite weird, but you could just have log7(those arguments you've got there) = log7(7^58), then just solve for the arguments

But that is not a log that you'd typically get asked to solve

[prisma badger]

ok

also

did i do this right

[hazy spindle]

It is quite messy, could you provide just the intervals for which it holds true?

[prisma badger]

x<-1/2 and x>2

also i checled the answer for this and it was 3 but i also got -5

[hazy spindle]

So you are saying that for x smaller than -1/2 it holds true?

[prisma badger]

the ones that are positive should be true because its > 0

[prisma badger]

yes

[hazy spindle]

Try plugging in -1 then. You are right with x>2, but not the other one.

[prisma badger]

they are both negitive

which equals positive

[hazy spindle]

????

Plug it into the inequation

[prisma badger]

(4(-1) + 2) = -2

-1 - 2 = -3

wait

[hazy spindle]

Into the original one

[prisma badger]

but i shouldnt need to

[hazy spindle]

Aight, do as you think, gl

[prisma badger]

wait a sec

how do i get the right answer then??

[vocal sleet]

@prisma badger Has your question been resolved?

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[heavy palm]

Am i on the correct path so far?

[heavy palm]

trig substitution

[silk osprey]

x = atan(x)?

use a different variable

[heavy palm]

yes

[silk osprey]

like u

or theta

[heavy palm]

ok

[silk osprey]

x = atan(u)

[heavy palm]

ok

👍

[heavy palm]

i cant do that?

uhh

[silk osprey]

the a^2 is being raised to the power of 3/2

since it’s in the parentheses

so it becomes a^3

(ab)^c = a^c * b^c

[heavy palm]

ok

[silk osprey]

$\int_0^a \frac{a\sec^2 u}{a^3\sec^3 u} \dd{u}$

[twin meteor]

knief

[silk osprey]

$\int_0^a \frac{1}{a^2\sec u} \dd{u}$

[twin meteor]

knief

[silk osprey]

do you see how i got that

[heavy palm]

i am confused on the bottom part of this image only, how is a^3/2 become a^3?

[silk osprey]

$(a^2\sec^2 u)^{\frac{3}{2}} = (a^2)^{3/2} (\sec^2 u)^{3/2}$

[twin meteor]

knief

[silk osprey]

multiply the powers

[heavy palm]

ah

sorry

[silk osprey]

no worries

[heavy palm]

and because of the condition that a>0 that means i make a abs right? |a|

[silk osprey]

where?

[heavy palm]

https://cdn.discordapp.com/attachments/903463353417072641/1331773359507116103/image.png?ex=6792d618&is=67918498&hm=3ed0e202ef4e77aa0623a03f902bf7eeabdb0bf0576f78bf3a865619c37b469e&

[silk osprey]

but where do you take |a|

oh wait i forgot to change the bounds lol

[heavy palm]

well just replace a with |a| no

[silk osprey]

x goes from 0 to a

honestly i never change the bounds i just find the anti derivative then convert back to x

[silk osprey]

ignore the bounds here ^

[silk osprey]

well a = |a|?

since a > 0

doesn’t make a difference

[heavy palm]

yeah

i see that now

[silk osprey]

we can just integrate now

[heavy palm]

and bunds just plug in a into x?

to change th ebounds

of what i made x right like

i made x = atan(theta or u)

[silk osprey]

$\int \frac{\cos u}{a^2} \dd{u}$

you can change the bounds if you’d like

i just find it to be more work usually

[heavy palm]

would it make a difference?

[silk osprey]

essentially no

but i guess it’s personal preference

i just made it a habit to find the anti derivative first because i’d get carried away doing that instead of caring about changing the bounds

[heavy palm]

also this should be 1/a^2(secu)

oh nvm

you already accounted

[twin meteor]

knief

[silk osprey]

no you were right

[heavy palm]

mbmb

[silk osprey]

i just saw the sec^2 up top and in my head saw a^2 as well

[heavy palm]

did you struggle with calc2

at all

?

just curious

[silk osprey]

no

[heavy palm]

okay

[silk osprey]

i took ap calc

i’m just tired lol

it’s easy to make mistakes in these for sure though

[heavy palm]

last question if you dont mind

a is a constant here correct

[silk osprey]

yes

[heavy palm]

no worries

[silk osprey]

$\frac{\sin u}{a^2}$

[twin meteor]

knief

[silk osprey]

then in trig sub you’ll usually have to draw a triangle

and relate sin u to x

x = atan(u) so tan(u) = x/a

[heavy palm]

is this different

[heavy palm]

so i dont just apply what i have in bounds?

i have to draw a triangle first

[silk osprey]

no i mean you definitely can here if you converted the bounds correctly but it’s never a bad idea to remember how to do this for expressing the anti derivative in terms of x

not all questions will be definite integrals

[heavy palm]

Yeah

For indefinite I assume I’ll have to do triangles

[silk osprey]

but if you’re comfortable with it then by all means

[silk osprey]

yea

[heavy palm]

So the only exceptions to where I don’t need a triangle

Would be definite integrals?

Just wanna clear it up

[silk osprey]

yes

[heavy palm]

Ok thank you

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[mighty stone]

help?

[velvet karma]

first step would be to eliminate the denominators

can you think of how we might do that

[vocal sleet]

@mighty stone Has your question been resolved?

[quiet echo]

what is the greated common multiple of (x+5) and x?

[mighty stone]

multiply everyside by (x)(x+5) ?

[velvet karma]

exactly!

[mighty stone]

so 8x=-3x+15=5x^2+25x?

[velvet karma]

im assuming u mean 8x - 3x not 8x =, if thats the case then you're right

except for a small error, it should be - 15 not + 15

[mighty stone]

wait why -15?

oh i see

[velvet karma]

distribute the negative

[mighty stone]

okay so i think i get it now

[velvet karma]

it becomes a -

[mighty stone]

thanks

[velvet karma]

np!

[mighty stone]

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[arctic bison]

Could someone tell me about this method, way of doing derivatives please

[arctic bison]

This is how i did. But this method seems easier

Maybe just tell me the formula or something, i will youtube the video on a person doing an example

(Sorry if noob question)

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@arctic bison Has your question been resolved?

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@arctic bison Has your question been resolved?

[vocal sleet]

@arctic bison Has your question been resolved?

[night zodiac]

I have no idea what you did to arrive at your second step. But if you have reasoning and proper justification, then yes, that is both easier and quicker.

Generally, most people would use a combination of the product rule and the quotient rule to solve this.

And that is what I did as well.

Doing the method you used is most likely not going to be allowed on an exam though; I can't pinpoint what rule—if there even is one—you pulled from to reach such a conclusion! Would you care to elaborate on your workings?

[vocal sleet]

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[crystal python]

how do i do this

[mental falcon]

what are the options? what are your thoughts?

[crystal python]

im not sure how to even approach this

the answer options are <0 =0 >0

[torpid olive]

what do u think is at f(-1)

[crystal python]

4

[mental falcon]

well you have the graph of f, so what do you think about its value at x = -1

if you are in a calculus class and you cant answer this one then you are in trouble

[mental falcon]

so is that >0 <0, = 0 ?

[crystal python]

0

[mental falcon]

so what was unclear about that one?

[crystal python]

but what about f'(-1) how would i get that

[mental falcon]

what does the derivative of the function represent geometrically?

[crystal python]

slope?

[mental falcon]

yeah slope \ whether the function is increasing or decreasing

so what is happening with that at x = -1?

[crystal python]

slope is increasing

i think

[mental falcon]

yeah, the slope seems positive

to be careful lets say the slope is positive, not that the slope is increasing because that sort of seems to imply youre talking about how the slope is changing not what it actually is

[crystal python]

okay

so since its increasing would we make that as >0

[mental falcon]

yep

positive*

[crystal python]

positve

yes

so what about f''

[mental falcon]

what do we know about a point of inflection and its definition?

since we have one there

[crystal python]

im not sure

[mental falcon]

ok so look up the definition of point of inflection in your book \ wherever you have definitions and see if that helps you answer

[crystal python]

so its where the concavity changes

a point on where it changes

[mental falcon]

yes, and what's the relationship between the second derivative and concavity?

and do you understand concavity visually, looking at the graph of a function?

[crystal python]

the second derivative determines the concavity

[mental falcon]

yep

so when a function is concave up the 2nd derivative is positive, when its concave down the 2nd derivative is negative

[crystal python]

ok well its also concave up right

[mental falcon]

looking at the graph of f, what is the concavity for example on the interval (-9, -4)? what about (-4, -1) ?

[crystal python]

(-9,-4) would be concave down and (-4,-1) would be up i think?

[mental falcon]

yeah exactly, we can just tell based on which direction the graph is "curving towards"

ok and at a point of inflection, what would we know about the second derivative?

[crystal python]

yes so for f'' (-1) it would be >0 then

[mental falcon]

hang on, what do we know about points of inflection and the 2nd derivative value there?

what IS a point of inflection?

[crystal python]

it is where it changes concavity

so that would mean its tagent line is 0

would that make it 0

[mental falcon]

no, it doesnt have to do with the tangent line

[crystal python]

dang

[mental falcon]

there can be a point of inflection with a non flat tangent line

[crystal python]

im lost man

[mental falcon]

so concavity doesnt directly describe anything about the tangent lines

to keep things simple

2nd derivative has to do with concavity

1st derivative has to do with slope of tangent lines

right?

so how come we brought up tangent lines when we were talking about concavity \ 2nd derivative?

so for example this graph, wouldn't you agree it has a point of inflection at 0? it goes from concave down to concave up? but it never has a horizontal tangent

[mental falcon]

the 2nd derivative is 0 yes, but im just saying that doesn't have to do with the tangent line

@crystal python are you still here??

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@crystal python Has your question been resolved?

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[jolly glacier]

idk how to approach or do this

i know from 0 to .25 you make it a and b

but how do i integrate

[unborn wind]

u can do substituition

or by parts

[jolly glacier]

can you walk me through it

for sub

[unborn wind]

u=(1-x)^1/2

[jolly glacier]

ohkay

and what about x^3

[mental egret]

what would x be in terms of u if you substitute this?

[unborn wind]

1-u^2

[jolly glacier]

-(u^2-1)

yes

[unborn wind]

now do cube

[jolly glacier]

ohhh

wait but i have a question

why is it u=(1-x)^1/2

[unborn wind]

because it makes integration simpler

[jolly glacier]

but how do you js know that

[unborn wind]

intuition

[jolly glacier]

bro 😭

[languid berry]

experience

theres patterns

just do more problems and youll start to get the intuition for it

[jolly glacier]

my quiz is tmr im cooked

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@jolly glacier Has your question been resolved?

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[primal hedge]

im kinda confused on what the right answer is for this

[edgy sapphire]

either look for the one that uses $7 correctly or see which one actually gives a cost

[primal hedge]

i dont understand what 7 is

c?

[edgy sapphire]

7 is the product price

any cost would look like C(...)

[primal hedge]

so is it C(7)?

[languid berry]

C(x) outputs a cost

[primal hedge]

is it the 3rd answer

dang i really dont understand then

is it supposed to =7?

oh its the first answer

is f(x) 1?

[deft oar]

No

F(x) can’t be a constant

What’s the relationship between h and g

[primal hedge]

im not really sure, they both share x?

[deft oar]

Let’s abstract this a bit

g(x) = symbol

How does that symbol appear on h(x)

Actually that’s a bad way to understand this

Let me try again

g(x) = T , f(g(x)) = h(x) , h(x) = 1/T, what is f doing to g(x)

[primal hedge]

oh its dividing it?

[deft oar]

Therefore f must be

[primal hedge]

im not sure

what makes something divide

[deft oar]

Think of a function, that’s all about dividing

A function that looks like h(x)

[languid berry]

can you write h(x) in terms of g(x)?

replace the function with just g(x)

[primal hedge]

i dont understand what you are referring to

replace what function with g(x)

[languid berry]

g(x) = x-9

h(x) = 1/(x-9)

what can you replace here

in h(x)

[primal hedge]

divided by g(x)?

[languid berry]

1/g(x), yes

so if f(g(x)) = 1/g(x), what is f(x)?

[primal hedge]

im not sure

i still dont understand anything going on TT

what are the steps im supposed to use to solve a problem like that

[languid berry]

replace g(x) with x

f(g(x)) = 1/g(x)

if u replace g(x) with x

what do u get

[primal hedge]

where do i get x from?

[languid berry]

just replace the letters g(x) with x

[primal hedge]

oh f(x)=1/g(x)

[deft oar]

So close

[languid berry]

you didnt replace the second g(x)

[primal hedge]

f(x)=1/x?

[languid berry]

bingo

[deft oar]

🎊🎊🎊

[primal hedge]

no clue what happened ngl

[deft oar]

You gotta review what a function is

You don’t understand what a function fundamentally represents

[primal hedge]

it represents a set of instructions for a given input right?

[deft oar]

Yes

[languid berry]

say you have some function f(x) = x + 1

you can plug any value in for x

for example

x = 2

then you have f(2) = 2 + 1

but you can also plug other symbols in for x

if x = z, you have f(z) = z + 1

or if x = g(z), you have f(g(z)) = g(z) + 1

[deft oar]

Input —> f does something —> output, you have to abstract it a bit, numbers are a surprising rarity in math

[languid berry]

doesnt seem like they’re at the stage in math where numbers really become a rarity, abstracting might just make it more confusing at this stage

typically starting with examples and concrete numbers makes it easier in the early stages

[primal hedge]

precalc 1 in college rn

[deft oar]

True. @primal hedge ignore what I said

[primal hedge]

i dont really understand what this means

is there not like concrete steps for this type of question

[deft oar]

Lowkey barely understand math myself

[deft oar]

See aurelianus’s example it’s really good

[primal hedge]

i understand what im looking at

why can g(x) be replaced with x

[deft oar]

x is just a placeholder for the value

Again

Say f(x) = 13x-1

f(random ahh symbol) = 13(random ahh symbol) -1

So in this context

The random ahh symbol is g(x)

But now we don’t want no random ahh symbol

We’re working backwards for that placeholder

So we swap it to, x,

To generalize it beyond the random ahh symbol

It’s like f(symbol)= 13(symbol)-1, or f(g(x)) = 13(g(x))-1

We just use x as a place holder because it’s nice

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[final cedar]

how did t^-3 become positive, its related to indices probably, but if it matters, im trying to solve a differential equations problem, x2y dx - (x3+y3) dy = 0

[final cedar]

💔

..

<@&286206848099549185> im sorry 😭

..

[somber portal]

!original

[vocal sleet]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[final cedar]

oh hi

question is x2y dx - (x3+y3) dy = 0

[somber portal]

as i said, post the original problem, with all the context

[final cedar]

uh, thats the original problem itself?

[somber portal]

There must be instructions

[final cedar]

"evaluate the diff eqns"

[somber portal]

also, i'm completely sure it's not written like that

[final cedar]

okay okay holup

just the powers

..

[somber portal]

okay, and you're asking about the solution that you were given here?
Can you show all of it?

[final cedar]

ah yesyes

all good till here

here the problem comes

how did t^-3 become t^3

thanks anywayy

[somber portal]

hm, i'm having trouble seeing that step, that would be true if $t=\frac{-1}{t}$

[twin meteor]

LordFelix

[final cedar]

yeah thanks anyway

[somber portal]

t= y/x

seems to me to be an errata, if you look at the next line it just replaces t for its value on the previous line

[final cedar]

uhhhhh

one second let me read that again

ohhhh it seems like a typoo

[final cedar]

yesyes

that step shouldve been 3t^3 in the denominator

thanksthankss

[somber portal]

yeah, i think so

[final cedar]

yhanks again

[somber portal]

np

[final cedar]

i do .close right?

[somber portal]

i suppose, if there's no more about this?

[final cedar]

yes

.close

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[cursive narwhal]

idk i cant do it

[dense cedar]

Compute the values of a nickel and dime

Then formulate the eqn

[vocal sleet]

@cursive narwhal Has your question been resolved?

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[frail violet]

How do I row reduce further from here

[hollow kite]

multiply first column by -5 add it to the fourth

?

[mental egret]

R2 -> R2 + 2R3/3```

[hollow kite]

that is if you want to get zeroes in the first row

[frail violet]

Lemme try this

[frail violet]

The goal is the get to reduced row echelon form

[hollow kite]

ohh

[frail violet]

thank you @mental egret

.close

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[somber portal]

!original

[vocal sleet]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[kindred obsidian]

how do I restructure this function in order to find where the function is >0 etc

I have to find f(x)>0 ,< 0, = 0 of this function

but I dont know how to restructure it

I tried to time it by (x-1)^6 for f(x)> 0

.close

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[oak magnet]

how can i simplify arsinh(2x*sqrt(x^2-1)) using t = arsinh(x) ?

[oak magnet]

.close

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[hexed onyx]

Top question

[river kettle]

,rcw

[hexed onyx]

Not sure how to start this, directions say no u sub, I overheard someone in recitation talking about a circle

[hexed onyx]

So I assume I can say x^2 + y^2 = 16 since 16 the only number there

[river kettle]

u can use trig sub or do it geometrically

[hexed onyx]

How would I do it geometrically cus I think thats what they wanted

[river kettle]

this is just a semi circle

[hexed onyx]

Bruh is it just 4 pi

[river kettle]

well whats formula for the area of a circle?

[hexed onyx]

Pi r^2

So 1/4 of 16 pi

?

[river kettle]

right

this is just

a quarter circle

and u know the radius is 4

[hexed onyx]

Ok yea I just need to learn to recognize these things

Thanks

How do we close this

[river kettle]

type .close

[hexed onyx]

.close

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[acoustic flower]

Hello! I need help again I’m really understanding how to do this problem (it’s translated from Italian so it might be a bit different idk)

“For his girlfriend’s birthday, Andrea buys her a bouquet of roses. Of these, the 3/4 are red, 1/3 of the remaining ones are yellow and the last 4 They are white. How many roses did Andrea give?”

[acoustic flower]

I know it’s probably easy but I suck at math so please don’t judge me!

[peak shard]

Its all good, we are all here to learn, its great that you are looking for help if you dont know how to solve it

!status

[vocal sleet]

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

[acoustic flower]

1

[peak shard]

first when you are giving such a problem try to make equations out of what you are given

[acoustic flower]

Icc I tried some but they weren’t leading me no where 😭 I tried stuff like 3/4 x 1/3 but after that I’m clueless

[peak shard]

have you used letters before to represent numbers?

like in general not for this problem

specifically

[acoustic flower]

I know stuff about letters in numbers but not really

[forest spade]

hi

[acoustic flower]

Yea no not really

[forest spade]

can i help?

[acoustic flower]

Any help is appreciated!!

[peak shard]

so lets see then why using letters to represent numbers can be useful okay?

[acoustic flower]

Okayy

[peak shard]

Let R be the total number of roses Andrea bought.

now R is just a number, but we dont know which one and our question is to find R

[forest spade]

1/4x2/3=1/6

[acoustic flower]

Right

[forest spade]

1/6x=4

solve for x

[acoustic flower]

Right sorry that kinda confused me 😭 I prefer going step by step if that’s okay

[forest spade]

ok sorry

[acoustic flower]

No worries

[forest spade]

im in highschool

[peak shard]

If the total number of roses he bought is the number R, then the total red roses he bought is 3/4 * R okay?

[forest spade]

Are you in highschool too?

what grade

[acoustic flower]

3/4 is like the amount of red roses though right?

[peak shard]

just go very slowly when you want to try to explain something

[acoustic flower]

Yes me too but I prefer not saying which grade

[forest spade]

just asking

[forest spade]

no worries

[peak shard]

No of the total number of Roses 3/4 are Red.

Andrea buys her a bouquet of roses. Of these, the 3/4 are red

[acoustic flower]

Right

[forest spade]

i have to go

bye

[peak shard]

so $3/4 \cdot R$ is the total number of red roses

[acoustic flower]

So that means maggiority of the roses are red

[twin meteor]

tobi

[acoustic flower]

Oh okay I get it

[peak shard]

yes

So then all the other remaining roses are $1/4 \cdot R$

[twin meteor]

tobi

[peak shard]

does that make sense to you

[acoustic flower]

Right

Yes

So if 3/4 of the total roses are red

That means 1/4 are the rest of the roses of different colour..?

[peak shard]

yes

[acoustic flower]

Okay understood

[peak shard]

of these Roses 1/3 are yellow

this means

$\frac{1}{3} \cdot \frac{R}{4}$

[twin meteor]

tobi

[acoustic flower]

Okay so

That means 1/12..?

[peak shard]

yes the total number of Yellow Roses is R/12

[acoustic flower]

Okayy got it

[peak shard]

So and then we get told we have 4 white roses

[acoustic flower]

Right

[peak shard]

and if we add up all the yellow, white and red roses

we get the total number of roses

yes?

[acoustic flower]

Yes

[peak shard]

then this means

$\frac{3}{4}R +\frac{1}{12}R + 4 = R$

[twin meteor]

tobi

[peak shard]

this equation must hold true

[acoustic flower]

Okay wait let me process this equation

Okay got it

[peak shard]

first it total number of red roses, second one is yellow and then white

and thats equal to the total number of Roses which is R

[acoustic flower]

Okay so We already know the amount of white roses so we have to calculate the amount of red and yellow roses?

To find the sum

[peak shard]

the only thing we need to do from our point now

[acoustic flower]

okay I’m understanding

[peak shard]

is we can solve this equation for R

do you know how to work with equations like these?

[acoustic flower]

Yess

Let me try

Soo

22/12?

3/4 + 1/12 + 4 and 12 is the dividible with 4 so then it’s 12 divided by 4 which is 3 times 3 which is 9 so basically 9 + 1 + 12

I hope that’s correct-

[peak shard]

you cant add the 3/4 and 4

[acoustic flower]

Ooh

Ic

Wait I’m confused now

[peak shard]

because

there is still the number R in front of it

[acoustic flower]

Oh right

But we don’t know what number is R

[peak shard]

okay so let me try to explain how to solve this

the basic idea is

[acoustic flower]

Okay!

[peak shard]

if way say

$x = 2$

[twin meteor]

tobi

[peak shard]

then $x + 2 = 2 + 2$

[twin meteor]

tobi

[peak shard]

is also a correct statement

[acoustic flower]

Yes because X is 2 so it’s basically 2+2

[peak shard]

well yes

but the reasoning here is that

I added something on both sides

then it has to still be equal

[acoustic flower]

Oh right

Cuz

We know on the right that 2 plus 2 needs to be equal to the one on the left

Which means it has to be 2 + 2 too right?

[peak shard]

well yes

so maybe that example was to easy then

let me give you something else

If we have

$x - 7 = 12$

[twin meteor]

tobi

[peak shard]

then we can add +7 on both sides

to find

[acoustic flower]

So

[peak shard]

$x = 19$

[twin meteor]

tobi

[acoustic flower]

Right

Because since it needs to be equal

We need to find a number that becomes 12 when we subtract it with 7

[peak shard]

yes so what we learn here is we are allowed to add and subtract any numbers if we do it on both sides

[acoustic flower]

Right?

[peak shard]

yes

[acoustic flower]

Okay got it

[peak shard]

and

we are allowed to do even more

if

$12x = 7$

[twin meteor]

tobi

[acoustic flower]

X would be -5..?

[peak shard]

12x means multiplied

[acoustic flower]

Ohh okay

[peak shard]

so we have

$12 \cdot x = 12x = 7$

[twin meteor]

tobi

[peak shard]

what operation do we need to do to get rid of the 12?

[acoustic flower]

Minus..?

[peak shard]

Minus would be if it was added

[acoustic flower]

Ohh

[peak shard]

so earlier we had + and we did - on both sides

[acoustic flower]

Uhh division?

[peak shard]

yes and do you know by what number we should devide?

[acoustic flower]

Right

[peak shard]

if we want x to be alone?

[acoustic flower]

Im going to say 12?

[peak shard]

yeah correct and why?

very nice

[acoustic flower]

Because this way we can get rid of the X..?

No sorry 12

[peak shard]

because 12/12 = 1

[acoustic flower]

Yes

[peak shard]

so this means

$12 \cdot x \cdot \frac{1}{12}= 7 \cdot \frac{1}{12}$

[twin meteor]

tobi

[peak shard]

and therefore

x = 7/12

[acoustic flower]

Righttt?

[peak shard]

was that understandable?

[acoustic flower]

Gimme a sec

Ooh right

Got it got it

[peak shard]

so we are allowed to add and subtract on both sides and also multiply both sides (but we have to do it to the whole equation! don't forget brackets)

[acoustic flower]

Right

[peak shard]

we are not allowed to multiply by 0 tho, because then we lose all information and would only get 0 = 0

which is true but not very helpfull

[acoustic flower]

Yess

[peak shard]

now we can use this to solve for R here

[acoustic flower]

Okay I understood this part

[peak shard]

$\frac{3}{4}R +\frac{1}{12}R + 4 = R$

[twin meteor]

tobi

[peak shard]

how we do this has the same concept as before, but just a little bit more complex

[acoustic flower]

Soo basically 3/4R plus 1/12R plus 4 are all equal to the total amount of roses?

Okayy

[peak shard]

yes that's what we came up with earlier

[acoustic flower]

Ooh I see

[peak shard]

first of all we can simplify 3/4 * R + 1/12 * R

[acoustic flower]

Okay so

We can do

12 : 3?

[peak shard]

do you know the distributive law?

[acoustic flower]

Like this?

Oh wait

Nvm it’s wrong I think

[peak shard]

first of all lets not forgot that you still have an R there

[acoustic flower]

Let me translate that since English isn’t my Og language

[peak shard]

it states

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

[twin meteor]

tobi

[acoustic flower]

Okay I kinda know it

[acoustic flower]

Yess

[peak shard]

we can use this

[acoustic flower]

This is familiari

Okayy

Soo

[peak shard]

$R = \frac{3}{4}R +\frac{1}{12}R + 4 = R \cdot \left(\frac{3}{4} +\frac{1}{12} \right) + 4$

[twin meteor]

tobi

[acoustic flower]

Okay yea I was about to say that

So basically

[peak shard]

nice

[acoustic flower]

We do R x all those fractions

[peak shard]

and now you can add up the fractions like you are used to

[acoustic flower]

Okay let me try

Okay so we solve the parenthesis first

Which leaves me at

R x 10/12 + 4

Oh no wait

We need to use the distributive law

RightV

So basically R x 3/4 and then R x 1/12

Or am I getting mixed up

[peak shard]

well

we used the distributive law to get the parenthesis in the first place

so we can add up the fractions

[acoustic flower]

Right

Okay so I was correct before

?

[peak shard]

yes we can go back but it would not be very helpfull

yes its R*10/12

[acoustic flower]

Oh ic

Got it

[peak shard]

also know as R*5/6

[acoustic flower]

Ooh okay

[peak shard]

it has a common factor of 2

[acoustic flower]

Right

Ohh okay I get it

10 : 2

And 12 : 2

To simply it

[peak shard]

the more mathematical reason why this works is this, you might have not seen this

$\frac{5}{6} = \frac{5}{6} \cdot 1$

[twin meteor]

tobi

[peak shard]

do you agree with me on that

I just multiplied with 1

[acoustic flower]

Yes

Because Multiplying by 1 doesn’t change the number

It remains the same

[peak shard]

$\frac{5}{6} = \frac{5}{6} \cdot 1 = \frac{5}{6} \cdot \frac{2}{2}$

[twin meteor]

tobi

[peak shard]

but 1 is also known as 2/2

[acoustic flower]

Oh okay

It still the same

[peak shard]

$\frac{5}{6} = \frac{5}{6} \cdot 1 = \frac{5}{6} \cdot \frac{2}{2} = \frac{5\cdot 2}{6\cdot 2} = \frac{10}{12}$

[twin meteor]

tobi

[acoustic flower]

Right it’s all still 5/6

Just in different ways

[peak shard]

we just multiplied by 1

thats the reason you can cancel out common factors

because multiplying by 1 doesn't change the result

[acoustic flower]

Right

[peak shard]

you probably already know that but it can be helpfull to think about stuff like that from time to time

and remind ourself why it works

[acoustic flower]

Okayy

Got it

[peak shard]

$R = R \cdot \frac{5}{6} + 4$

[twin meteor]

tobi

[peak shard]

so we get this

which looks much nicer

[acoustic flower]

Ohhh okay

Yea it does

So we would have to do

R x 5/6 since multiplication comes first

[peak shard]

we want to have everything with R on one side and all the pure numbers on the other side

[acoustic flower]

Basically get the R to the left side?

[peak shard]

that's the first idea after simplifying

yeah

[acoustic flower]

Ohh okay but how do we do that

[peak shard]

what do we have to do to achieve that

[peak shard]

good question

we accomplished that we are allowed to add numbers on both sides and multiply by anything that's not a zero right?

[acoustic flower]

Yea

[peak shard]

so now if we want R to be gone in the right side

we only want the +4 to be left

how do you get rid of 5/6 * R

[acoustic flower]

But what about the 5/6

Oh

Hmm

We use 1..?

No wait

Let me think

[peak shard]

maybe not so easy if you have never done something like that

you will get used to it the more you do it

[acoustic flower]

Honestly I don’t know

[peak shard]

all good

lets think about it like this

you want 5/6 * R to be zero right

[acoustic flower]

Yes

[peak shard]

if we subtract 5/6 * R then

5/6 * R - 5/6 * R = 0

[acoustic flower]

Ooh okay

Since we can add numbers

[peak shard]

yes and

5/6 * R

is just a number

the whole expression is just a single number

[acoustic flower]

Right

[peak shard]

so we subtract 5/6 * R on both sides

$R - R \cdot \frac{5}{6}= R \cdot \frac{5}{6} - R \cdot \frac{5}{6} + 4$

[twin meteor]

tobi

[peak shard]

$R - R \cdot \frac{5}{6} = 4$

[twin meteor]

tobi

[acoustic flower]

Right so the one on the right becomes 0

And on the left the R is removed..?

[peak shard]

yes thats the reason we did it

[acoustic flower]

Okay so basically it’s just 4 on the left that’s left

[peak shard]

well now we only have R,s on one side

and numbers on the other side

[acoustic flower]

Ooh right

[peak shard]

now you can again use the distribute law

[acoustic flower]

But what do we do with just these numbers left to helps us

Okay

[peak shard]

remember what we want to find

R = ...

because R is the number of Roses bought

$R - R \cdot \frac{5}{6} = 4 = R \cdot 1 - R \cdot \frac{5}{6}$

[twin meteor]

tobi

[peak shard]

R = R x 1

[acoustic flower]

Basically just R=R

[peak shard]

yes but with wrting R as R x 1

we can now use the distribute law

the thing is we dont know what R is, but we know its a number and we know the distribute law works for numbers

$4 = R \cdot \left(1 - \frac{5}{6} \right)$

[twin meteor]

tobi

[acoustic flower]

R= R x ( 4

Oh wait nvm

Okay wait

[peak shard]

do you see what I did here?

[acoustic flower]

I think so

[peak shard]

I used the distribute law, which tells us how to do brackets and remove them

[acoustic flower]

Basically

If we solve the thing in the bracket and then do x R it’s gonna be the same as 4..?

[peak shard]

yes

what is the number inside the bracket?

hint: ||write 1 as 6/6||

[acoustic flower]

Okay gimme a sec

1/6?

[peak shard]

yeah :D

[acoustic flower]

So 1/6 x R is 4?

[peak shard]

$R \cdot \frac{1}{6} = 4$

yes

[twin meteor]

tobi

[acoustic flower]

Hmm

How do we do calculate that then?

[peak shard]

you might remember something like this from before

you basically want R to be alone for that 1/6 should turn into a 1, because R * 1 = R

[acoustic flower]

Right

[peak shard]

how can you achieve 1/6 becoming a 1?

what operation do you need to do

[acoustic flower]

Okay wait

0?

Because X 0 the number becomes 1..?

No wait

Omg

Im soo dumb

[peak shard]

nah you are good, you realized it

[acoustic flower]

Okay wait

[peak shard]

Im basically asking

1/6 times what number equals 1

[acoustic flower]

Oh god

This is hard

😭

[acoustic flower]

Right

[peak shard]

lets say you have 1/2 if you multiply by 2 you get 1

[acoustic flower]

Oh wait

1/6

1/6 x 1/6 ?

No wait

Hold on

Let me write this

Would it be 6?

[peak shard]

yes

because 6 * 1/6 or 6/6 = 1

[acoustic flower]

Right got it

Writing it down is helping me more

On my book

So

R becomes 6?

Which becomes 1=4?

[peak shard]

1 = 4

is wrong

[acoustic flower]

Oh right

[peak shard]

but

[acoustic flower]

Because it isn’t equal

[peak shard]

what we do

is we multiply both sides by 6

$R \cdot \frac{1}{6} \cdot 6 = 4\cdot 6$

[twin meteor]

tobi

[acoustic flower]

Okay wait

Let’s say R is still 6 right

?

[peak shard]

R is not equal to 6

we don't know what R is jet

but we multiply by 6 on both sides

[acoustic flower]

Oh alr

[peak shard]

to get rid of the 1/6

like you said 1/6 times 6 is equal to 1

but we cant just do that on one side

so we have to do to it on both sides

[peak shard]

yes! :D

[acoustic flower]

Okay got it

[peak shard]

so

$R = 24$

[twin meteor]

tobi

[acoustic flower]

Basically

R would be 24?

To make both sides equal?

[peak shard]

and R was the total number of Roses they bought

[peak shard]

yes

[acoustic flower]

Which means

The total is 24??

[peak shard]

yes

[acoustic flower]

And if we want to know the amount of individual flowers

I would do

Omg yes

It’s adding up

!!!

[peak shard]

nice

[acoustic flower]

Omg my brain is exploding

Luckily I took notes

Omg wait

This is soo cool

[peak shard]

so I hope this might have thought you, how writing numbers you don't know jet as letters, and the solving for them doing nothing but plus minus, multiplying and dividing on both sides

is something usefull

[acoustic flower]

Ur soo right!!

Im gonna try doing that more often

[peak shard]

also

if you want to get used to it

try solving stuff like

$3x+12 = 17x - 5$

[twin meteor]

tobi

[acoustic flower]

Ooh okay

[peak shard]

and then find what the number x must be so the equation is true

[acoustic flower]

Im gonna try doing that

Mabye later because it’s really late for me rn haha 🥲

But I’ll keep it in mind

[peak shard]

and with fractions like before it just gets a bit messier, but I showed you the basic idea, of-course you would need to solve many more of these equations to get a better feel for it

thats the idea how you try to solve these questions

[peak shard]

all good! :)

[acoustic flower]

Thank you soo much for spending soo much time for me and being patient with me😅💗

I truly appreciate it a lot

[peak shard]

thanks :)

you can do .close if you are done

[acoustic flower]

Okay!!

.close

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Use .reopen if this was a mistake.

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[urban zinc]

hey guys

[urban zinc]

So i need a bit of help with this

it basically says

If m is the number of elements of the set

then:
a. m = 3
b. m = 0
c. m = 1
d. m = 2
e. none of the above answer choices is correct

problem is, i have made it 5 times already

it always gives me M= {2,3}, so m=2

but the correct answer is c.m=1

can somebody please help me

heres my work

please help me

<@&286206848099549185>

[edgy sapphire]

that's the right setup and solution, but you have to plug 2 and 3 back into the original problem to check if they make sense

since polynomials can be negative and negatives don't really work for C

[urban zinc]

so replace n here

in 2 scenariios?

2 and 3

[edgy sapphire]

oh I guess they both aren't negative hmm

seems like d is right

[urban zinc]

the "good answers sheet" show it is C

c.m=1

[trim tree]

have you tried if n = 2,3 actually equals 1 if you plug if into the binomial coefficient?

[urban zinc]

you mean here? at c? @trim tree